# Whats the output voltage of my filter capacitor for my LM338 voltage regulator

#### Kim Obordo

Joined Jun 21, 2018
2
So here's the deal: I made a power supply with a step down transformer (60Hz) 220v-24v operating at 1amp, with a full bridge rectifier(1N4001). i have a 1000uF filter capacitor (50v). and an LM338T adjustable voltage regulator. I need it to output 2-30v. It works...
my professor taught us that our filter capacitor's output is defined by the formula:
Vrms=(current)/(4(sqroot of 3) (frequency)(capacitance))
VDC= (rectified voltage) - (current)/4(frequency)(capacitance)

as far as I know the voltage regulator needs a bigger input voltage than what it can put out.
but calculating for the capacitors output dc voltage: (I assumed that the diodes had0.7v as their voltage drop)

VDC= (24(sqroot of 2) - 0.7v - 0.7v) - ( (1amp)/(4)(60hz)(1000uF))

i get 28.37v for the regulators input (the caps output) which doesn't make sense when I can easily get 32v as its output.

Am I doing something wrong here?

#### AlbertHall

Joined Jun 4, 2014
12,328
I haven't checked the arithmetic but were you drawing 1A from the output while you were getting 32V output?

#### MrAl

Joined Jun 17, 2014
11,253
So here's the deal: I made a power supply with a step down transformer (60Hz) 220v-24v operating at 1amp, with a full bridge rectifier(1N4001). i have a 1000uF filter capacitor (50v). and an LM338T adjustable voltage regulator. I need it to output 2-30v. It works...
my professor taught us that our filter capacitor's output is defined by the formula:
Vrms=(current)/(4(sqroot of 3) (frequency)(capacitance))
VDC= (rectified voltage) - (current)/4(frequency)(capacitance)

as far as I know the voltage regulator needs a bigger input voltage than what it can put out.
but calculating for the capacitors output dc voltage: (I assumed that the diodes had0.7v as their voltage drop)

VDC= (24(sqroot of 2) - 0.7v - 0.7v) - ( (1amp)/(4)(60hz)(1000uF))

i get 28.37v for the regulators input (the caps output) which doesn't make sense when I can easily get 32v as its output.

Am I doing something wrong here?
Hello there,

There are a few points to be made here that have to do with both accuracy of measurement and how measurement instruments make measurements.

First, we have the error in the primary voltage, which might be higher than thought.
Second, we have the error in output voltage which could be higher than thought, which leads to a higher than thought peak voltage.
Third, we have the formula which is only an approximation subject to some deviation, and is actually the minimum DC voltage for the output which contains ripple.

All this taken together means that your meter might be measuring the average voltage which will not show you the same reading as the VDC voltage you are calculating because the calculation yields the voltage at the valley of the ripple and it's just an approximation anyway. And, because the differences are so little, it only takes a little difference to make it look like you are getting 32v output when really it is less, or you may actually get 32v output if the line or output voltages for the transformer are actually a little higher. It only takes a little, so it may look better than the calculation yields, or it may actually be better because of other errors.

To see what is happening you could look at the output with a scope, and there you may see some ripple even after the regulator. If you dont, then you got lucky with the line voltage and transformer ratio, and the cap may even be a little higher in capacitance which would make the ripple valley voltage higher and thus allow that 32v output too. It could also be combination of effects that work in your favor.

So next maybe check with a scope. The average voltage is around 31v going by the assumed perfect input and output and calculation voltages, so we are not far from seeing 32v output if something isnt quite what we think it is.
Once you see the output with the scope, you can tell what it is that is causing this to look like it is working, or actually work despite the calculation which we are basing on perfect knowledge of variables which in real life we almost never have complete knowledge of.

#### Picbuster

Joined Dec 2, 2013
1,047
MrAI is correct.
First of all you measure the 'open' voltage at the bridge. (open is no load).
this should as stated 24 x 1.41 32V approx.
A load will produce ripple this is defined by the value of running current,circuit resistance, capacitor and frequency.
The discharge of the capacitor is causing the ripple.
That implies that the 'open voltage' is reduced by the effect of voltage ripple.
What will you meter show you? I don't know but is measure device depended.
Again MrALI gives the correct solution use a scope.

Picbuster

#### AlbertHall

Joined Jun 4, 2014
12,328
First of all you measure the 'open' voltage at the bridge. (open is no load).
this should as stated 24 x 1.41 32V approx.
There is another complication here because if the transformer is rated for 24V at 1A, its output voltage with no load will be higher than 24V. How much higher depends on the particular transformer specification.
Hence my quetion in post #2.

#### Kim Obordo

Joined Jun 21, 2018
2
Hello there,

There are a few points to be made here that have to do with both accuracy of measurement and how measurement instruments make measurements.

First, we have the error in the primary voltage, which might be higher than thought.
Second, we have the error in output voltage which could be higher than thought, which leads to a higher than thought peak voltage.
Third, we have the formula which is only an approximation subject to some deviation, and is actually the minimum DC voltage for the output which contains ripple.

All this taken together means that your meter might be measuring the average voltage which will not show you the same reading as the VDC voltage you are calculating because the calculation yields the voltage at the valley of the ripple and it's just an approximation anyway. And, because the differences are so little, it only takes a little difference to make it look like you are getting 32v output when really it is less, or you may actually get 32v output if the line or output voltages for the transformer are actually a little higher. It only takes a little, so it may look better than the calculation yields, or it may actually be better because of other errors.

To see what is happening you could look at the output with a scope, and there you may see some ripple even after the regulator. If you dont, then you got lucky with the line voltage and transformer ratio, and the cap may even be a little higher in capacitance which would make the ripple valley voltage higher and thus allow that 32v output too. It could also be combination of effects that work in your favor.

So next maybe check with a scope. The average voltage is around 31v going by the assumed perfect input and output and calculation voltages, so we are not far from seeing 32v output if something isnt quite what we think it is.
Once you see the output with the scope, you can tell what it is that is causing this to look like it is working, or actually work despite the calculation which we are basing on perfect knowledge of variables which in real life we almost never have complete knowledge of.

I forgot to mention it, I did use an Oscilloscope on its parts from the bridge to the voltage regulator. The regulator outputs a flat dc signal.

shoot me down if I'm wrong but does the filter capacitor really have a voltage drop?... that is does it really take down the rectified signal a notch before passing it on to the regulator? cause every power supply tutorial I found (including the legitimate ones) show that the filter capacitor shouldn't right, it should just charge up to the applied voltage and discharge as the rectified voltage dips down to zero...

I mean the cap is in parallel with the rectifier and the voltage regulator so that would mean that they should share the same DC voltage right? and by that the regulator should only receive an input of 24(1.414...)v-0.7v-0.7v= 32.54v which would make a lot more sense when i get 32v as its output.

so the Filter caps Vdc output (the one my prof. defined earlier) would be its discharging voltage (i mean its voltage just before it gets charged back up again) and shouldn't matter. Yes the cap attenuates the signal but it shouldn't change the signals Vpk(or as far as I know it as the voltage that goes into the regulator.

Or at least thats what I think its true right now... so if anyone sees a flaw please... I need your help

#### ebp

Joined Feb 8, 2018
2,332
Accurately calculating the behavior of the filter is astoundingly complex.

In practical circuits the filter capacitors will somewhat reduce the peak voltage because the current at the peak is high and the source impedance is not zero.

Current through the rectifier flows only when the instantaneous voltage is higher than the capacitor voltage. This means there are "pulses" of current, vaguely sinusoidal in shape, that flow starting before and ending after the peak of the voltage from the input source. The larger the capacitor, the narrower the pulse and hence the higher the peak current and RMS current - but the average current remains exactly the same if there is a voltage regulator down-stream. Because the source impedance is not zero, those high peak currents will drop the peak voltage seen at the terminals of the source, with the rest of the "theoretical" voltage appearing across the internal impedance of the source.

Transformer voltage is typically specified at full rated load. For small transformers, the effective source impedance (mostly winding resistance) can be quite high, so the open circuit voltage may be a good deal (10 to 20% is not uncommon for small xfmrs) higher than the full-load voltage.

Simulation is a great way of experimenting with filtered rectifier circuits. You can play with source impedance (resistance in series with an ideal source, though most sim packages allow you to spec the source impedance), capacitor values, etc. and get both waveform plots and numerical values for peak, average and RMS currents and voltages. Lots of people at AAC use LTSpice from Linear Technology (now part of Analog Devices). I've never used it (don't do electronics anymore) but it looks very powerful and is ... FREE!

#### crutschow

Joined Mar 14, 2008
33,983
Also it's important to note that the DC output current needs to be derated at least 60% from the transformer's RMS output current rating, due to the high peak RMS currents that a full-wave bridge rectifier-capacitor takes from the transformer.
Thus you should draw no more than 0.6Adc from your 1Arms rated transformer, otherwise it can overheat.

Another way to look at it is from a transformer power output point-of-view.
The rectified output is at a voltage ≈1.4 times the RMS voltage so, for the same DC current output as its RMS (or VA) rating, you would drawing 1.4 times the power from the transformer over what it's rated for.

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#### MrAl

Joined Jun 17, 2014
11,253
I forgot to mention it, I did use an Oscilloscope on its parts from the bridge to the voltage regulator. The regulator outputs a flat dc signal.

shoot me down if I'm wrong but does the filter capacitor really have a voltage drop?... that is does it really take down the rectified signal a notch before passing it on to the regulator? cause every power supply tutorial I found (including the legitimate ones) show that the filter capacitor shouldn't right, it should just charge up to the applied voltage and discharge as the rectified voltage dips down to zero...

I mean the cap is in parallel with the rectifier and the voltage regulator so that would mean that they should share the same DC voltage right? and by that the regulator should only receive an input of 24(1.414...)v-0.7v-0.7v= 32.54v which would make a lot more sense when i get 32v as its output.

so the Filter caps Vdc output (the one my prof. defined earlier) would be its discharging voltage (i mean its voltage just before it gets charged back up again) and shouldn't matter. Yes the cap attenuates the signal but it shouldn't change the signals Vpk(or as far as I know it as the voltage that goes into the regulator.

Or at least thats what I think its true right now... so if anyone sees a flaw please... I need your help

Hello again,

Well if you read my previous reply i mentioned several errors and the way to tell WHICH error is more likely you should check with a scope. Since you did that, that tells you that the voltage you are measuring on the output is the true flat DC voltage and that was one of the possibilities i had mentioned. That means that the measuring instrument used to measure voltage is not at fault, and that in turn means that it is something else.
That something else could be a higher than normal line voltage, or a higher than thought secondary output voltage from the transformer. That would the the next place to check hopefully using the same test equipment.

The error we are talking about here isnt that large, maybe 1 or 2 volts, so it's possible that the transformer was wound to allow for a higher load current, which would mean the output with only 1 amp could be higher than 24v. In fact, it could be 25v or higher.

So next you might check the output of the transformer to see what the voltage really is. Peak voltage would be good to measure and check for any waveform flat topping.

In cases like this when you have a question of theory vs practice it makes sense to measure as much as you can and try to account for all the voltage drops and what each source voltage actually is. You can then find out where the difference lies. You can even look across the diodes to see how much they actually drop at the maximum current point.

I realize that if you get 32vdc out of the regulator then you must have more than 32v in, so now the idea is to find out how that is actually happening through the practice of making measurements.

#### Picbuster

Joined Dec 2, 2013
1,047
Hello again,

Well if you read my previous reply i mentioned several errors and the way to tell WHICH error is more likely you should check with a scope. Since you did that, that tells you that the voltage you are measuring on the output is the true flat DC voltage and that was one of the possibilities i had mentioned. That means that the measuring instrument used to measure voltage is not at fault, and that in turn means that it is something else.
That something else could be a higher than normal line voltage, or a higher than thought secondary output voltage from the transformer. That would the the next place to check hopefully using the same test equipment.

The error we are talking about here isnt that large, maybe 1 or 2 volts, so it's possible that the transformer was wound to allow for a higher load current, which would mean the output with only 1 amp could be higher than 24v. In fact, it could be 25v or higher.

So next you might check the output of the transformer to see what the voltage really is. Peak voltage would be good to measure and check for any waveform flat topping.

In cases like this when you have a question of theory vs practice it makes sense to measure as much as you can and try to account for all the voltage drops and what each source voltage actually is. You can then find out where the difference lies. You can even look across the diodes to see how much they actually drop at the maximum current point.

I realize that if you get 32vdc out of the regulator then you must have more than 32v in, so now the idea is to find out how that is actually happening through the practice of making measurements.
Fist off all you are correct , out of a practical view however; its a student and they should work from an ideal situation with ideal components.
From the transformer is goes to a rectifier bridge loading a capacitor to its maximum voltage defined U x 1.41.
This capacitor does deliver the voltage during the time that the U sinus is lower than U x 1.41.
The time that this happens is frequency depended.
The energy supplied by the value of the cap.
The discharge depends on the resistance of the circuit and has a direct relation to the current.
The statement current defines ripple is correct.
An approximation for the discharge
Udischarge = Umax / e ^ (t/ZC)
e natural log
t one period time ( frequency)
Z impedance
C capacitor value

When the transformer can't supply the current you will run into an undefined situation.
Voltage will drop over the internal transformer resistance transformer saturation might heat up the core.
Wire insolation will melt and it will produce the smoke signal I am dying.
This also will prove that a transformer works on smoke when that comes out it stops working.
Stay always far away from the max specs.

Picbuster

#### MrAl

Joined Jun 17, 2014
11,253
Fist off all you are correct , out of a practical view however; its a student and they should work from an ideal situation with ideal components.
Picbuster
Hi,

Well it looks like the transformer is non ideal
It is supposed to be 24v and it looks like it could be higher, like 25v or even more
Alternately, the input line voltage is higher than thought.
This puts it into the realm of practical measurement unless we can show the reason for this theoretically. The only reason i can think of is that the transformer has a tolerance and it happens to be positive, or the line tolerance (which always exists) happens to be positive at the time of the measurement of the output.

Since the line can vary by 10 percent that puts low line at 108vac RMS and high line at 132vac RMS, so we may just be seeing a high line situation. That would be rare however. That would fall into the category of theory though if that's what you want to see.

Only a real life measurement can tell us which one (or both) it is however, although we can theorize (as i am doing right now) that it can be either one or both. We will never get the result from an ideal line and ideal transformer though as given.

#### ebp

Joined Feb 8, 2018
2,332
I have absolutely zero doubt that the lightly loaded voltage from the transformer will be at least a few percent higher than the specified voltage. Good transformer manufactures are careful to optimize the amount of copper used to get the best compromise between performance and cost for a given window area in the bobbin for a particular core stack.

The magnetic flux in the core is the same at no load as it is at full load, so you won't saturate the core by overloading the transformer.

When a linear voltage regulator is used, the load on the filter capacitor looks like a constant current because the voltage across the actual load is fixed and therefore the current through it is fixed. If a switch mode converter is used for regulation, then the load on the capacitor is constant power which implies a negative resistance characteristic - if the voltage rises,the current falls.

Transformer current derating, as crutschow pointed out, is essential when capacitive filtering is used, however to blindly say that you should stay well away from the "maximum" can just result in a larger and more expensive transformer. [EDIT: for clarification, "however to blindly ..." was directed at picbuster's statement "Stay always far away from the max specs.", not at anything crutschow said.]

Transformer "regulation" imperfection due to winding resistance reduces the peak current into the filter capacitor and increases the conduction time, reducing the ratio of RMS to average current.

Circuits like this seem relatively simple, but when they are examined closely and all the necessary allowances for non-ideal components are made, they are surprisingly complex.

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#### crutschow

Joined Mar 14, 2008
33,983
Transformer current derating, as crutschow pointed out, is essential when capacitive filtering is used, however to blindly say that you should stay well away from the "maximum" can just result in a larger and more expensive transformer.
I didn't say to "stay well away from the "maximum"", I said you should not exceed the maximum.
Good engineering practice is to not operate a component over its maximum rating.

So rather than "blindly" saying that, what would be your unblinded recommendation?
Certainly you can momentarily operate a transformer beyond its rating, but the user must be well aware what "momentary" means.
Otherwise not using a "larger and more expensive transformer" can result in a burned out transformer.

#### ebp

Joined Feb 8, 2018
2,332
Sorry, crutschow, my comment was directed at picbuster's remark "Stay always far away from the max specs." What you said is perfectly correct, and failure to follow your advice will result in a burned up transformer - which is hardly economical!

Like so many things, if you want "optimum" careful analysis is required.

#### MrAl

Joined Jun 17, 2014
11,253
I have absolutely zero doubt that the lightly loaded voltage from the transformer will be at least a few percent higher than the specified voltage. Good transformer manufactures are careful to optimize the amount of copper used to get the best compromise between performance and cost for a given window area in the bobbin for a particular core stack.
Yes, but there are so many things that can be in err here that i beg for actual measurements so we know for SURE what it is. That is what makes the difference between guessing and knowing...measurements.

The magnetic flux in the core is the same at no load as it is at full load, so you won't saturate the core by overloading the transformer.
We can still see flat topping however for several reasons. Again i believe a measurement is the best way to know for sure.

Circuits like this seem relatively simple, but when they are examined closely and all the necessary allowances for non-ideal components are made, they are surprisingly complex
The exact solution is a somewhat simpler but still rather complex partial differential equation, but to actually get exact solutions we need to know virtually EVERYTHING about the ENTIRE circuit, including line impedance, which usually is not bothered with. Even the diodes dont behave like we normally think they should with peak voltages that could be much above expected values.

So the bottom line is, if we dont get the measurements we never know the true situation. AFTER we get the measurements, we should be able to apply theory and get much closer to the real world results.

#### ebp

Joined Feb 8, 2018
2,332
Mr Al, I did not require you permission several weeks ago when you so generously granted it to me to believe something contrary to your incorrect assertion. I do not require your approval of my comments now.

Once again moderators, I ask for a "complete and utter contempt" smiley.

#### crutschow

Joined Mar 14, 2008
33,983
my comment was directed at picbuster's remark
Okay, sorry for the misunderstanding. You made the comment right after you mentioned my posting, so I thought you were referring to that.

#### MrAl

Joined Jun 17, 2014
11,253
Mr Al, I did not require you permission several weeks ago when you so generously granted it to me to believe something contrary to your incorrect assertion. I do not require your approval of my comments now.

Once again moderators, I ask for a "complete and utter contempt" smiley.
Hello again,

Not exactly sure what you mean here but if you are referring to a place where another member agrees or something that's just something that people naturally do. People agree and disagree that's just natural. If you can explain yourself a little more then i might know exactly what you are trying to say.
I hope it is not about and i quote, "Your best statement so far", which i said because your statement was so very correct and a very good observation, so i cant see how you could take that badly.

Perhaps we also need an emoticon for "complete and utterly baffled"