# Full wave rectifier with center-tap transformer and output stage of Filter and Regulator

#### huntersmile

Joined Jun 10, 2017
34
btw, it is standard practice to quote AC voltages in RMS value.

Hence mains voltage @ 120VAC RMS going to a step down 10:1 transformer will produce 12VAC RMS.
Then you convert to 12 x 1.414 = 17V amplitude.

Since these are both multiplication you end up with the same result.

Just wanted to let you know.
Alright noteddddd!
how about my other questions? do you have any idea?
I know the Vpeak at the capacitor voltage, then use ripple factor= Vrms/Vaverage right?
the Vrms is it just Vpeak/square root (3) since it is a sawtooth waveform? ==> I get Vaverage, then what to do? How to relate it to the Vp-p (peak to peak ripple) so that I can use the equation sqrt(2)*f*C*Vp-p= I load = 101mA?

#### MrAl

Joined Jun 17, 2014
8,868
Ok I think it is fine, they rectify peak voltage yes, then the average here means the average of the ripple voltage.
So I think I am fine to use that /pi equation.

Now would you please allow me to confirm whether it is radical 2 or just 2?
I found at some websites that it is just 2 : http://www.visionics.a.se/html/curriculum/Experiments/FW Rectifier/Full Wave Rectifier1.html

And the I, since we are in normal load condition, that means I = Izk (of zener) + Iload = 1mA+10.2mA=11.2mA right? 10.2 is from 5.1 divided by 500Ohm.

Then, after get C above, I now can get R from here (use I=11.2mA):
I*R= (Vrectified-dc) - 5.1V = 10.8 - 5.1
R=.....
Correct?

Thank you!

Hello there,

The average is the average using your given equation with the integral. That provides you with the average DC output voltage, as an approximation (ignores the rectifier diode drop). You integrate Vs*sin(w*t) from 0 to 2*pi with respect to wt (taking note to break that up into two integrals). You then work the rest as if that was really the average.

The small input resistor drops the input voltage by a very small amount. For example, if it was 1 ohm and you had 10ma input current then the voltage would drop by 10mv.

You can put this all into an equation that tells you what happens from the input to the very output.

#### huntersmile

Joined Jun 10, 2017
34
Hello there,

The average is the average using your given equation with the integral. That provides you with the average DC output voltage, as an approximation (ignores the rectifier diode drop). You integrate Vs*sin(w*t) from 0 to 2*pi with respect to wt (taking note to break that up into two integrals). You then work the rest as if that was really the average.

The small input resistor drops the input voltage by a very small amount. For example, if it was 1 ohm and you had 10ma input current then the voltage would drop by 10mv.

You can put this all into an equation that tells you what happens from the input to the very output.
Hello!
EHHHHHHH???Did you mean Vc,ave= 2*Vmax/pi this one, where Vmax is the peak voltage at the rectifier output(=secondary=16.97)?
Just now #12 told me that I must not use this one and I started to believe him for the reason that, this average is the average of voltage after the rectifier, before the filter.
But now I have filter, so I guess my average voltage is higher right? Am I correct?

#### MrChips

Joined Oct 2, 2009
25,928
This is a classic assignment in any introductory course in analog electronics.
The solution is a very complex one. However, we can arrive at satisfactory answers if we make some simple assumptions.

Let's begin.
Let us assume that your input is an AC voltage of 12VAC RMS = 17VAC amplitude from the output of the transformer.
Furthermore, we will assume that the transformer is a centre-tap transformer and you are using two rectifiers to produce a full-wave rectified signal.
Hence your peak amplitude will be 17V/2 = 8.5V
Subtract one diode forward voltage of 0.7V = 8.5 - 0.7 = 7.8V
This is the peak DC value from the power supply when the supply is lightly loaded.

The next assumption: we will assume that this is the average output voltage and will call this V.
Now we apply a load resistor R.
The current through the load is I = V/R.

The next assumption: we assume that this current is being supplied by the reservoir capacitor C over the complete half-cycle.
The period of this half-cycle is t = 1/2f where f is your AC line frequency.

The charge delivered by the capacitor is ΔQ = I x t = I x 1/2f = C x ΔV

Hence the voltage on the capacitor will fall by ΔV = I x 1/2fC = I/2fC

ΔV is the peak-to-peak ripple voltage = I/2fC

Now you can calculate your average DC voltage and percent ripple.

I will leave it up to you to calculate the peak diode current.
Warning: Again, this is very complicated and you have to make more assumptions.

Good luck!

#### huntersmile

Joined Jun 10, 2017
34
This is a classic assignment in any introductory course in analog electronics.
The solution is a very complex one. However, we can arrive at satisfactory answers if we make some simple assumptions.

Let's begin.
Let us assume that your input is an AC voltage of 12VAC RMS = 17VAC amplitude from the output of the transformer.
Furthermore, we will assume that the transformer is a centre-tap transformer and you are using two rectifiers to produce a full-wave rectified signal.
Hence your peak amplitude will be 17V/2 = 8.5V
Subtract one diode forward voltage of 0.7V = 8.5 - 0.7 = 7.8V
This is the peak DC value from the power supply when the supply is lightly loaded.

The next assumption: we will assume that this is the average output voltage and will call this V.
Now we apply a load resistor R.
The current through the load is I = V/R.

The next assumption: we assume that this current is being supplied by the reservoir capacitor C over the complete half-cycle.
The period of this half-cycle is t = 1/2f where f is your AC line frequency.

The charge delivered by the capacitor is ΔQ = I x t = I x 1/2f = C x ΔV

Hence the voltage on the capacitor will fall by ΔV = I x 1/2fC = I/2fC

ΔV is the peak-to-peak ripple voltage = I/2fC

Now you can calculate your average DC voltage and percent ripple.

I will leave it up to you to calculate the peak diode current.
Warning: Again, this is very complicated and you have to make more assumptions.

Good luck!
OK.... Thank you very much for your help..
Now I am trying to find the peak diode current as well as the peak surge current.... If you want to give me hint, please do so, hahaha.

#### #12

Joined Nov 30, 2010
18,223
By definition ripple factor= Vripplep-p/Vaverage right?
I don't know. I went to a cheap electronics school 50 years ago. They didn't teach me what's in your school book.
What do you mean by: "Average ripple factor? Peak ripple factor? Peak to peak ripple factor? I don't know. "
I mean, "I don't know."
I am worried that I make mistake.
I'm not. You're still in school. You have a teacher. Your customer is not going to send the police.
This is how you nail down exactly what the teacher wants and what the circuit needs. You try the best you know how today and on Monday you will know better.
Just now #12 told me that I must not use this one
I did not say that. I said you have to continue without me if you want to use a page of Calculus which I can't read.

Go ahead with these 2 Misters. They are both more educated than I am and it is past my bedtime.

#### MrChips

Joined Oct 2, 2009
25,928
OK.... Thank you very much for your help..
Now I am trying to find the peak diode current as well as the peak surge current.... If you want to give me hint, please do so, hahaha.
The hint is, all the current supplied to the load happens during the full half-cycle period.
However, the diode only conducts for a fraction of that period. The total charge drained from the reservoir capacitor must be supplied by the diode when it is in conduction mode.

#### huntersmile

Joined Jun 10, 2017
34
The hint is, all the current supplied to the load happens during the full half-cycle period.
However, the diode only conducts for a fraction of that period. The total charge drained from the reservoir capacitor must be supplied by the diode when it is in conduction mode.

Can't I just approximate it as ILmax/2 since every diode only conducts half of the cycle?

#### huntersmile

Joined Jun 10, 2017
34
I don't know. I went to a cheap electronics school 50 years ago. They didn't teach me what's in your school book.

I mean, "I don't know."

I'm not. You're still in school. You have a teacher. Your customer is not going to send the police.
This is how you nail down exactly what the teacher wants and what the circuit needs. You try the best you know how today and on Monday you will know better.

I did not say that. I said you have to continue without me if you want to use a page of Calculus which I can't read.

Go ahead with these 2 Misters. They are both more educated than I am and it is past my bedtime.
Thank you #12,,,,
you have been saving my day... glad that there are people like you who care and help others....

#### MrChips

Joined Oct 2, 2009
25,928
Can't I just approximate it as ILmax/2 since every diode only conducts half of the cycle?
No.
The voltage on anode of the diode is following a sine wave. The voltage on the cathode is hanging up there, an exponential decay of the C and R load. The diode stops conducting when the sine wave drops below the exponential decay (taking the diode forward voltage into account).

#### MrAl

Joined Jun 17, 2014
8,868
Hello!
EHHHHHHH???Did you mean Vc,ave= 2*Vmax/pi this one, where Vmax is the peak voltage at the rectifier output(=secondary=16.97)?
Just now #12 told me that I must not use this one and I started to believe him for the reason that, this average is the average of voltage after the rectifier, before the filter.
But now I have filter, so I guess my average voltage is higher right? Am I correct?
Hello there,

Well that's a very good point to bring up by both you and #12.

However, with these school problems and many times in real life the filter is assumed to be designed such that with the cap and load you actually get the average DC voltage as mentioned using the integral, and also the diode drops are ignored as suggested here. These things combined with the exact solution shown in the attachment tell us that we are dealing with an approximation, and because of the information already given for this problem it looks like the average voltage assumption is a good one and also that the diode drops are to be ignored (also not good for a low voltage supply but more about this in a moment).

We might now consider the alternatives.

First we have the exact solution for any full wave rectifier (four diodes or two diodes) as shown in the attachment. This requires a very lengthy calculation that most of us dont want to have to do, as well as including the equation for the diodes so we know exactly how much they drop throughout the entire half or full cycle (not a constant voltage drop).

Next, if we dont use the exact solution then we have to figure out how to estimate the drop by the diodes, which is quite impossible without using the exact equation for the diodes. Since we are ignoring the diode drops that means we are deep into approximation land because the diode drops are extremely dynamic. So what this means is we dont have a very good idea what the output would be, and because the diodes have such a dynamic voltage drop the output capacitor, if too large, will keep the average voltage fairly close to the peak. That's not something we want because the diodes are stressed too much during the peaks, so we want a reasonable cap value.

Next, ignoring the diode drops we do not have a capacitor value to work with, which means we have to choose our own capacitor vlaue. Now what makes more sense: choosing a cap that keeps the output close to the peaks, choosing a cap that allows a lot of ripple so we need a large overhead voltage, or choosing a cap value that gives us the average output DC voltage? I think you can guess the answer to this.

Choosing a cap value that gives us the average output DC voltage seems like a good choice, as that provides a cap value that is not too high and not too low, and also keeps the price of the cap (for production runs) at a decent level.

Now you could use a larger cap, but what would be the point. If you do that though you also wont be able to use that integral formula, and there will be more variance because of the way the diodes behave with increased peak current. So in short the problem gets a lot harder if you dont assume the average output DC as given by the integral.

If you really want to go all the way, then take a look at the equations given in the attachment. That's the only way to get it right, and even then you have to know the line impedance which usually isnt known exactly. So we are working with approximations and that means we have to choose one.

In conclusion, the output of a rectifier may actually be close to the peak value of the sine, but we dont want to design iit that way as the average voltage design is better for several reasons. Things change drastically if we add an input choke, but we dont want to do that here.

The exact solution (for any rectifier with little modification, and includes cap ESR):

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#### #12

Joined Nov 30, 2010
18,223
This is a spectacular example of how crazy a beginner can get by trying for perfection at every possible aspect of the calculations while an experienced designer can get this circuit working in 10 minutes with about 4 equations.

Perfection is not the goal. Every circuit is a great, huge, mass of compromises. Nothing you build in the real world will be perfectly what the math said it will be. In this case you are trying to get the math accuracy down to less than a tenth of a volt while ignoring the fact that the series resistor and the zener diode will compensate, even if your math is wrong by a whole volt.

It might feel very good to be able to calculate the theoretical answer down to a thousandth of a volt, after 10 pages of calculations that take a week to do, but you haven' t even considered the fact that every place you plug this into a wall socket, the power line voltage will be different from what you expected by as much as 5%. What's the use of being able to calculate to a thousandth of a volt when the power line voltage can be not what you expected by more than 8 volts at the peaks?

I am telling you the practical reality of circuits because that is the nature of who I am. I am very different from the text books, but I'm trying to save you thousands of hours of agonizing perfectionism by tempering the quest for knowledge with a cold splash of reality. Study hard. Learn good. Then realize that perfection doesn't matter.

#### MrAl

Joined Jun 17, 2014
8,868
This is a spectacular example of how crazy a beginner can get by trying for perfection at every possible aspect of the calculations while an experienced designer can get this circuit working in 10 minutes with about 4 equations.

Perfection is not the goal. Every circuit is a great, huge, mass of compromises. Nothing you build in the real world will be perfectly what the math said it will be. In this case you are trying to get the math accuracy down to less than a tenth of a volt while ignoring the fact that the series resistor and the zener diode will compensate, even if your math is wrong by a whole volt.

It might feel very good to be able to calculate the theoretical answer down to a thousandth of a volt, after 10 pages of calculations that take a week to do, but you haven' t even considered the fact that every place you plug this into a wall socket, the power line voltage will be different from what you expected by as much as 5%. What's the use of being able to calculate to a thousandth of a volt when the power line voltage can be not what you expected by more than 8 volts at the peaks?

I am telling you the practical reality of circuits because that is the nature of who I am. I am very different from the text books, but I'm trying to save you thousands of hours of agonizing perfectionism by tempering the quest for knowledge with a cold splash of reality. Study hard. Learn good. Then realize that perfection doesn't matter.
Hello there,

Sorry to say but you dont seem to be able to appreciate the value of cold hard theory. Also sorry to say that is a typical view from someone who only wants practical results. It's not that practical results dont mean anything or dont matter, they certainly do, but having the perfectionist theory is better than having the practical view alone because we can always degrade the theory to include the practical, but we will find it hard or impossible to upgrade the practical to the theory because the practical leaves out details that the theory always includes. In short, theory subsumes the practical. For example, taking that perfect solution and estimating some values leads us to a very very practical formula. But try going the other way, try finding that set of equations without going into detail. That's like saying you would be happy to use a bent ruler than a straight ruler in order to measure the length of your carpet. Out of all possible rulers, we'd want the straightest one so we can at least hope to get the right measurement. That shows that the most perfect theory is the best ruler. We can use a bent ruler, but only if that is the only one available because otherwise it would just be unreasonable.

What this means to us in a real life situation is that we can make better judgements about things to come, and we can always handle the variables that come up. For the line voltage variation scenario, in the theoretical case we can make detailed judgements about what happens when the line voltage varies, and in the practical case we can do the same thing just with less accuracy. So there we see we can deal with the situation to some degree in both cases but to say we have uncertainty and therefore we should not require the more detailed calculation is a mistake because we may want to someday analyze something about the circuit that is not included in the practical calculation. This is especially true in developing a new theory, or proving that the practical calculation works over a range of variables. By studying the variation of variables in the perfect equation we can prove or disprove any practical calculation.
A simple example is that it is impossible to study the effects of cap ESR on the ripple voltage without having a model (formula) that includes the cap ESR. How much does the ripple voltage change when the cap ESR doubles. We dont care if the input line voltage changes by 10 percent because that is not as significant to the calculation we need, although we could look at that in combination with the ESR to find out how much that changes the ripple too. Can we estimate this? Yes, we can, but we still need to include the ESR in the calculation.

Add to that the reason i posted the solution was to show that the perfect calculation requires knowing a lot of variables and since we dont want to go through that extreme in most cases we have to accept some form of an approximation. This in turn leads to finding a reasonable approximation from possibly a host of possibilities. This then leads to realizing that there are a lot of possibilities but not every one is reasonable so we try to weed out the less reasonable ones based on any other facts we happen to have on hand. The result i came up with was that the average voltage would be a reasonable assumption.

Now it is also reasonable to ask if this result is actually a great idea or just a good mediocre idea. That's where the theory comes in, and the better it is the better our conclusion will be.

One of the biggest problems with trying to ignore perfectionist theory is that formulas that estimate results are often created that work over only a range of the variables. That means that those kinds of formulas are difficult or impossible to use for advancing the theory to a new level without going all the way back to the basic concepts to try to figure out why the author estimated some parameters and how to get the more exact values back into the equations. So you see perfect theory can be a very good thing

There is an example of a full wave rectifier on the web somewhere that does NOT take into account the cap ESR. It is at least one full page long and probably would take up two pages or more in a regular book. The derivation was done by some professor and the result is analytical. We usually dont need that kind of perfection it's true, but when we do need it we need it and cant get it directly from a superficial estimation.

We can look into the point of using the average value DC as the work starting point. We can examine what the power losses are by using various cap values with some chosen load resistance, and see what the effects of varying the secondary voltage and cap value brings in. We should be able to find an optimum value for each thing in equation, or a set of optimum values if there is a range there. Would be interesting.

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#### MrChips

Joined Oct 2, 2009
25,928
This is the classic argument of theory vs practical, or scientist vs engineer.
I am both a scientist and an engineer and hence can appreciate both approaches.

However, how much time and effort are you willing to expend for something if even 10% deviation makes no difference to the outcome?

Take, for example, the forward voltage of the simple rectifier. Are you going to apply the Shockley diode equation with temperature and saturation current parameters?

Same thing goes for the zener regulator circuit. What is the correct theoretical zener voltage at what current and what temperature?
From a pragmatic perspective, wouldn't it be simpler and more expedient to give a 5.1V zener a knee voltage of 5.1V?

#12

#### #12

Joined Nov 30, 2010
18,223
Sorry to say but you don't seem to be able to appreciate the value of cold hard theory.
Wrong.
I know lots of theory and I can do math to useless numbers of digits after the decimal point. I just don't recommend doing that to design a circuit that interfaces with a big, sloppy, power grid for its input.

You want to know the speed limit of a digital circuit? You better know the math and do it, but the communication between digital chips is in a very protected and predictable environment. Even then, you plug it in before you send it to production, and you usually find out your first iteration has limitations and sometimes, outright flaws...after you did the most rigorous analysis you know how!

To quote a local signature line: In theory, there is no difference between theory and reality. In reality, there is.

#### MrAl

Joined Jun 17, 2014
8,868
This is the classic argument of theory vs practical, or scientist vs engineer.
I am both a scientist and an engineer and hence can appreciate both approaches.

However, how much time and effort are you willing to expend for something if even 10% deviation makes no difference to the outcome?

Take, for example, the forward voltage of the simple rectifier. Are you going to apply the Shockley diode equation with temperature and saturation current parameters?

Same thing goes for the zener regulator circuit. What is the correct theoretical zener voltage at what current and what temperature?
From a pragmatic perspective, wouldn't it be simpler and more expedient to give a 5.1V zener a knee voltage of 5.1V?

Hi,

Some good points, but this should not be that hard to figure out. Sometimes a certain parameter matters, sometimes id does not. Sometimes you want to include a parameter so that in the future you can work with that when questions come up about how it affects other things.

A good case in point is with the boost regulator and even some linear regulators with PNP output stage. If we dont consider the ESR of the output filter cap, we could get a design that oscillates, and that is because the ESR is known to have a profound effect on the stability. This is a well known issue. Also, since the computer is now a common tool of both scientist and engineer alike. it makes sense to include as many parameters as possible so we can call functions and gets results right away without having to change the circuit every time we want an answer.
For example, if we have a function like this that does not include ESR:
y=F(t,v,i)

we have to go into the circuit and modify the cap, then start making approximations as to how the ESR affects the outcome. But if someone before us took the time to include the cap ESR we'd call a function like this:
y=F(t,v,i,ESR)

and we'd have the answer, or we might have to call it two times to see what happens when the ESR change.

Furthermore, if the function is analytic, we can do lots of cool stuff with it like differentiate it and thus use it in more advanced formulas.

Also, if someone does take the time to create that function y=F(t,v,i,ESR) then someone else after that can follow their work and perhaps extend it to include some other parameter of interest, like say frequency: y=F(t,v,i,ESR,F). If they have the original function , then can extend it without having to go back and derive the expressions for the function that includes the ESR.

This is why we see papers written about getting exact results. Yes we also see papers that write about approximations, but for the advancement of theory we usually have to go back to the exact results so we can properly compare notes. If one person uses one approximation and antoher person uses a different approximation, they both think the other is wrong

Yes it is kinda like the Theory vs the Practical debate, and they both have their place, but that doesnt mean we just eliminate the theory just because it's a little harder to do. Some people actually prefer the challenge too, and the outcome can be very beautiful. It's like uncovering Nature down to the tiniest detail. No, it actually is uncovering Nature in as much detail as we possible can.

Remember we can always go backward and substitute approximations when we move into the practical, but now we can also study those approximations to see how good they are and when they work and when they fail.

There are actually some frontiers we cant even make good measurements in so we have to actually rely solely on theory. Without it we woudl have no good measuring tool.

#### MrAl

Joined Jun 17, 2014
8,868
Wrong.
I know lots of theory and I can do math to useless numbers of digits after the decimal point. I just don't recommend doing that to design a circuit that interfaces with a big, sloppy, power grid for its input.

You want to know the speed limit of a digital circuit? You better know the math and do it, but the communication between digital chips is in a very protected and predictable environment. Even then, you plug it in before you send it to production, and you usually find out your first iteration has limitations and sometimes, outright flaws...after you did the most rigorous analysis you know how!

To quote a local signature line: In theory, there is no difference between theory and reality. In reality, there is.
Hi,

I dont know who said that because it is outright wrong. Theory is made by man, reality is made by God if you believe in one or just Nature if you dont. We try to get as close to reality as we can with theory, and sometimes we can get pretty close.

I didnt say you didnt know theory, i said you dont seem to be able to appreciate it fully.

If you look back you'll see i presented the exact solution so we can use it as a measurement tool, not so that we have to solve it. It was used fro comparison to what an approximation would be. I agree that many times we use approximations in the local here and now case. It's in the more universal case we use the theory so we can get answers to problems that are more advanced in nature than a simple everyday design.

#### MrChips

Joined Oct 2, 2009
25,928
Hi,

Some good points, but this should not be that hard to figure out. Sometimes a certain parameter matters, sometimes id does not. Sometimes you want to include a parameter so that in the future you can work with that when questions come up about how it affects other things.

A good case in point is with the boost regulator and even some linear regulators with PNP output stage. If we dont consider the ESR of the output filter cap, we could get a design that oscillates, and that is because the ESR is known to have a profound effect on the stability. This is a well known issue. Also, since the computer is now a common tool of both scientist and engineer alike. it makes sense to include as many parameters as possible so we can call functions and gets results right away without having to change the circuit every time we want an answer.
For example, if we have a function like this that does not include ESR:
y=F(t,v,i)

we have to go into the circuit and modify the cap, then start making approximations as to how the ESR affects the outcome. But if someone before us took the time to include the cap ESR we'd call a function like this:
y=F(t,v,i,ESR)

and we'd have the answer, or we might have to call it two times to see what happens when the ESR change.

Furthermore, if the function is analytic, we can do lots of cool stuff with it like differentiate it and thus use it in more advanced formulas.

Also, if someone does take the time to create that function y=F(t,v,i,ESR) then someone else after that can follow their work and perhaps extend it to include some other parameter of interest, like say frequency: y=F(t,v,i,ESR,F). If they have the original function , then can extend it without having to go back and derive the expressions for the function that includes the ESR.

This is why we see papers written about getting exact results. Yes we also see papers that write about approximations, but for the advancement of theory we usually have to go back to the exact results so we can properly compare notes. If one person uses one approximation and antoher person uses a different approximation, they both think the other is wrong

Yes it is kinda like the Theory vs the Practical debate, and they both have their place, but that doesnt mean we just eliminate the theory just because it's a little harder to do. Some people actually prefer the challenge too, and the outcome can be very beautiful. It's like uncovering Nature down to the tiniest detail. No, it actually is uncovering Nature in as much detail as we possible can.

Remember we can always go backward and substitute approximations when we move into the practical, but now we can also study those approximations to see how good they are and when they work and when they fail.

There are actually some frontiers we cant even make good measurements in so we have to actually rely solely on theory. Without it we woudl have no good measuring tool.
To each his own.
I still use resistors with 10% tolerance and capacitors with 20%. When I wire up a monostable or 555-timer I discard the formulas and go with what my o'scope tells me.

#12

#### #12

Joined Nov 30, 2010
18,223
I didn't say you didn't know theory, i said you don't seem to be able to appreciate it fully.
I simply disagree. I appreciate theory just fine, and use it daily. I just don't design power supplies down to the last millivolt unless I'm doing a reference voltage.

Based on an old QC datasheet which I still have, in 1975, I designed a 28 volt linear supply that was accurate within .0002 volts from 0 to 4 amps of load and power line voltage from 105 VAC RMS to 125 VAC RMS, but that's not competitive in the retail market and that's not what I teach to beginners. If you think .000714% accuracy across line and load is showing no appreciation for theory we will just have to agree to disagree.

#### MrAl

Joined Jun 17, 2014
8,868
To each his own.
I still use resistors with 10% tolerance and capacitors with 20%. When I wire up a monostable or 555-timer I discard the formulas and go with what my o'scope tells me.
Hi,

Perhaps, but when you measure those values you usually want the best equipment you can afford, which usually means the best accuracy. You want the best tool for the job in almost every case, except those where you dont have it or cant get it then you accept a less accurate tool.