# Full-Wave Rectifier DC Output Calculation Error

#### Danlar81

Joined Apr 19, 2019
33
Hi Guys,

This is my first post and was hoping to get some feedback with regards to the following problem.
I am calculating the avg DC output of a Full-Wave bridge rectifier with a 1K ohm load, using a 10 mic filter capacitor from a 14.5 50 Hz RMS AC source.
I have uploaded a copy of my working with regards to this problem.
The issue I have is that there is approx 2.1 Volts discrepancy between what I have calculated and my measured values from the build.
I also have simulated the build using LTSpice and the output value from the simulation is very similar to my measured vale from the build.
Calculated = 13.07 VDC
Measured = 15.13 VDC
Simulated = 14.908 VDC
If someone would't mind having a look over my working to identify an error I would be most appreciative as I cannot see it, or alternatively explain the discrepancy that would also be equally as welcome.
Regards
Dan

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#### crutschow

Joined Mar 14, 2008
23,544
The ripple time from the peak until it decays to meet the next peak is around a quarter cycle (about 6ms for this actual ripple).
This means the peak voltage will thus decay for about 0.6 time-constant, or about 0.45 of the peak voltage = 0.45 * 19.1v = 8.6V of decay.
The ripple minimum voltage below the peak is thus 19.1V - 8.6V =10.5V.
Using the approximation of a triangle shaped ripple voltage, the average is then half-way between the minimum and maximum or (19.1V +10.5V) / 2 =14.8V.

So you made a mistake in the droop time for the time-constant calculation, and a mistake in how you then calculated the minimum voltage during that time.
Not sure about how you then calculated the average.

Incidentally, that's a 10μF capacitor not a 10 mic capacitor (mic is a common abbreviation for microphone). #### Danlar81

Joined Apr 19, 2019
33
Hello Crutschow,
I'm a bit confused where you you obtained 0.6 time-constant and 0.45 of the peak voltage. Could you please explain this further? Obviously this is the correct approach as you figures are very close to the measured and simulated values.
I would very much like to see an example if that is ok.

This is how I have been instructed to calculate these values from my course.
Vc = Vm(1-e-t/RC)
Where Vc = ripple voltage of capacitor and Vm is the peak secondary voltage minus the diode drops of 1.4.

I worked out the time constant RC by multiplying 1k by 10μF which gave me a result of 0.01 seconds or 10 mS.
The period was worked out as follows 1/2f = 1/100 = 0.01 seconds or 10 mS.

The average DC out was found as per below
Vavg = Vm - (Vc/2)

Yes I was aware of the μ representation not done correctly, I initially could not find the symbols but thanks for the correction in any case.

Regards
Dan

#### ericgibbs

Joined Jan 29, 2010
8,887
hi Dan,
Look thru this PDF.
E

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#### Danlar81

Joined Apr 19, 2019
33
Thank you Eric,

Yes I have seen this during google searches. I think if I understand where Crutschow gets the 0.6 ms time-constant. It should work out. I used his ripple value of 8.6V in the Vavg formula above and it works out as his did.

I just understand where 0.6 comes from.

Regards
Dan

#### ericgibbs

Joined Jan 29, 2010
8,887
hi Dan,
Are you now satisfied that you follow the maths in post #2.?
E

#### Danlar81

Joined Apr 19, 2019
33
Hi Eric,

I miss-worded my previous post. I don't understand where 0.6 time-constant comes from or 0.45 of the peak voltage.
This circuit is a filtered Full-Wave bridge rectifier.
I think if I can see an example where/how the 0.6 time constant and 0.45 of peak voltage comes it would clarify my understanding.

Regards
Dan

#### ericgibbs

Joined Jan 29, 2010
8,887
hi,
OK, does this LTSim of the circuit help you visualise the waveforms.?
I have run it for 4 half cycles, then allowed Vc to decay exponentially.

E
Expanded time scale. 2nd image

#### Danlar81

Joined Apr 19, 2019
33
Hi Eric,
Yes thank you.
I see the discharge time is approx 7 mS and that the peak to peak ripple or ΔVc is approx 8.5 Volts.

I was using the following formula to find the change of voltage over the capacitor.
ΔVc = Vm(1-e-t/RC)
But I now see that RC in this formula does not account for the charge pulse before the next time period.

Is there a formula that can explain this mathematically so I can calculate ΔVc without the use of the time curve?

Regards
Dan

#### crutschow

Joined Mar 14, 2008
23,544
What I meant was 0.6 of a time constant which is 6ms/10ms.
This gives the 0.45 exponential discharge from the peak value.

Make better sense now?

#### ericgibbs

Joined Jan 29, 2010
8,887
hi Dan,
I will look thru my doc's see if I can find a 'full' formula.
Most users adopt the approximation method aka rule of thumb in finding suitable values for reservoir caps.
E
BTW:
Most smoothing caps are electrolytic and have a very wide value tolerance, some are +/-20% > +/-40% !
So we tend to err on the 'high' side when choosing caps.

Interested to hear what my peers think about this problem. #### Danlar81

Joined Apr 19, 2019
33
Hi Crutschow,

Yes that makes sense and thank you for clarifying, but I really need to obtain this via a formula as opposed to viewing on the time curve.
If you know of one please let me know.

Regards
Dan

#### Danlar81

Joined Apr 19, 2019
33
Hi Eric,

Thanks for the assistance. If you manage to find a formula please pass it my way.

Thanks
Dan

#### Danlar81

Joined Apr 19, 2019
33
Hi Guys,

I can use the formula
ΔVc = Vm(1-e-t/RC)
I don't know why it didn't click sooner, it will become
ΔVc = 19.1(1-e-0.6)
ΔVc = 8.62
Now I can find the avg with the other formula
Vavg = 19.11 - (8.62/2)
Vavg = 14.8

I apologize for most likely making it harder than required. I really appreciate your help.
Regards
Dan

#### ericgibbs

Joined Jan 29, 2010
8,887

#### MrAl

Joined Jun 17, 2014
6,649
Hi Guys,

This is my first post and was hoping to get some feedback with regards to the following problem.
I am calculating the avg DC output of a Full-Wave bridge rectifier with a 1K ohm load, using a 10 mic filter capacitor from a 14.5 50 Hz RMS AC source.
I have uploaded a copy of my working with regards to this problem.
The issue I have is that there is approx 2.1 Volts discrepancy between what I have calculated and my measured values from the build.
I also have simulated the build using LTSpice and the output value from the simulation is very similar to my measured vale from the build.
Calculated = 13.07 VDC
Measured = 15.13 VDC
Simulated = 14.908 VDC
If someone would't mind having a look over my working to identify an error I would be most appreciative as I cannot see it, or alternatively explain the discrepancy that would also be equally as welcome.
Regards
Dan

Hello there,

There is no simple hard and fast rule that works with every full wave rectifier circuit. That is because the load vs capacitor value vs peak diode drop changes for every circuit. So there is no 0.5*Vpk or 0.6*Vpk or anything like that unless certain other requirements are met and they often simply are not met.

The exact solution is more complicated and is shown in the attachment. We can see that the solution requires a partial differential equation. However, we can approximate when the load is light such as for this problem with load of 1k and input of 14.5 volts AC RMS.

The approximate solution to the partial diff equation for this circuit is:
(Vpk-2*Vd)*e^(-t/RC)=sin(2*pi*f*t+ph)*(Vpk-2*Vd)

where:
Vpk is the peak sine voltage input,
Vd is one diode drop,
RC is the RC time constant here 10uf*1k,
f is the frequency 50Hz,
ph is -pi/2 (-90 degrees),
pi is 3.1415926...
t is time in seconds.

Immediately we see one thing nice here, and that is the peak and diode drop occurs as a factor of both sides, so the equation reduces to:
e^(-t/RC)=sin(2*pi*f*t+ph)

and since ph=-pi/2 we end up with:
e^(-t/RC)=-cos(2*pi*f*t)

and of course we could then insert the values for f and RC:
e^(-t/0.01)=-cos(100*pi*t)

and now we solve for the first solution for t and get:
t1=0.0067082998427697

We are now about 1/2 way there.

Next, we average the value of the voltage between the true peak (Vpk-2*Vd) and the voltage of the -cosine at t=t1, and add that to the average of the -cosine between t1 and t=0.01, and we get:
Vavg=14.64033

Note especially that we had to take two averages, one where we average over the droop, and one where we average over the part ot the cosine wave that makes up the remainder of the half cycle. The total average is then the sum of the two.

One of the reasons i went into such depth (see attachment) is because i noticed that there were solutions to limited component rectifier circuits but there was no general solution, so i went and looked for one and came up with the equations in the attachment. Those equations can lead you to an exact value as long as you know all the variables like the diode curve, but you may not need that level of accuracy so the above solution would be ok for now.

You may also want to note that the approximation requires knowledge of the diode drop and that varies quite a bit with load. You'd have to look at the curve of the diode to know this information. With light load we can assume 0.65 volts, but with heavy load we need to use the curve of the diode because the drop could be twice that just for one diode and that of course changes the real peak drastically.
The reason we see a lower computed average may be because the diodes are not dropping much voltage once the cap becomes charged. This would be typical for very light loads. With Vd=0.4 for example we see an average of about 15.02 volts.
It could also be because of imperfect source voltage measurements.

Also note that the attachment mentions one particular circuit with cap ESR which is often left out of other academic solutions, but that first equation as simple as it looks actually outlines a strategy that will work for ANY rectifier circuit no matter how complex and provide exact results given some circuits will require a numerical solution as did this one. Last edited:
• RockBottom

#### MrChips

Joined Oct 2, 2009
19,428
As Al says, the exact solution is very complex. All formulas you find will be approximations with given assumptions.
For example, if C is very large and current I is small, the average voltage is simply Vpk (no diode voltage drop).

Approximations will usually assume low ripple voltages. Hence one can approximate the exponential decay on the discharge portion with a simple saw-tooth wave. The simple formula for ripple voltage ΔV is
ΔV = I / (2fC)

This assumes that I is relatively low and C is relatively high.

C = 10μF is way too low and puts you outside of this model.

Finally, as eric points out, electrolytic capacitors have wide margins. When testing with real components, you can bargain on a new electrolytic capacitor having +20% on its nominal value.

#### Zeeus

Joined Apr 17, 2019
468
As Al says, the exact solution is very complex. All formulas you find will be approximations with given assumptions.
For example, if C is very large and current I is small, the average voltage is simply Vpk (no diode voltage drop).

Approximations will usually assume low ripple voltages. Hence one can approximate the exponential decay on the discharge portion with a simple saw-tooth wave. The simple formula for ripple voltage ΔV is
ΔV = I / (2fC)

This assumes that I is relatively low and C is relatively high.

C = 10μF is way too low and puts you outside of this model.

Finally, as eric points out, electrolytic capacitors have wide margins. When testing with real components, you can bargain on a new electrolytic capacitor having +20% on its nominal value.
Approx. Formula yh please why is 10uF way too low

#### MrChips

Joined Oct 2, 2009
19,428
Approx. Formula yh please why is 10uF way too low
Plug 10μF into the formula and see what you get for ΔV.

Or calculate time-constant τ = RC and compare that with T = 1/2f.

#### MrAl

Joined Jun 17, 2014
6,649
Approx. Formula yh please why is 10uF way too low
Hello again,

Here are some graphs of the cap voltage, input sine voltage, and rectified voltage.
You can see the solution is where the black dot is located.
The droop is measured from the peak of the rectified voltage (red).

In the top graph, using 10uf, we see the solution (Vc,t) arrive at about 1/2 of the peak voltage meaning that the ripple is very high like 50 percent.

In the bottom graph, using 100uf, we see the solution arrive at a much higher value meaning the ripple might only be around 10 percent. So the larger cap holds the voltage up longer than the smaller cap can that makes the average higher and most of the time that is a desirable feature.

As the electrolytic caps age, the capacitance goes down among other problems and the ripple again becomes too large and that often trips undervoltage detection circuits and turns the power supply off. This is typical of computer power supplies as well as TV's and other stuff.