Tony your way of troubleshooting and explaining to your subscribers is excellent regardless of one's proficiency. Every time Mr. Carlton would build a kit on his channel prices of components would shoot up, so it's probably a good thing that you didn't get sources and emphasize purchasing said kits.
The only thing omitted was the most important part... WHY is the RMS is equivalent to the DC. And that can be explained with the oscilloscope by offsetting the AC Sinewave, so that it sits completely above the Zero Volt centreline. So now we have 5V RMS AC which is peaking at 5 x 1.414 = 7.07V Peak-to-Peak. Now "fill in" the section between zero volts line and the sinewave line. In other words, look at the amount of area (power) under the curve of the sinewave... If you calculate the entire filled area, it will be the same amount of area (power) as a continuous 5V DC (which is a straight horizontal line). The formula for other shapes of waveforms is different to the 0.707/1.414 ratios used for Sinewaves. The RMS of a Square Wave with equal 50:50 pulse widths is 50% of the Peak-to-Peak, because effectively the power is only occurring for half the time. So, as Tony says the RMS of an AC voltage is the amount of power that will do the same amount of work as the stated DC voltage.
Well done my friend,from my point of view is nice to come back to the basics.the learning experience we get from your videos is something you very rare find in this platform TK Sir
Nice to have a primer. I have a request. When you move from basics to actual circuits, make sure to cover the role of every component, each resistor and capacitor, even if it's totally clear to you. In my experience, at some point text writers lose patience and skip a few steps earlier than enemies of the electron like me are capable of treating as obvious. Things like pull-up resistors, filter capacitors, current limiting thisahs and thatahs -- at some point enough of these things show up at once that I begin to get lost and can follow only the gist. A particular example is the aforementioned current limiting resistors, which I always just equate with "space heater" and wonder why some less wasteful thing couldn't be used instead. It goes on like that. As an example, it does matter about that resistor right now. What's it *&(*&(*$R# for? And what does the initialism RMS expand to? You never once said the words "Root Mean Square." That's dumbing down the audience, and annoying people who dislike unexpanded acronyms and initialisms.
That is what I was thinking. The meter is in AC which removes the DC component, it doesn't know it is there. You would measure the DC first, and add it to the AC RMS reading for a total.
Thanks. I was hoping to understand why the digital scope would be "wrong". Is there value in the way the meter was doing it? (Even if that's all it could do for technical reasons.) Why wouldn't the scope offer both, or maybe have a toggle button, for example?
A great book that compliments this lecture series is "Practical Electronics for Inventors". (auth: Scherz & Monk) . I have the 4th edition. The title is a bit misleading. You don't have to "invent" anything to benefit from this book. It's now 900+ pages (softcover) and covers MANY of the subjects and measurements that Tony has talked about. I highly recommend it as an Electronics 101-201 book. Doesn't cover tubes though.
When you get to describing the bridge rectifier could you perhaps explain why some audio amplifiers have ceramic capacitors in parallel with each diode in the rectifier? I see this mostly on amplified computer speakers.
Tony, this is an excellent video! Best I have ever seen to explain AC and DC. Do the electons in an AC circuit actually travel a distance or is it just a shift in polarity going positive and negative, such as a magnetic shift in polarity? 73, AA4EZ
I've always understood that in ac the electrons move forwards and then reverse backwards. In effect they end up more or less where they started bearing in mind that if the current is interupted part way in a cycle they will be displaced slightly.
@@jonka1 That has been my understanding also, that the electrons just kinda shift back and forth without really "traveling" any distance, as in an electron "flow". It isn't really a flow of electrons going from one charged pole to the opposite pole, but a transfer of that energy from one electron to the next, with that energy being "used" by the load, if one is present. Does that fit in with your understanding?
@@raymondlewis2055 I'm not certain that I am up to speed with that idea. I can see that true AC (not varying DC) would fit but as to what happens with DC, I'm not sure. It makes sense that electrons would flow onwards but I've never explored this. BTW I have 55 years of hobby electronics behind me and stuff seems to work out well.
I measured the output power of my 1965 Magnavox console with 8 ohm dummy loads at 1KHz and got 20 volts PP at the onset of clipping, which would be 10 volts peak AC. RMS would be peak X .707 = 7.07 volts RMS, that squared then divided by 8 = 6.25 watts RMS per channel.
Tony, great explanation. I'm wondering, does the frequency of the AC signal change the average (RMS) value? Your example shows 60 Hz. What if the signal was 60,000 Hz?
The RMS value of any waveform at any frequency will be the equivalent to its DC value. Both the DC and RMS values are capable of performing the same amount of work. Only a sine wave will have an RMS value that is 0.707 (which is 1/√2) of the peak. For non-sine waves you need to calculate the square root of the mean of the squared magnitude of the signal. Lower cost meters can't accurately measure the RMS value of non-sinusoidal waveforms. A meter that is "True RMS" can accurately measure the RMS of non-sinusoidal waveforms, as long as the frequency is within the range of the meter. ;)
Some still call pulsating DC, or AC with a DC offset, as AC. In a debate on another channel I tried pointing out crossing over zero point and changing polarity to qualify as true AC. His definition of DC was a constant current without any variations or change. I had to say we have to agree to disagree and part ways, our training in electrical school was different and we were getting nowhere in a debate. Have you Tony, or anyone out there run into this?
According to Tony's definition (in this video), it changes between "Pulsed DC" and "AC" when you move the zero volt reference point. In actual fact, it makes absolutely no difference whatsoever... all the reference points are arbitrary anyway.
@@zulumax1 - you need to read my comment properly before arguing. Also, research the meaning of the term "arbitary", when used in the context that I wrote it..
I think Tony’s scope is correct and Tony is wrong on this point. The scope (correctly) reports AC with DC offset - NOT DC as Tony claims. The wave form is sinusoidal = AC, in my opinion. If the scope were fed the output of a full-wave rectifier, I’ll bet that it would correctly interpret it as (pulsating) DC. So, in this case, I feel Tony muddles the issue for newbies by introducing a waveform not commonly seen (at least by me). It would be more helpful, my opinion, to show the student an AC waveform compared to the output of either, or both, a half-wave and a full-wave rectifier which is universally interpreted as DC by both analog and digital meters, let alone DC motors and the like. Having said all that, I can certainly empathize with the difficulty of explaining this to people who are unfamiliar with the distinction between AC and DC. I spent two months this summer trying to explain how rectifiers worked to groups of would-be electricians in our local community college. It was an exercise in frustration (mine), it apparently is a difficult concept to grasp if you don’t get it early on.
(@3:29) - Whoa pony , whoa! You absolutely can have voltage without current. A battery sitting there has voltage, but (for all practical purposes) no current. Likewise, a dead short will have current, but no voltage across it. Okay, giddy-up, pony! (Or, should I say, “Tony”!)
A battery, capacitor, etc. can have a potential charge on it. This is not the same as voltage and current. Voltage and current are the factors used to determine how much work is being done. The Work is defined by its electrical power and power is the product of Voltage and Current. In other words, P=EI. If either I or E = 0, then P would equal zero. You must have Voltage AND Current to have power. Even a dead short has a very small amount of resistance, which would equal a very small voltage. The exception is in the case of superconductors, where there is zero resistance and very high current, but not how we are used to looking at it. The energy in an MRI cold-head magnet is very high, but not infinite. You can't really measure voltage or current (unless you are ramping the magnet up or down, in which there is a certain amount of voltage & current in the resistor in the ramp circuit) and you need to use quantum mechanics to explain what's going on inside the coil. When you start talking about Cooper pairs and the quantum state of a superconductor, the whole concept of current flow and voltage is no longer relevant. The energy still produces a resultant magnetic field, which is still subject to Maxwell's equations. I don't really know as much about this stuff as some of my friends do, as I don't work that much with it and they do every day. So that's the best explanation the village idiot can give ;) Good to hear from you William! I always enjoy your comments!
@@xraytonyb - Actually, it is: here’s the definition from Wikipedia (search: voltage) “Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points.” (In this case, between the terminals of your battery, or your bench power supply.) It can also be thought of as the force required to separate unlike charges, or the force needed to keep them separated. Just because one isn’t measuring the voltage across a battery, doesn’t mean the voltage goes away - that would royally suck! And, yes, a big, copper bus bar stuck across a car battery will quickly drain the battery, and it’ll get really hot (as it’s internal ESR is higher than that of the bus bar) and maybe burst, catch fire, or explode. The voltage drop across the short, though, will be negligible compared to the nominal open-circuit voltage. Does anyone know what the ESR of a typical lead-acid car battery is, anyway? That being said, though, those are edge-cases, and only the open-circuit case is normally encountered (until you forget to use an isolation transformer on a hot-chassis set, and you go to measure the voltage somewhere in the circuit with your ‘scope - oops)!
@@HazeAnderson True - that’s why batteries left on the store shelves for decades are no good when you get them home. Of course, measuring said leakage isn’t quite as easy as it is with capacitors. Of course, I should have been more specific; there’s no current “external to the battery” (since there’s no complete circuit).
Some digital meters do not measure RMS voltage correctly. I believe the old analog meters like a Simpson 260 read RMS as a function of their mechanical nature due to inertia of the mechanical movement.
I like the shorter videos. It gives time to "digest" what you've said. When you post the longer videos thing tend to "overflow" and get lost. Keep in mind that average attention span is around 25 minutes or so. As for this series, I like the way you present things. I have an advanced class Ham license so what you are saying is familiar to me. But there are people who are just getting into electronics and this way they don't get scared off by all the tech jargon.
18 millivolts shouldn't be enough to trip the protection circuit. I would start by checking the power supplies as well as checking for DC offset BEFORE the protect relay.
Great explanation of R.M.S. and why we use it. Could you explain why that doesn't work with a non-sinusoidal wave? (Or maybe that's for a different video). (I'm not trying to show off - I know it doesn't work, but I couldn't explain why properly).
The RMS of ANY waveform will always equal the same DC value. 1 volt RMS will do the same amount of work as 1 volt DC. The issue is that the √2 formula only applies to a pure sine wave, if you are calculating the RMS value. For non-sine waves you need to calculate the square root of the mean of the squared magnitude of the signal, which is a bit more tricky. True RMS meters (like the fluke 187 I was using) will measure the actual RMS value of an AC signal regardless of the waveform. The meter still has a limit on the maximum frequency it can accurately measure.
@@xraytonyb Thanks for your detailed reply. I did mean the square root formula wouldn't work on none sinusoidal waveforms. As you say any waveform has an RMS value. But I haven't learned the math. I didn't know a true RMS meter would show the RMS of any waveform (ltd by frequency constraints) but I guess the electronics works the same.
Just been to a website which shows the formulae for different waveform RMS values. Thought I should learn at least the squarewave formula as I repair SMPS units for a living 🤣
@@johncoops6897 LOL yes it is. Kind of dumb of me not to realise that. I think I've been fixing SMPS for too long and stopped thinking about the theory.
I got a intake like that input for a bike kit system a little more advanced than this since you need to draw charge at once on top used a dynamo solar AC 5v one amp charger... so upto 20v at around half a amp constant unless slow to charge a high power front, indicator and dule brake gps speedometer horn want to app it learning 😁 curcit drawn prototype in works to make sure time should fit all 50mm frames may get 25 maybe!. Had to do a few things to get it working with load drop over!. And stick a inductor dule line should help with temp maybe???😃😄😁😆😅 5 Smiles generally to current.
AC/DC is still such a popular rock band because they still use tube amps. 🤪
That is the clearest demonstration of DC vs RMS AC that I have come across. Thank you very much for putting this together.
Yes i totally agree. Well done Tony!
This is the level of detail it takes to get these concepts into peoples' heads. Very well explained!
Fantastic Tony; I so look forward to future videos in this series
I find your voice soothing but authoritive. Thanks Tony.
Tony your way of troubleshooting and explaining to your subscribers is excellent
regardless of one's proficiency.
Every time Mr. Carlton would build a kit on his channel prices of components
would shoot up, so it's probably a good thing that you didn't get sources and
emphasize purchasing said kits.
The only thing omitted was the most important part... WHY is the RMS is equivalent to the DC.
And that can be explained with the oscilloscope by offsetting the AC Sinewave, so that it sits completely above the Zero Volt centreline.
So now we have 5V RMS AC which is peaking at 5 x 1.414 = 7.07V Peak-to-Peak. Now "fill in" the section between zero volts line and the sinewave line. In other words, look at the amount of area (power) under the curve of the sinewave... If you calculate the entire filled area, it will be the same amount of area (power) as a continuous 5V DC (which is a straight horizontal line).
The formula for other shapes of waveforms is different to the 0.707/1.414 ratios used for Sinewaves. The RMS of a Square Wave with equal 50:50 pulse widths is 50% of the Peak-to-Peak, because effectively the power is only occurring for half the time.
So, as Tony says the RMS of an AC voltage is the amount of power that will do the same amount of work as the stated DC voltage.
thank you for making this video x-ray tony
Well done my friend,from my point of view is nice to come back to the basics.the learning experience we get from your videos is something you very rare find in this platform TK Sir
Nice to have a primer. I have a request. When you move from basics to actual circuits, make sure to cover the role of every component, each resistor and capacitor, even if it's totally clear to you. In my experience, at some point text writers lose patience and skip a few steps earlier than enemies of the electron like me are capable of treating as obvious. Things like pull-up resistors, filter capacitors, current limiting thisahs and thatahs -- at some point enough of these things show up at once that I begin to get lost and can follow only the gist. A particular example is the aforementioned current limiting resistors, which I always just equate with "space heater" and wonder why some less wasteful thing couldn't be used instead. It goes on like that.
As an example, it does matter about that resistor right now. What's it *&(*&(*$R# for?
And what does the initialism RMS expand to? You never once said the words "Root Mean Square." That's dumbing down the audience, and annoying people who dislike unexpanded acronyms and initialisms.
What's it *&(*&(*$R# for?
That doesn't even make sense.... at some point we readers just lose patience and stop reading.
Just watched part 2. Thanks.
20:26 - I would say that the 'scope is correct and that "True RMS" meter is lying! It's not taking into account the DC offset!
That is what I was thinking. The meter is in AC which removes the DC component, it doesn't know it is there. You would measure the DC first, and add it to the AC RMS reading for a total.
Thanks. I was hoping to understand why the digital scope would be "wrong". Is there value in the way the meter was doing it? (Even if that's all it could do for technical reasons.) Why wouldn't the scope offer both, or maybe have a toggle button, for example?
I'm ready for my test Mr. Xray
Great stuff, thanks.
Great lab demo of rms! Much better than all the trig mumbo-haha.🙃
Excellent tut!
A great book that compliments this lecture series is "Practical Electronics for Inventors". (auth: Scherz & Monk) . I have the 4th edition. The title is a bit misleading. You don't have to "invent" anything to benefit from this book. It's now 900+ pages (softcover) and covers MANY of the subjects and measurements that Tony has talked about. I highly recommend it as an Electronics 101-201 book. Doesn't cover tubes though.
When you get to describing the bridge rectifier could you perhaps explain why some audio amplifiers have ceramic capacitors in parallel with each diode in the rectifier? I see this mostly on amplified computer speakers.
Thanks!
Tony, this is an excellent video! Best I have ever seen to explain AC and DC. Do the electons in an AC circuit actually travel a distance or is it just a shift in polarity going positive and negative, such as a magnetic shift in polarity? 73, AA4EZ
Even in DC, the electrons themselves move relatively slowly, transferring their energy to those downstream. In AC.
I've always understood that in ac the electrons move forwards and then reverse backwards. In effect they end up more or less where they started bearing in mind that if the current is interupted part way in a cycle they will be displaced slightly.
@@jonka1 That has been my understanding also, that the electrons just kinda shift back and forth without really "traveling" any distance, as in an electron "flow". It isn't really a flow of electrons going from one charged pole to the opposite pole, but a transfer of that energy from one electron to the next, with that energy being "used" by the load, if one is present. Does that fit in with your understanding?
@@raymondlewis2055 I'm not certain that I am up to speed with that idea. I can see that true AC (not varying DC) would fit but as to what happens with DC, I'm not sure. It makes sense that electrons would flow onwards but I've never explored this. BTW I have 55 years of hobby electronics behind me and stuff seems to work out well.
I measured the output power of my 1965 Magnavox console with 8 ohm dummy loads at 1KHz and got 20 volts PP at the onset of clipping, which would be 10 volts peak AC.
RMS would be peak X .707 = 7.07 volts RMS, that squared then divided by 8 = 6.25 watts RMS per channel.
Tony, great explanation. I'm wondering, does the frequency of the AC signal change the average (RMS) value? Your example shows 60 Hz. What if the signal was 60,000 Hz?
Wouldn't change the RMS value. Some meters might not be able to read it accurately though. Some cheaper meters quit reading accurately at a few khz.
The RMS value of any waveform at any frequency will be the equivalent to its DC value. Both the DC and RMS values are capable of performing the same amount of work. Only a sine wave will have an RMS value that is 0.707 (which is 1/√2) of the peak. For non-sine waves you need to calculate the square root of the mean of the squared magnitude of the signal. Lower cost meters can't accurately measure the RMS value of non-sinusoidal waveforms. A meter that is "True RMS" can accurately measure the RMS of non-sinusoidal waveforms, as long as the frequency is within the range of the meter. ;)
Some still call pulsating DC, or AC with a DC offset, as AC.
In a debate on another channel I tried pointing out crossing over zero point and changing polarity to qualify as true AC. His definition of DC was a constant current without any variations or change. I had to say we have to agree to disagree and part ways, our training in electrical school was different and we were getting nowhere in a debate. Have you Tony, or anyone out there run into this?
Yes. Many times
According to Tony's definition (in this video), it changes between "Pulsed DC" and "AC" when you move the zero volt reference point. In actual fact, it makes absolutely no difference whatsoever... all the reference points are arbitrary anyway.
@@johncoops6897
When there is no current flow, that is the zero reference point, it is not arbitrary.
@@zulumax1 - you need to read my comment properly before arguing. Also, research the meaning of the term "arbitary", when used in the context that I wrote it..
I think Tony’s scope is correct and Tony is wrong on this point. The scope (correctly) reports AC with DC offset - NOT DC as Tony claims. The wave form is sinusoidal = AC, in my opinion. If the scope were fed the output of a full-wave rectifier, I’ll bet that it would correctly interpret it as (pulsating) DC. So, in this case, I feel Tony muddles the issue for newbies by introducing a waveform not commonly seen (at least by me). It would be more helpful, my opinion, to show the student an AC waveform compared to the output of either, or both, a half-wave and a full-wave rectifier which is universally interpreted as DC by both analog and digital meters, let alone DC motors and the like.
Having said all that, I can certainly empathize with the difficulty of explaining this to people who are unfamiliar with the distinction between AC and DC. I spent two months this summer trying to explain how rectifiers worked to groups of would-be electricians in our local community college. It was an exercise in frustration (mine), it apparently is a difficult concept to grasp if you don’t get it early on.
(@3:29) - Whoa pony , whoa! You absolutely can have voltage without current. A battery sitting there has voltage, but (for all practical purposes) no current. Likewise, a dead short will have current, but no voltage across it. Okay, giddy-up, pony! (Or, should I say, “Tony”!)
A battery, capacitor, etc. can have a potential charge on it. This is not the same as voltage and current. Voltage and current are the factors used to determine how much work is being done. The Work is defined by its electrical power and power is the product of Voltage and Current. In other words, P=EI. If either I or E = 0, then P would equal zero. You must have Voltage AND Current to have power. Even a dead short has a very small amount of resistance, which would equal a very small voltage. The exception is in the case of superconductors, where there is zero resistance and very high current, but not how we are used to looking at it. The energy in an MRI cold-head magnet is very high, but not infinite. You can't really measure voltage or current (unless you are ramping the magnet up or down, in which there is a certain amount of voltage & current in the resistor in the ramp circuit) and you need to use quantum mechanics to explain what's going on inside the coil. When you start talking about Cooper pairs and the quantum state of a superconductor, the whole concept of current flow and voltage is no longer relevant. The energy still produces a resultant magnetic field, which is still subject to Maxwell's equations. I don't really know as much about this stuff as some of my friends do, as I don't work that much with it and they do every day. So that's the best explanation the village idiot can give ;) Good to hear from you William! I always enjoy your comments!
@@xraytonyb - Actually, it is: here’s the definition from Wikipedia (search: voltage) “Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points.” (In this case, between the terminals of your battery, or your bench power supply.) It can also be thought of as the force required to separate unlike charges, or the force needed to keep them separated. Just because one isn’t measuring the voltage across a battery, doesn’t mean the voltage goes away - that would royally suck! And, yes, a big, copper bus bar stuck across a car battery will quickly drain the battery, and it’ll get really hot (as it’s internal ESR is higher than that of the bus bar) and maybe burst, catch fire, or explode. The voltage drop across the short, though, will be negligible compared to the nominal open-circuit voltage. Does anyone know what the ESR of a typical lead-acid car battery is, anyway?
That being said, though, those are edge-cases, and only the open-circuit case is normally encountered (until you forget to use an isolation transformer on a hot-chassis set, and you go to measure the voltage somewhere in the circuit with your ‘scope - oops)!
@@williamsquires3070 a battery just sitting there most definitely has current. Why? Because it has internal resistance. No battery lasts forever.
@@HazeAnderson True - that’s why batteries left on the store shelves for decades are no good when you get them home. Of course, measuring said leakage isn’t quite as easy as it is with capacitors. Of course, I should have been more specific; there’s no current “external to the battery” (since there’s no complete circuit).
Some digital meters do not measure RMS voltage correctly. I believe the old analog meters like a Simpson 260 read RMS as a function of their mechanical nature due to inertia of the mechanical movement.
I like the shorter videos. It gives time to "digest" what you've said. When you post the longer videos thing tend to "overflow" and get lost. Keep in mind that average attention span is around 25 minutes or so. As for this series, I like the way you present things. I have an advanced class Ham license so what you are saying is familiar to me. But there are people who are just getting into electronics and this way they don't get scared off by all the tech jargon.
A have a sansui au719 having protect issues have 18.77 mv at idle is this a sign of a problem
18 millivolts shouldn't be enough to trip the protection circuit. I would start by checking the power supplies as well as checking for DC offset BEFORE the protect relay.
Great explanation of R.M.S. and why we use it. Could you explain why that doesn't work with a non-sinusoidal wave? (Or maybe that's for a different video). (I'm not trying to show off - I know it doesn't work, but I couldn't explain why properly).
The RMS of ANY waveform will always equal the same DC value. 1 volt RMS will do the same amount of work as 1 volt DC. The issue is that the √2 formula only applies to a pure sine wave, if you are calculating the RMS value. For non-sine waves you need to calculate the square root of the mean of the squared magnitude of the signal, which is a bit more tricky. True RMS meters (like the fluke 187 I was using) will measure the actual RMS value of an AC signal regardless of the waveform. The meter still has a limit on the maximum frequency it can accurately measure.
@@xraytonyb Thanks for your detailed reply. I did mean the square root formula wouldn't work on none sinusoidal waveforms. As you say any waveform has an RMS value. But I haven't learned the math. I didn't know a true RMS meter would show the RMS of any waveform (ltd by frequency constraints) but I guess the electronics works the same.
Just been to a website which shows the formulae for different waveform RMS values. Thought I should learn at least the squarewave formula as I repair SMPS units for a living 🤣
@@radio-ged4626 - The square wave formula is the Mark/Space Ratio.
@@johncoops6897 LOL yes it is. Kind of dumb of me not to realise that. I think I've been fixing SMPS for too long and stopped thinking about the theory.
I got a intake like that input for a bike kit system a little more advanced than this since you need to draw charge at once on top used a dynamo solar AC 5v one amp charger... so upto 20v at around half a amp constant unless slow to charge a high power front, indicator and dule brake gps speedometer horn want to app it learning 😁 curcit drawn prototype in works to make sure time should fit all 50mm frames may get 25 maybe!. Had to do a few things to get it working with load drop over!. And stick a inductor dule line should help with temp maybe???😃😄😁😆😅 5 Smiles generally to current.
Beware your eyesight may fail.
Come back when the drugs wear off, Troy.
Hi Tony, I believe your kit is a 6P1 push pull not 6P14.
You're right. It is a 6P1 output.
Xraytonyb the Amp kit is cool