Eduardo and David, you're both correct. The purpose of that arranjment is to assure that the voltage after the 3.9k resistor is constant (0.7V, for silicon PN junction) they should be connected in parallel.
2 back to back diodes never conduct any current. What is said at 12:00 is wrong! The diodes are effectively an open circuit. However, Having these diodes in reverse parallel, that would fit the given definition.
Agreed, this looks like an error in the circuit. It would still function just at a higher voltage because the drop isn't there. However, you could use two back-to-back (not reverse parallel) zener diodes if you wanted a higher drop that still worked for both polarities. When examining the prototype PCB you made, @xraytonyb, it also looks like they were installed back-to-back and not reverse parallel, just like on the schematic. Hopefully we aren't missing something here! hehe. Thanks for these videos, they are excellent!
The two diodes should be in parallel, not in series. If they are in series there will be no current through the two diodes. The microcontroller based 'Mtester V2.68' that I bought tells me the Vbe of the transistor. All I have to do is write down the Vbe of the transistors and find two that are very close or equal. I use the edge of a piece of corrugated cardboard to hold the transistors. I put each transistor into one of the holes of the corrugation. I can write the Vbe below it.
I have found that the best way to test them without the heat from your hand affecting the reading is to put every one of them in a breadboard. By the time you put the last one in the first one has cooled and you can now test them without touching them. I've also found out that you should test them all at the same time because the slightest change in room temperature significantly changes the reading.
Good design! That's a neat way of matching transistors. If you want the collectors to be within .6 V of ground, you need the diodes in parallel, one with anode up, the other with anode down. As you've drawn it, in series, there will never be any current in the diodes. Does the match change if you change collector current? Do you have to do the test close to the current you will be using in your amplifier?
The transistors heat up, they have 11V because the diodes are wrong. They should be connected anti - paralell, and then the Vbe readings will be much more stable.
indeed, on top of that the reference transistor heats up and the DUT needs more time to stabilize. HFE is highly dependent on temperature of the junction.
There is indeed nearly 12V across the transistor Vce (actually 12V + Vbe - 0.94V across the 3.9K resistor). But with Ic ≂ 0.1mA, that's only 1.2mW. The 2N2222 has a thermal resistance of 2.28mW/°C, so they will heat up by just 0.5°C. Since the circuit is only measuring the Vbe of the transistors and nothing else, the effect of self-heating is around 2mV/°C x 0.5°C = 1mV. Not enough to worry about.
Greetings Mr. Tony, I really like your videos, I always learn something . In this case I was searching for a transistor matching circuit, and came across your videos. I watched all three of them, carefully and with curiosity. The bridge you recommend is a very effective and cheap way to match BJT, no doubts, but I think you have made a mistake in the method you want to balance it. In order for me to show to you my point, I will make an example. Let's get for simplicity's sake this values for the circuit: Supply V = 10V, Resistor R1=8k, R2=9k and the Potentiometer = 3k. In this case the balance point of the bridge (shorting out Collector-emitter) will be achieved when the potentiometer is at 66.6% (or 33.3% depending on the side we consider). In this case each branch is thoroughly compensated 8k+2k = 9k+1k=10k and in each branch will flow the same Current of 10V/10k=1mA. But if we measure the voltage across the pot, IS NOT 0V as you have explained in order to balance the bridge, but is 1V. (2k*1mA=2V one side of the pot and 1k*1mA=1V in the other, 2-1=1V across) For this reason please consider making a video and explaining in that this bridge can not be balanced zeroing the voltage across the pot as you have described. Waiting for an answer Kind regards Jurgen
Nice video! Along with the 3.3k current limit, those back to back diodes should be 1-watt ~6.3v zener type (this establishes a Vref.) For TO-126 / 220 devices, the resistors can go 3 to 5 times smaller for better matching at the higher currents.
I don't see how the two diodes back toback in series are doing anything. They are effectively an open circuit until/unless the voltage exceeds their breakdown voltage???
I built my own solution - I call it the 'sillyvolt' meter. It combines a precision ultra-low-voltage power supply (1mV-4.5V) with similarally precise dual-channel ammeters and voltmeters. It's also a curve tracer, if you plug it into a computer to record the readings and draw the graph. It can be used to very accurately measure component characteristics. I actually built it to test suspicious parts off eBay and detect counterfeits. Happy to share the design if anyone wants it.
Diodes do nothing ! The input is +12 & -12v so 24v across the inputs . The base of the transistors is at 0v ( 12v up from the -12v input ) So you have a 12v bias on the 2 emitter followers .
The way the diodes are connected at the + rail, they do nothing. Maybe you wanted to make then ANTIPARALLEL. That way they would always produce about 0.6 - 0.7 V, positive or negative. The tester would work fine to determine the V_BE balance even at this reduced collector voltage. But another issue is the -2 mV/degC temperature dependency of silicon transistors. You really should do something to match the temperatures and especially avoid adding your finger heat to the result. Maybe using tweezers or some other tool. But best to mount them on a single heat sink. The operation at 120 microampere current is typical for high gain low noise transistors in front end differential pair.. Other transistors have their optimum (highest gain) points at as much as 100 mA and even higher for actual power transistors. But this device does not try to match the gains. And that is fine, the V_BE match is the essential parameter for use in differential amplifiers.
I don't think xraytonyb actually measured with a DMM to see if the Vc voltage was limited. I agree the back to back diodes are doing nothing. What is the test base current? What parameters are actually being matched, Vbe, Hfe, collector leakage? Good try at cobbling.
@@tomgeorge3726 The test base current will be whatever is needed to drive around 0.11mA through the transistor depending on the Hfe. There's 100% negative feedback at the emitter, so the only thing being measured is the Vbe of each transistor.
Yes correct. I noticed the same. Those two back-to-back diodes in series won't conduct any current or clip any voltage. Unless he used two Zener diodes back-to-back in series at different clipping voltage, for example 0.6V + 2V (Zener voltage) = 2.6 VDC. I also think he meant to connect them in parallel and each cathode to the other's anode. (I have made much worse mistakes in electronics... it happens).
@@kylesmithiii6150 A 2V Zener has a very poor "knee" voltage and behaves rather like a 100R resistor at the current we're looking at. They make pretty poor shunt regulators.
Problem is, you measure at a very low current. This will probably work with small transistors, but larger power transistors must be tested at much higher C-E currents. Also the series back to back diodes will not drop any voltage, unless you exceed the reverse breakdown voltage of one of the diodes plus the forward voltage of the second one. That is around 700V rms for the 1N4007.
Diodes are wrong. They should be parallel to limit to 0.7 volts in both directions. But a BJT can't operate with only 0.7 volts, so I don't understand. As it is, your diodes aren't doing anything... Unless they are both Zener diodes then they would become crude voltage regulators. Having both bases tied to ground doesn't make sense to me either. An NPN wouldn't conduct at all like this. Having them both forward biased at some level for comparison seems like it would be required for comparison.
I came here for saying that those diodes do nothing at the circuit. But you've done already. Anyway, bases tied to ground are correct. Transistors are forward biased because emitters are below (or above, in case of PNPs) ground due to the use of a split power supply. This is a sort of common-base arrangement and it is legit.
Just a thought, but if I understand correctly, even a minor difference in temperature between any 2 transistors that are actually exactly matched at the same temperature, will cause them to read/ compare differently. So wouldn't handling each transistor that you want to test for a match, cause it to warm up enough to make at least some difference in how it reads, or in this case to how it compares? I'd have thought that would be especially so, considering that once the "reference" transistor is in place, it isn't being touched again, so is more likely to be at an ambient temperature, compared to the one that has just been inserted. If that is the case, wouldn't it be a better and more accurate test, to wait for let's say 10 minutes, before actually performing the test, to give the 2 transistors a chance to equalise?
You are correct. Usually, it takes a couple minutes for the two transistors to stabilize. It is usually less than 10 minutes, but it does take some time. Check out my series on the Pioneer Spec 1 preamp. I do a much better job explaining how this little transistor matching device works.
Hi this may help . You have 0 V , 12 V & 24 V input ( -12 0 +12 V on the map ) . The base of the 2 transistors are on the 12 volt line ( or your 0 volt line ). So the 2 diodes do nothing & could be removed !
Hello Tony, thank you very much for sharing your knowledge, it is very entertaining and motivating. I wanted to ask you how it would be for the case of J-FET or MOSFET transistors
I have seen the back to back diode before. To me it seems like there is no way for current to go in either direct, like if you put to check valves back to back would prevent water from flowing in either directions. This has always baffled me. Could someone explain how 2 diodes that are back to back and facing opposite direction let anything through. Was thinking if the diodes were separated you qould get the .6 drop. But back to back either one would block not letting anything through. Iam sorry if iam missing something, iam still learning. So could you please explain how any current gets through when there back to back? Great Video and thanks in advance.
Generally, they won't let anything through. A common trick is to use two back-to-back zener diodes, in which case one will go into reverse conduction and they'll clamp the voltage at the forward voltage drop plus the reverse breakdown voltage regardless of its polarity. Those seem to be ordinary diodes though.
It's simply a mistake. If you want to clamp the collectors to 0.7V from ground, then you put the two diodes in anti-parallel, not in series. By that I mean you have anode 1 connected to cathode 2 and anode 2 connected to cathode 1. The purpose of such an arrangement is to limit the voltage across the transistor to about 1.3V and minimise the effects of self-heating from the power dissipated in it. However, in this circuit, even with 12V across the transistor there is still only 0.1mA through it, so the self-heating will contribute about 0.5°C or 1mV of change to the Vbe which is being measured. I seriously doubt that it is noticeable.
Would the thermal drift be more stable if a momentary switch is put in series with one rail or the other so that the reference transistor is not heated continuously campared to the dut?
This method is matching on Vbe characteristics isn't it, what about all the other characteristics like Vce, Vcb and Ic ? Surely this is only good for checking a replacement transistor (maybe diff manufacturer) against the original of same transistor number.
question sir, the supply voltages are +12 DC GND -12V DC similar to a supply of a simple class AB audio amplifier kind like positive rail 0V GND reference and a negative rail right ? like this +12/0/-12 DCV
I build music-synthesizer circuits' where (Vbe) is matched in 'like' transistors'. I generally test with 'diode-check' and that works' well,also.Most Companies' matched/tested within 2-3mv.,in general.
Just put a resistor between collector and Vcc and another between Vcc to base to bais the transistor on at some middle of the road current level. Read the voltage across the collector resistor or collector to ground. Just keep swapping the transistors in and out until you get two that have the same voltage drop. Seems a lot simpler to me. It would be interesting to let the matched pair sit drawing current for 5 or so minutes to see they have the same thermo properties.
Hi, what do you think about the PEAK ATLAS DCA75 PRO. You can hook it up to a PC and draw caracteristic lines that way. It's about 100 Bucks ... worth the money? Many Thanks for a response :-)
Besides the diodes being wrong, it's too complicated because of the +/- power supply. The same thing could be accomplished with a single polarity power supply and a dpdt switch.
a good PCB making yes definitely, don't worried we all know that we make first a "working circuit first" then later to be transfer to layout design is all good sir :)
There is now a board available base on this exact concept but UA-cam is so strict that I am afraid to give you the exact info. They don't allow referring to things outside UA-cam. I tried to add a link somewhere else and poof gone.
Hi sir ! I have a simple question about the hFE of the final transistors of an NSA amplifier operating in the parallel SEPP circuit . Particularly I am talking about Pioneer A9 ,which is working in parallel SEPP with a high frequency DC servo bias. So the question is should the end power transistor have the same hFE? Likes the the class letter and likes a digit number to prevent the distortion?
I can see merits of this circuit...but...is it not actually matching Vbe primarily and gain only secondarily? What if you loaded the connected together emitters with a single current source at your desired bias point (say 1 ma), grounded both bases and measured the difference across identical collector resistors? Wouldn’t that actually compare gain regardless of vbe? You could easily switch in various bias currents and see the match on several quickly. I also suggest the two devices under test be thermally bonded together
Hey mate, i dont if you've if it's already done, but if give me a BOM and the desired board dimensions, i'll make the gerber files and send em to you so you can order them from a PCB manufacturer
A question of curiosity: Was it any chance to compare the matched transistor pairs found with this nice little bridge and your Atlas or curve trace please? As the circuit accuracy is limited to about 0.5 meV voltage difference, it would be interesting to know the corresponding gain difference for a matched pair. Anyway thank very much for the nice videos.
I actually matched up two sets of transistors using just the little device and a multimeter (no scope). I then broke out the curve tracer and tested all four transistors. The curve tracer confirmed what I tested on the device and even proved the one matched set had slightly higher gain than the other! Of course, I could calculate the actual gain on the curve tracer, but for just looking for matches, this little device worked great.
I prefer matching transistors to the circuit in which they have to run. Matched transistors mostly run in differential amplifiers or current mirrors. So I used measure several transistors to match with their hfe as close as possible. Then I prepare a test circuit which is as close as possible to the application circuit running under the same current and voltage conditions and test the transistors to get the best match (minimum differential voltage in a differential amplifier or minimum current deviation for the current mirror. BTW: it is not surpricing to see different hfe at different test devices. hfe depends on collector current, collector / emitter voltage (and also on temperature).
Hi tonyb...I need to know the best way to clean a Yamaha c-2a volume and tone controls...I purchased one and hoave a lot of noise on all the vrs..Thanks .
It depends on what you mean by "levels". If you mean current gain at a particular collector current and collector voltage, then just measure it with a couple of resistors and a multimeter.
I think it does not matter what a voltage is applied to collectors. Transistors will conduct as much current as it's needed to have 0.7V Vbe (0 at B, and -0.7 at E). Everything is determined by emitter resistors.
The purpose of reducing the voltage at the collectors is to minimise the voltage across the transistors and hence minimise the self-heating. However, 12V @ 0.1mA is only going to increase the junction temperature by 0.5°C above ambient, so it's irrelevant. What you're seeing in the video is the effect of handling the transistors, which then have to _cool down_ to 0.5°C above ambient, not warm up.
de basis van de transistors liggen op 0v aan de massa , je kan die diodes weglaten , ze doen niets. om echt te werken zou je via een transistor en een zennerdiode de basis op een constante spanning kunnen brengen waarbij de stroom opgenomen de spanning niet wijzigt, daarbij liggen de collectors altijd op dezelfde spanning!!! hier zouden ook weerstanden tussen moeten staan om stroomverschillen te detecteren.
All you are doing is determining the DC bias collector current for that particular circuit. With obvious differences in hfe and different operating voltages of different circuits, how does that match two transistors with nearly the same collector current at one bias point, i.e. at different bias points they could have different characteristics from one transistor to the next It's been a long time since my engineering days, so forgive me if I'm way off base (no pun intended).
This is an old video and I've done a lot more with this design in later videos. There are two factors that are being tested in this circuit. First is the Vbe. As the transistors are configured in the circuit, if both transistors have equal Vbe properties, they will cancel out and the net result should be zero volts while the transistors are "on". This is the fuzzy line in the waveform. The non-fuzzy line is when the transistors are turned off. Matched Vbe transistors will display a line centered at 0 volts. Unmatched will show either a positive or negative offset, depending on which transistor has the higher Vbe. The second thing we are looking at is the on and off pulses, when the transistors begin to conduct and when they turn off. Since transition frequency. Hfe and miller capacitance all interact with one another, two unmatched transistors will have different slew rates and/or different gain. This will show up as unequal amplitude pulses at turn on/ turn off of the transistor. Equal amplitude pulses indicate a matched pair. I compared the results from this tester to my full size curve tracer and they agreed with one another 100% of the time, when testing small signal transistors. The curve tracer will give the actual values of each transistor, while this device only can tell if they are matched or not. This also isn't good for higher powered transistors, as they need more current than this little device can provide.
@@xraytonyb Yes, we always designed using curve tracing, finding the hfe, proper dc bias point on the curve, Rc, Re, and base circuit. Not to say your circuit doesn't work, it's just hard to understand why it does, but I like your explanation; and enjoy your videos. Thanks.
The Keithley is just the bench multimeter I'm using in the video. It was NOS military surplus. I got it online a while back for around $150. You don't need this. Any meter that can read DC millivolts will work. I even have a little meter I got for free with a coupon from Harbor Freight that has a millivolt scale and will work just fine.
I think you're not really matching the resistance in each branch. You're matching the resistance ratio, not the total resistance. If you match the ratio, you're magnifying the error instead of reducing it. You should rethink your calibration procedure. Or maybe use a resistor network with a good matching coefficient. Another problem I see with your method is the lack of thermal coupling between the DUTs. Thermal effects are notoriously big in semiconductors. Especially if you keep one transistor plugged in your circuit (and heating up) and keep swapping the other one (not heating up as much). I think it would be hard to be sure two transistors are really matched if thermal effects weren't accounted for.
As long as the supply voltage to both transistors are the same, and the parts can handle the voltage, it does not matter. I use a slightly different circuit with a 9V battery.
$15 multimeter, diode test. You are also handling the bjt with your hand, changing it's temperature, thus inducing error in the measurement... Your method also takes too long to find if it matches.
Good video! I'll use this circuit. But in 14:10, 'how to calibrate' is wrong. If we want the same current thru the transistors, the voltage in the potentiometer will be 0V only in the middle, but the reason to the potentiometer be there is to compensate the difference between the resistors, therefore the potentiometer shouldn't be in the middle. I'm waiting for the next videos.
@@envisionelectronics It really annoys me when so called experts misinform their viewers! I have a huge reference library of electronics books I refer to when I need to! I've been a bench technician for nearly 60 years! If you are technician or electronic engineer, good luck to you in the future!
@@janakapriyadarshana5835 If you really want to clamp the collectors to about 0.7V above ground you put the two diodes in _parallel_ with anode 1 connected to cathode 2 and anode 2 connected to cathode 1. As show in the schematic, the two diodes do precisely nothing and the collector voltages will be close to +11V when testing NPN and close to -11V when testing PNP. Fortunately, that's irrelevant and you might as well omit the diodes and the 3.9K resistor as they do nothing.
I don't think you really understand what is happening with this circuit. As it is, it is comparing the "non-conducting" resistance of both transistors.
Built his other design. Tested with an SSM2212. Perfect match. I build a lot of modular synth stuff and the Tony circuit works great. ua-cam.com/video/J6lUs1LI2QI/v-deo.html
I always question UA-camrs if their knowledge is sufficient to build a device that actually works. Reading the comments can give insight whether I continue watching or to move on. I'll move on.
Can the same thing be made up for MOSFETs and can you build one of those and show us I’m starting construction on this little unit you just spilled but I don’t have the knowledge to create from nothing
Yes, you would be measuring the difference in gate voltages required to drive around 0.1mA through the FETs. It would work for JFETs just the same. But a word of caution: for BJTs, matching the Vbe will be a good indication of tracking across varying currents, but for FETs, matching the Vgs at 0.1mA doesn't necessarily mean they will track well as the current varies around that value. It's so much more difficult to get a differential pair of discrete FETs to match that almost everybody will use an FET-input opamp instead.
Dang, I never thought about finding matching transistors in this way! I see this circuit as another excellent use of the Wheatstone bridge.
The diodes in series will never conduct as one will be blocked always. Am I wrong?
That would be my interpretation also. Either there should be a connection in the middle or zener diodes should be used.
Eduardo and David, you're both correct.
The purpose of that arranjment is to assure that the voltage after the 3.9k resistor is constant (0.7V, for silicon PN junction) they should be connected in parallel.
2 back to back diodes never conduct any current.
What is said at 12:00 is wrong!
The diodes are effectively an open circuit.
However, Having these diodes in reverse parallel, that would fit the given definition.
Agreed, this looks like an error in the circuit. It would still function just at a higher voltage because the drop isn't there. However, you could use two back-to-back (not reverse parallel) zener diodes if you wanted a higher drop that still worked for both polarities.
When examining the prototype PCB you made, @xraytonyb, it also looks like they were installed back-to-back and not reverse parallel, just like on the schematic.
Hopefully we aren't missing something here! hehe. Thanks for these videos, they are excellent!
Good catch, man!
The two diodes should be in parallel, not in series. If they are in series there will be no current through the two diodes.
The microcontroller based 'Mtester V2.68' that I bought tells me the Vbe of the transistor. All I have to do is write down the Vbe of the transistors and find two that are very close or equal. I use the edge of a piece of corrugated cardboard to hold the transistors. I put each transistor into one of the holes of the corrugation. I can write the Vbe below it.
I have found that the best way to test them without the heat from your hand affecting the reading is to put every one of them in a breadboard. By the time you put the last one in the first one has cooled and you can now test them without touching them. I've also found out that you should test them all at the same time because the slightest change in room temperature significantly changes the reading.
Good design! That's a neat way of matching transistors. If you want the collectors to be within .6 V of ground, you need the diodes in parallel, one with anode up, the other with anode down. As you've drawn it, in series, there will never be any current in the diodes. Does the match change if you change collector current? Do you have to do the test close to the current you will be using in your amplifier?
I have for years professionally used the transistor substitution manual and found great matches that way.
The transistors heat up, they have 11V because the diodes are wrong. They should be connected anti - paralell, and then the Vbe readings will be much more stable.
indeed, on top of that the reference transistor heats up and the DUT needs more time to stabilize. HFE is highly dependent on temperature of the junction.
There is indeed nearly 12V across the transistor Vce (actually 12V + Vbe - 0.94V across the 3.9K resistor). But with Ic ≂ 0.1mA, that's only 1.2mW. The 2N2222 has a thermal resistance of 2.28mW/°C, so they will heat up by just 0.5°C. Since the circuit is only measuring the Vbe of the transistors and nothing else, the effect of self-heating is around 2mV/°C x 0.5°C = 1mV. Not enough to worry about.
Greetings Mr. Tony, I really like your videos, I always learn something . In this case I was searching for a transistor matching circuit, and came across your videos. I watched all three of them, carefully and with curiosity. The bridge you recommend is a very effective and cheap way to match BJT, no doubts, but I think you have made a mistake in the method you want to balance it.
In order for me to show to you my point, I will make an example. Let's get for simplicity's sake this values for the circuit: Supply V = 10V, Resistor R1=8k, R2=9k and the Potentiometer = 3k. In this case the balance point of the bridge (shorting out Collector-emitter) will be achieved when the potentiometer is at 66.6% (or 33.3% depending on the side we consider). In this case each branch is thoroughly compensated 8k+2k = 9k+1k=10k and in each branch will flow the same Current of 10V/10k=1mA. But if we measure the voltage across the pot, IS NOT 0V as you have explained in order to balance the bridge, but is 1V. (2k*1mA=2V one side of the pot and 1k*1mA=1V in the other, 2-1=1V across)
For this reason please consider making a video and explaining in that this bridge can not be balanced zeroing the voltage across the pot as you have described.
Waiting for an answer
Kind regards Jurgen
Tony,
I like the disclaimer/warning about the circuit. LMAO! You should have added... "No Transistor was harmed in the making of this circuit."
Nice video! Along with the 3.3k current limit, those back to back diodes should be 1-watt ~6.3v zener type (this establishes a Vref.) For TO-126 / 220 devices, the resistors can go 3 to 5 times smaller for better matching at the higher currents.
I don't see how the two diodes back toback in series are doing anything. They are effectively an open circuit until/unless the voltage exceeds their breakdown voltage???
Our was a mistake. I corrected it on the next video.
Whose should have been zeners diodes.
I built my own solution - I call it the 'sillyvolt' meter. It combines a precision ultra-low-voltage power supply (1mV-4.5V) with similarally precise dual-channel ammeters and voltmeters. It's also a curve tracer, if you plug it into a computer to record the readings and draw the graph. It can be used to very accurately measure component characteristics. I actually built it to test suspicious parts off eBay and detect counterfeits. Happy to share the design if anyone wants it.
ipfs.io/ipfs/Qmd8GXUEkw63J8SBmkS5thnoDTyJTyp1yM6SfzEHvLZp3M/
Vyl Bird. Yes, very much so.
Link provided. Enjoy. Doesn't show the metering part, but that's just three module boards wired together and it says which chip you need.
Vyl Bird I would be interested in the design Also. Could you send me the link please
Diodes do nothing !
The input is +12 & -12v so 24v across the inputs .
The base of the transistors is at 0v ( 12v up from the -12v input )
So you have a 12v bias on the 2 emitter followers .
The way the diodes are connected at the + rail, they do nothing. Maybe you wanted to make then ANTIPARALLEL. That way they would always produce about 0.6 - 0.7 V, positive or negative. The tester would work fine to determine the V_BE balance even at this reduced collector voltage.
But another issue is the -2 mV/degC temperature dependency of silicon transistors. You really should do something to match the temperatures and especially avoid adding your finger heat to the result. Maybe using tweezers or some other tool. But best to mount them on a single heat sink.
The operation at 120 microampere current is typical for high gain low noise transistors in front end differential pair.. Other transistors have their optimum (highest gain) points at as much as 100 mA and even higher for actual power transistors. But this device does not try to match the gains. And that is fine, the V_BE match is the essential parameter for use in differential amplifiers.
Those diodes were my WTF moment too.
I don't think xraytonyb actually measured with a DMM to see if the Vc voltage was limited. I agree the back to back diodes are doing nothing.
What is the test base current?
What parameters are actually being matched, Vbe, Hfe, collector leakage?
Good try at cobbling.
@@tomgeorge3726 The test base current will be whatever is needed to drive around 0.11mA through the transistor depending on the Hfe. There's 100% negative feedback at the emitter, so the only thing being measured is the Vbe of each transistor.
Very nice little test circuit you came up with. I will for sure build myself one. Many thanks for the video and the good idea.
If the first diode block than there is no way for the second diode conduct to ground.
Pretty much, I think they should have been drawn in an anti-parallel configuration.
Yes, back to back like meter protection diodes.
u are correct should be in parallel
Yes correct. I noticed the same. Those two back-to-back diodes in series won't conduct any current or clip any voltage. Unless he used two Zener diodes back-to-back in series at different clipping voltage, for example 0.6V + 2V (Zener voltage) = 2.6 VDC. I also think he meant to connect them in parallel and each cathode to the other's anode. (I have made much worse mistakes in electronics... it happens).
@@kylesmithiii6150 A 2V Zener has a very poor "knee" voltage and behaves rather like a 100R resistor at the current we're looking at. They make pretty poor shunt regulators.
Problem is, you measure at a very low current. This will probably work with small transistors, but larger power transistors must be tested at much higher C-E currents. Also the series back to back diodes will not drop any voltage, unless you exceed the reverse breakdown voltage of one of the diodes plus the forward voltage of the second one. That is around 700V rms for the 1N4007.
Very simple and best of all accurate and inexpensive. thanks for sharing!
Good idea, but back to back diodes are going to be always off, so you may use two zener back to back diode to get some known voltage.
You're correct. As others have mentioned, they must be Zeners but I think it's weird that he doesn't know their symbol.
Diodes are wrong. They should be parallel to limit to 0.7 volts in both directions. But a BJT can't operate with only 0.7 volts, so I don't understand.
As it is, your diodes aren't doing anything... Unless they are both Zener diodes then they would become crude voltage regulators.
Having both bases tied to ground doesn't make sense to me either. An NPN wouldn't conduct at all like this.
Having them both forward biased at some level for comparison seems like it would be required for comparison.
I came here for saying that those diodes do nothing at the circuit. But you've done already. Anyway, bases tied to ground are correct. Transistors are forward biased because emitters are below (or above, in case of PNPs) ground due to the use of a split power supply. This is a sort of common-base arrangement and it is legit.
Just a thought, but if I understand correctly, even a minor difference in temperature between any 2 transistors that are actually exactly matched at the same temperature, will cause them to read/ compare differently. So wouldn't handling each transistor that you want to test for a match, cause it to warm up enough to make at least some difference in how it reads, or in this case to how it compares? I'd have thought that would be especially so, considering that once the "reference" transistor is in place, it isn't being touched again, so is more likely to be at an ambient temperature, compared to the one that has just been inserted. If that is the case, wouldn't it be a better and more accurate test, to wait for let's say 10 minutes, before actually performing the test, to give the 2 transistors a chance to equalise?
You are correct. Usually, it takes a couple minutes for the two transistors to stabilize. It is usually less than 10 minutes, but it does take some time. Check out my series on the Pioneer Spec 1 preamp. I do a much better job explaining how this little transistor matching device works.
@@xraytonyb Thanks for reply Tony. I'm still learning, so it's good to know I've understood it correctly.
I'm still learning too ;)
Nice work, enjoyed this greatly, thanks for sharing.
Would be interesting to identify a matched pair, and then test them for hfe readings on the PEAK meter and on the ATmega328 component tester.
Hi this may help .
You have 0 V , 12 V & 24 V input ( -12 0 +12 V on the map ) .
The base of the 2 transistors are on the 12 volt line
( or your 0 volt line ).
So the 2 diodes do nothing & could be removed !
Accuracy would improve if the transistors were thermally coupled with some thermal compound and a clamp.
thats right!
I have a Peak Atlas after a lot of time on the first variety you showed
I have found it to be very good but yours looks much better
Hello Tony, thank you very much for sharing your knowledge, it is very entertaining and motivating. I wanted to ask you how it would be for the case of J-FET or MOSFET transistors
If you record the values of the rejects, you can match the reject values to each other to get more matches :)
Thanks. It a practical thing with the Wheatstone bridge concept.
Except those diodes which are wrong, you could buid a current mirror and measure the voltage difference between thise transistors.
I have seen the back to back diode before. To me it seems like there is no way for current to go in either direct, like if you put to check valves back to back would prevent water from flowing in either directions. This has always baffled me. Could someone explain how 2 diodes that are back to back and facing opposite direction let anything through. Was thinking if the diodes were separated you qould get the .6 drop. But back to back either one would block not letting anything through. Iam sorry if iam missing something, iam still learning. So could you please explain how any current gets through when there back to back? Great Video and thanks in advance.
Learning Electronics Cheaply back to back and you connect in the middle
Generally, they won't let anything through. A common trick is to use two back-to-back zener diodes, in which case one will go into reverse conduction and they'll clamp the voltage at the forward voltage drop plus the reverse breakdown voltage regardless of its polarity. Those seem to be ordinary diodes though.
I think the confusion is that it doesn't appear there is any connection in the middle being used in this circuit.
It's simply a mistake. If you want to clamp the collectors to 0.7V from ground, then you put the two diodes in anti-parallel, not in series. By that I mean you have anode 1 connected to cathode 2 and anode 2 connected to cathode 1. The purpose of such an arrangement is to limit the voltage across the transistor to about 1.3V and minimise the effects of self-heating from the power dissipated in it. However, in this circuit, even with 12V across the transistor there is still only 0.1mA through it, so the self-heating will contribute about 0.5°C or 1mV of change to the Vbe which is being measured. I seriously doubt that it is noticeable.
Would the thermal drift be more stable if a momentary switch is put in series with one rail or the other so that the reference transistor is not heated continuously campared to the dut?
Unless one of your two diodes are a zener would they do anything?? One would have to have a breakover voltage lower than 12 volts.
This method is matching on Vbe characteristics isn't it, what about all the other characteristics like Vce, Vcb and Ic ?
Surely this is only good for checking a replacement transistor (maybe diff manufacturer) against the original of same transistor number.
Great video, which diode do you use?
question sir, the supply voltages are +12 DC GND -12V DC similar to a supply of a simple class AB audio amplifier kind like positive rail 0V GND reference and a negative rail right ? like this +12/0/-12 DCV
yes
Your diodes need to be in parallel.
Or did you mean 2 x zeners, after all 12V inout is crazy high, 1.5v would be better otherwise.
You are 100% correct
clever circuit !! thanks
I build music-synthesizer circuits' where (Vbe) is matched in 'like' transistors'. I generally test with 'diode-check' and that works' well,also.Most Companies' matched/tested within 2-3mv.,in general.
the diodes are unable to let any current through the way they are connected.
He made a mistake, the should be in antiparallel, not in series to start with.
How do you ascertain the the diodes match exactly? If you get the transistors in wrong are they going to smoke?
You're doing a great job by the way!
Just put a resistor between collector and Vcc and another between Vcc to base to bais the transistor on at some middle of the road current level. Read the voltage across the collector resistor or collector to ground. Just keep swapping the transistors in and out until you get two that have the same voltage drop. Seems a lot simpler to me. It would be interesting to let the matched pair sit drawing current for 5 or so minutes to see they have the same thermo properties.
Hi, what do you think about the PEAK ATLAS DCA75 PRO. You can hook it up to a PC and draw caracteristic lines that way. It's about 100 Bucks ... worth the money? Many Thanks for a response :-)
Besides the diodes being wrong, it's too complicated because of the +/- power supply.
The same thing could be accomplished with a single polarity power supply and a dpdt switch.
Going back to the diodes, they are zeners, no doubt. Also if you put it BEFORE the switch you only need one diode :)
sir, you help me a lot you deserve a thumbs up and a sub
a good PCB making yes definitely, don't worried we all know that we make first a "working circuit first" then later to be transfer to layout design is all good sir :)
There is now a board available base on this exact concept but UA-cam is so strict that I am afraid to give you the exact info. They don't allow referring to things outside UA-cam. I tried to add a link somewhere else and poof gone.
Hi sir ! I have a simple question about the hFE of the final transistors of an NSA amplifier operating in the parallel SEPP circuit . Particularly I am talking about Pioneer A9 ,which is working in parallel SEPP with a high frequency DC servo bias.
So the question is should the end power transistor have the same hFE? Likes the the class letter and likes a digit number to prevent the distortion?
I can see merits of this circuit...but...is it not actually matching Vbe primarily and gain only secondarily?
What if you loaded the connected together emitters with a single current source at your desired bias point (say 1 ma), grounded both bases and measured the difference across identical collector resistors? Wouldn’t that actually compare gain regardless of vbe?
You could easily switch in various bias currents and see the match on several quickly.
I also suggest the two devices under test be thermally bonded together
Wouldn't that be close to what a curve tracer does ?
It's measuring Vbe at 0.1mA and nothing else. The Hfe doesn't come into it at all.
Hey mate, i dont if you've if it's already done, but if give me a BOM and the desired board dimensions, i'll make the gerber files and send em to you so you can order them from a PCB manufacturer
A question of curiosity: Was it any chance to compare the matched transistor pairs found with this nice little bridge and your Atlas or curve trace please? As the circuit accuracy is limited to about 0.5 meV voltage difference, it would be interesting to know the corresponding gain difference for a matched pair. Anyway thank very much for the nice videos.
I actually matched up two sets of transistors using just the little device and a multimeter (no scope). I then broke out the curve tracer and tested all four transistors. The curve tracer confirmed what I tested on the device and even proved the one matched set had slightly higher gain than the other! Of course, I could calculate the actual gain on the curve tracer, but for just looking for matches, this little device worked great.
There is a jig were to pair transistors in complementary way? PNP NPN working pushpull configuration.. Regards
Thanks, I need to build one soon... Good man.
I prefer matching transistors to the circuit in which they have to run. Matched transistors mostly run in differential amplifiers or current mirrors. So I used measure several transistors to match with their hfe as close as possible. Then I prepare a test circuit which is as close as possible to the application circuit running under the same current and voltage conditions and test the transistors to get the best match (minimum differential voltage in a differential amplifier or minimum current deviation for the current mirror.
BTW: it is not surpricing to see different hfe at different test devices. hfe depends on collector current, collector / emitter voltage (and also on temperature).
Do you know of the prebuilt kit or device that will allow for transistor matching?
Hi tonyb...I need to know the best way to clean a Yamaha c-2a volume and tone controls...I purchased one and hoave a lot of noise on all the vrs..Thanks .
Would using some sort of heater help? Hair dryer or something?
so if you're trying to match the levels of two complimentary transistors, will this work? That's mainly what I'm interested in at the moment.
It depends on what you mean by "levels". If you mean current gain at a particular collector current and collector voltage, then just measure it with a couple of resistors and a multimeter.
Sorry to say, but you got it all wrong. You are matching now Vbe instead of Hfe.
And also the two diodes will never conduct, they need to be in parallel insted
I think it does not matter what a voltage is applied to collectors. Transistors will conduct as much current as it's needed to have 0.7V Vbe (0 at B, and -0.7 at E). Everything is determined by emitter resistors.
The purpose of reducing the voltage at the collectors is to minimise the voltage across the transistors and hence minimise the self-heating. However, 12V @ 0.1mA is only going to increase the junction temperature by 0.5°C above ambient, so it's irrelevant. What you're seeing in the video is the effect of handling the transistors, which then have to _cool down_ to 0.5°C above ambient, not warm up.
Is it your fingers warming the transistor causes such wild fluctuation?????
Awesome stuff! keep it coming. This is a wondermethod for us poor peons.
Ummmmmm.....are those diodes sposeta be in PARALLEL!?!?!?
a ZIF socket would make plugging in new transistors a breeze.
Thumbs up done is that the vbe of the transistor sir? tnx
Superb!
de basis van de transistors liggen op 0v aan de massa , je kan die diodes weglaten , ze doen niets. om echt te werken zou je via een transistor en een zennerdiode de basis op een constante spanning kunnen brengen waarbij de stroom opgenomen de spanning niet wijzigt, daarbij liggen de collectors altijd op dezelfde spanning!!! hier zouden ook weerstanden tussen moeten staan om stroomverschillen te detecteren.
All you are doing is determining the DC bias collector current for that particular circuit. With obvious differences in hfe and different operating voltages of different circuits, how does that match two transistors with nearly the same collector current at one bias point, i.e. at different bias points they could have different characteristics from one transistor to the next It's been a long time since my engineering days, so forgive me if I'm way off base (no pun intended).
This is an old video and I've done a lot more with this design in later videos. There are two factors that are being tested in this circuit. First is the Vbe. As the transistors are configured in the circuit, if both transistors have equal Vbe properties, they will cancel out and the net result should be zero volts while the transistors are "on". This is the fuzzy line in the waveform. The non-fuzzy line is when the transistors are turned off. Matched Vbe transistors will display a line centered at 0 volts. Unmatched will show either a positive or negative offset, depending on which transistor has the higher Vbe. The second thing we are looking at is the on and off pulses, when the transistors begin to conduct and when they turn off. Since transition frequency. Hfe and miller capacitance all interact with one another, two unmatched transistors will have different slew rates and/or different gain. This will show up as unequal amplitude pulses at turn on/ turn off of the transistor. Equal amplitude pulses indicate a matched pair. I compared the results from this tester to my full size curve tracer and they agreed with one another 100% of the time, when testing small signal transistors. The curve tracer will give the actual values of each transistor, while this device only can tell if they are matched or not. This also isn't good for higher powered transistors, as they need more current than this little device can provide.
@@xraytonyb Yes, we always designed using curve tracing, finding the hfe, proper dc bias point on the curve, Rc, Re, and base circuit. Not to say your circuit doesn't work, it's just hard to understand why it does, but I like your explanation; and enjoy your videos. Thanks.
interesting stuff.
it should work with mosfets to?
@21:37 that was hot hot hot
1.5x speed is fine
Excellent demonstration
How much does that KEITHLEY 197 cost? Can you just use a multimeter?
The Keithley is just the bench multimeter I'm using in the video. It was NOS military surplus. I got it online a while back for around $150. You don't need this. Any meter that can read DC millivolts will work. I even have a little meter I got for free with a coupon from Harbor Freight that has a millivolt scale and will work just fine.
nice video, thanks tony
I think you're not really matching the resistance in each branch. You're matching the resistance ratio, not the total resistance. If you match the ratio, you're magnifying the error instead of reducing it. You should rethink your calibration procedure. Or maybe use a resistor network with a good matching coefficient.
Another problem I see with your method is the lack of thermal coupling between the DUTs. Thermal effects are notoriously big in semiconductors. Especially if you keep one transistor plugged in your circuit (and heating up) and keep swapping the other one (not heating up as much). I think it would be hard to be sure two transistors are really matched if thermal effects weren't accounted for.
Sorry to keep raining on you, but +/=12V, ...don't you mean 12/gnd which give you +/- 12V depending on switch position???
Does it needs to be precisely 12.6v or 12v will do .
As long as the supply voltage to both transistors are the same, and the parts can handle the voltage, it does not matter. I use a slightly different circuit with a 9V battery.
Cool! I think you just invented something!
Please make the drawing bigger!
It is not messy at all..but Simple Designs work best in my book..
Video over 1/2 hour on a simple thing is likely not simple. Ill watch anyways.
$15 multimeter, diode test. You are also handling the bjt with your hand, changing it's temperature, thus inducing error in the measurement...
Your method also takes too long to find if it matches.
Ahhh yes I thought it would be a Wheatstone Bridge!
Simple and inspiring. Just sent you an email.
Your voice sound like as john audio tech!?
Good video! I'll use this circuit. But in 14:10, 'how to calibrate' is wrong. If we want the same current thru the transistors, the voltage in the potentiometer will be 0V only in the middle, but the reason to the potentiometer be there is to compensate the difference between the resistors, therefore the potentiometer shouldn't be in the middle. I'm waiting for the next videos.
You need a 0 center meter, the diodes in the collector leads don't provide any bias!! You don't know what you're talking about!!
B..But he’s on UA-cam! He must be right!!
@@envisionelectronics It really annoys me when so called experts misinform their viewers! I have a huge reference library of electronics books I refer to when I need to! I've been a bench technician for nearly 60 years! If you are technician or electronic engineer, good luck to you in the future!
Pls advise us how you propose to do this so that no one would mislead .....
@@janakapriyadarshana5835 If you really want to clamp the collectors to about 0.7V above ground you put the two diodes in _parallel_ with anode 1 connected to cathode 2 and anode 2 connected to cathode 1. As show in the schematic, the two diodes do precisely nothing and the collector voltages will be close to +11V when testing NPN and close to -11V when testing PNP. Fortunately, that's irrelevant and you might as well omit the diodes and the 3.9K resistor as they do nothing.
@@RaulHernandez-lg5nw 60 years? woow, so you must be around 85 now...
E X C E L L A N T
"simple"
I don't think you really understand what is happening with this circuit.
As it is, it is comparing the "non-conducting" resistance of both transistors.
Waaay over my head. Too bad someone does not sell a kit to make a breadboard checker. They would sell a ton.
Sorry red lettering on a black background can not be seen!
Built his other design. Tested with an SSM2212. Perfect match. I build a lot of modular synth stuff and the Tony circuit works great. ua-cam.com/video/J6lUs1LI2QI/v-deo.html
PEAK is crazy over priced for what you get.
This is called the xraytony bridge.
I always question UA-camrs if their knowledge is sufficient to build a device that actually works. Reading the comments can give insight whether I continue watching or to move on. I'll move on.
This statement makes no sense. The content creator is completely separate from the people commenting.
@@MrMersh-ts7jl That's okay , he's probably already blown up several devices since he "moved on" 😂
In most applications, It is not necessary to match transistors.
In Audio applications, it is.
In some amps it is and in some its not.
Can the same thing be made up for MOSFETs and can you build one of those and show us I’m starting construction on this little unit you just spilled but I don’t have the knowledge to create from nothing
Yes, you would be measuring the difference in gate voltages required to drive around 0.1mA through the FETs. It would work for JFETs just the same. But a word of caution: for BJTs, matching the Vbe will be a good indication of tracking across varying currents, but for FETs, matching the Vgs at 0.1mA doesn't necessarily mean they will track well as the current varies around that value. It's so much more difficult to get a differential pair of discrete FETs to match that almost everybody will use an FET-input opamp instead.
@@RexxSchneider 👍