Thanks, this solves my problem! I was trying to use the test circuit from the datasheet, taking output directly off the cathode. You explained very well how to do it.
If you hooked this (or similar) circuit up to a battery in order to protect it from draining too low, would this protection circuit drain the battery much faster? By that I mean, will the circuit itself consume a lot of energy? I'm asking because I was thinking of hooking something up to a 12v solar battery in my vehicle. It sounds like a very good idea to set up a circuit similar to what you present in the video in order to keep from ruining the battery. However, if the protection circuit itself is eating a ton of juice then maybe it isn't a good solution since it would significantly reduce how long said device could run.
Hi Lewis, many thanks for the educational videos. Please can I ask a couple of questions: is the 9.3v at base of Q1 defined by the base-emitter drop (ie 10v - 0.7v) or by the set-point of the TL431A? Secondly, why would there be roughly 5-6mA going through the 330r resistor? I calculate Vdrop / Resistance = (10-9.3) / 330 = circa 2mA. Thanks again, Sean
if the input drops , so will the drive voltage for TL431 witch will then shut the transistor off maintaining your 10V say , if it is 10V . Now if voltage drops below 10V plus voltage drop of all additional circuitry you will lose the output voltage . Shunt regulators work only if the voltage is higher on the input then required by the load . For situations where you need higher voltage on the output then input , you use DC to DC boost converters / regulators . They have their own set of problems such as noise etc.. but there is no perfect solution anyways .
im planning to use this tl431's as overvoltage protection on agm/lead acid batteries ..and wonder if theyll cause a problem.. i mean adjust it to 14.5 volts and connect outpit/leak to transistor/mofset therere after it reaches 14.5/max voltage agm batterues it will swith on a lef or something else to waste excess woltage thats going tocause pwercharge... in theory this seems like going to work ;)
Oh sorry I don't the major difference between of both the use cases in my application I used the shunt connection for controlling the led dimmer of 0 to 10V.
Can you do a test circuit with 2 TL431 control a 2 mosfet to shunt a 2 battery 12v in serial to maintain the charger for both battery with out deferent in voltage charging exactly like BMS ?
Hmmm . Not that what I mean . I mean let we say that we have 2 serial resistance R1 and R2 and I need to use TL431 with transistor to keep a voltage at across a resistance R1 at a certain level even if the voltage increased in the terminal of the R1 R2 . So what ever the voltage in the input of the R1+R2 the R1 will maintain the voltage that we selected
is this resistor (between base and +) really necessary? if I want make simple voltage protection ( after solar panel)... Should I use some calculation to consider Transistor Veb drop?
@@LewisLoflin I will try use on practice shunt regulator... I want to charge Ni-Zn accumulator (nominal 1.6V, charge 1.9V ) by 4V solar panel. I find TLV431 with Vfb 1.25V, but current limit is low, so I need PNP transistor to conduct more current than this IC. So I need hold (1.9+0.2 Schottky diode ) 2.1 V. As I know solar panels are not scary for short circuit, only external heat decries efficiency. P.S. for this task (shunt external voltages) I can use few, serial connected, ordinary diodes, but I want try some precision way
does the voltage divider (R1 & R2) needs to be behind Rin? ... like in the drawing. I used the circuit to create 5V from 12V without Rin and it seams to work fine and the load is neglectable, although the output is lower than calculated.
It can't work without Rin because that drops the excess voltage. I don't know what you mean by " create 5V from 12V" because this is a current, not a voltage regulator. The output voltage is the load resistance times the input current or R*I.
@@LewisLoflin Well, in the first circuit that uses the TL431, i used them as a positive / negative voltage regulator to create +/- 5V from +/-12V and now I'm designing another circuit with them for the same purpose. Better to use LM317? the AT27C256 requires 6.5V and 13.5V, i have various zeners but not these values.
@@LewisLoflin Thnx, I'm making a programmer for them using a Microchip Fubarino. gonna use transformer of 2 x 15V, a bridge rectifier en than a 7815. so the LM317 will have some 1.5 V head room to make 13.5 V, that's what the datasheet says.
Most of the remaining current will pass through Q1, as it has no current resistance. Thus, the Q1 must be over scaled to support all the remaining current when there is no load connected to the circuit.
Having R1 adjustable is generally bad practice especially for a beginner. Sure, it has to be done to get down to 2.5V but there is a risk. Not much over 2.5V the reference pin starts conducting heavily and begins to destroy the 431 when it gets over 10ma. Good design can prevent this by the proper selection of resistors to prevent that in any case. It is quite easy for a beginner to destroy one should the load be removed. A resistor between the wiper and the reference pin to limit current might be appropriate for a beginner.
As the load draws more current, the overall available current and voltage drop. That is detected by R1/R2 voltage divider, which allows the TL431 to change its current draw. That changes the voltage drop across R3, changes the current through Q1.
In this particular case ( 8:52), it seems that the transistor is useless since the TL431 can handle up to 150 mA, and we will have less than that: 300 mA total, 10 V/44000 Ohm = 0.2 mA through the tension divider, 167 mA through the lamp, 133 mA left, max, through the TL431. BUT the transistor is required if the lamp is ... not there (burn out, open circuit) since then, without it, the 300 mA will have to go through the TL431 and that is the double of what it is safe for it.
Thank you. In general, I'd run the TL451at no more than 30-50 mA or less. While the device is stable generating extra heat will effect drift as it does in semiconductor devices.
Thanks, this solves my problem! I was trying to use the test circuit from the datasheet, taking output directly off the cathode. You explained very well how to do it.
i have been struggling over a month to get 5v stable output for my iot projects, this might be the answer. Thanks lewis for this wonderful circuit.
try 7805
Thank you for this! I needed this! It helped a lot in understanding TL431!
If you hooked this (or similar) circuit up to a battery in order to protect it from draining too low, would this protection circuit drain the battery much faster? By that I mean, will the circuit itself consume a lot of energy?
I'm asking because I was thinking of hooking something up to a 12v solar battery in my vehicle. It sounds like a very good idea to set up a circuit similar to what you present in the video in order to keep from ruining the battery. However, if the protection circuit itself is eating a ton of juice then maybe it isn't a good solution since it would significantly reduce how long said device could run.
How do I calculate the right value of R, and why is the other resistor 330? Seems like magic to me.
Hi Lewis, many thanks for the educational videos. Please can I ask a couple of questions: is the 9.3v at base of Q1 defined by the base-emitter drop (ie 10v - 0.7v) or by the set-point of the TL431A? Secondly, why would there be roughly 5-6mA going through the 330r resistor? I calculate Vdrop / Resistance = (10-9.3) / 330 = circa 2mA. Thanks again, Sean
Is there any reference current?
👍👍 great yt discovery, new sub indeed! Thanks for sharing!
Useful information. Thank0you.
Thank sir,,dan terimakasih atas terjemahannya,,
that was great>>> thank you very much ;)
What happens if the input voltage drops? Will this effect the output current? even if there is still enough excess current for the transistor?
if the input drops , so will the drive voltage for TL431 witch will then shut the transistor off maintaining your 10V say , if it is 10V .
Now if voltage drops below 10V plus voltage drop of all additional circuitry you will lose the output voltage .
Shunt regulators work only if the voltage is higher on the input then required by the load .
For situations where you need higher voltage on the output then input , you use DC to DC boost converters / regulators .
They have their own set of problems such as noise etc.. but there is no perfect solution anyways .
im planning to use this tl431's as overvoltage protection on agm/lead acid batteries ..and wonder if theyll cause a problem.. i mean adjust it to 14.5 volts and connect outpit/leak to transistor/mofset therere after it reaches 14.5/max voltage agm batterues it will swith on a lef or something else to waste excess woltage thats going tocause pwercharge... in theory this seems like going to work ;)
on the sensor about work video
2:47 why the current pass through the potentiometer isn't taken into account?
Because R1 and R2 are using high value resistor, just enough to draw 1 to 10 mA.
Thanks for posting. When do you use practically this circuit. What is the benefit?
This will be use for controlling the LED dimmer circuit.
@@shubhamingle1743 Yes but why with shunt paralel and not in serie?
Oh sorry I don't the major difference between of both the use cases in my application I used the shunt connection for controlling the led dimmer of 0 to 10V.
Do we able to give proctection to this shunt regulator psu if the pnp transistor failed ( there was a short between colector and emiter) ?
It would burn our RIN, but one could add a fuse before RIN.
@@LewisLoflinthanks
The formula shown on the circuit diagram is "It = Ik =Ice + IL". But I think it "It - Ik = Ice + IL ". Is my understanding correct?
You are correct.
Can you do a test circuit with 2 TL431 control a 2 mosfet to shunt a 2 battery 12v in serial to maintain the charger for both battery with out deferent in voltage charging exactly like BMS ?
Not sure what you are asking. A MOSFET is totally different from a bipolar transistor. One is voltage -operated the other current operated.
Hmmm . Not that what I mean . I mean let we say that we have 2 serial resistance R1 and R2 and I need to use TL431 with transistor to keep a voltage at across a resistance R1 at a certain level even if the voltage increased in the terminal of the R1 R2 . So what ever the voltage in the input of the R1+R2 the R1 will maintain the voltage that we selected
Good luck sir 🎉
is this resistor (between base and +) really necessary? if I want make simple voltage protection ( after solar panel)... Should I use some calculation to consider Transistor Veb drop?
Yes. It helps set the current for the transistor and the TL431. What do you mean by voltage protection?
@@LewisLoflin I will try use on practice shunt regulator... I want to charge Ni-Zn accumulator (nominal 1.6V, charge 1.9V ) by 4V solar panel. I find TLV431 with Vfb 1.25V, but current limit is low, so I need PNP transistor to conduct more current than this IC. So I need hold (1.9+0.2 Schottky diode ) 2.1 V. As I know solar panels are not scary for short circuit, only external heat decries efficiency. P.S. for this task (shunt external voltages) I can use few, serial connected, ordinary diodes, but I want try some precision way
Why dont you just connet variable resistor as voltage divider to Q1 to control it?
it will vary with in put voltage tl 431 never change irrespective of input voltage
OK, I built this and it is not working. The output stays at 16V no matter what I twiddle on the potentiometer. What could be wrong?
Which one did you build the current regulator or voltage regulator? What was the load?
does the voltage divider (R1 & R2) needs to be behind Rin? ... like in the drawing.
I used the circuit to create 5V from 12V without Rin and it seams to work fine and the load is neglectable, although the output is lower than calculated.
It can't work without Rin because that drops the excess voltage. I don't know what you mean by " create 5V from 12V" because this is a current, not a voltage regulator. The output voltage is the load resistance times the input current or R*I.
@@LewisLoflin Well, in the first circuit that uses the TL431, i used them as a positive / negative voltage regulator to create +/- 5V from +/-12V and now I'm designing another circuit with them for the same purpose.
Better to use LM317? the AT27C256 requires 6.5V and 13.5V, i have various zeners but not these values.
@@AnalogDude_ I have programmed those and used 12-13V and 5V.
Use an LM317.
@@LewisLoflin Thnx, I'm making a programmer for them using a Microchip Fubarino. gonna use transformer of 2 x 15V, a bridge rectifier en than a 7815. so the LM317 will have some 1.5 V head room to make 13.5 V, that's what the datasheet says.
The current passing through R1 and R2 has not been taken care why?
Not sure what you are asking. The current through R1, R2 is small compared to Q1.
@@LewisLoflin ok. Being very small , current passing through R1 and R2 has not been taken care of.
What happens when the load is disconnected?
Most of the remaining current will pass through Q1, as it has no current resistance. Thus, the Q1 must be over scaled to support all the remaining current when there is no load connected to the circuit.
How many volt of voltage clamp of TL431?
I don't understand the question.
Phúc Phan they are variable zener.
aplication circuit for balancing all types of batteries!
Having R1 adjustable is generally bad practice especially for a beginner. Sure, it has to be done to get down to 2.5V but there is a risk. Not much over 2.5V the reference pin starts conducting heavily and begins to destroy the 431 when it gets over 10ma. Good design can prevent this by the proper selection of resistors to prevent that in any case. It is quite easy for a beginner to destroy one should the load be removed. A resistor between the wiper and the reference pin to limit current might be appropriate for a beginner.
Thanks
Cool !!!
You did not explain WHY the draw is reduced by the transistor when the load draw increases.
As the load draws more current, the overall available current and voltage drop. That is detected by R1/R2 voltage divider, which allows the TL431 to change its current draw. That changes the voltage drop across R3, changes the current through Q1.
In this particular case ( 8:52), it seems that the transistor is useless since the TL431 can handle up to 150 mA, and we will have less than that: 300 mA total, 10 V/44000 Ohm = 0.2 mA through the tension divider, 167 mA through the lamp, 133 mA left, max, through the TL431. BUT the transistor is required if the lamp is ... not there (burn out, open circuit) since then, without it, the 300 mA will have to go through the TL431 and that is the double of what it is safe for it.
This is a demo on how the circuit works. The limit in the TL431 is 100 mA according to the spec sheet. thanks.
You are right, 100 mA is the maximum recommended; 150 mA is the absolute maximum under stress.
Thank you. In general, I'd run the TL451at no more than 30-50 mA or less. While the device is stable generating extra heat will effect drift as it does in semiconductor devices.
Ясно
the most inefficient way to explain things.