# Help with a simple Full Wave Bridge Rectifier Problem

Thread Starter

#### Kuturan

Joined Jan 11, 2020
11
My problem involves a full wave bridge rectifier with no capacitor or transformer attached to a 20Vrms 60Hz AC source.

I am trying to find Peak Inverse Voltage (PIV) across the diodes alongside DC voltage across the load. In this case, the problem asks that I assume that the diodes are not ideal and in forward bias, thus resulting in a 0.7V voltage drop across each one.

Here is my work so far.

Vout = (20 * 1.414) - 1.4 = 28.28 - 1.4 = 26.88V

V(DC) across the load = (2*(Vout)) / pi = (2*26.88) / pi = 53.76 / pi = 17.11V

PIV = Vin - 0.7 = Vout + 0.7 = 26.88V + 0.7V = 27.58V

However, these results do not match up with the simulations that I've done with an equivalent circuit, with virtual diodes at 25 degrees celsius and all other configurations set to their defaults.

I would like to know what I am missing in regards to my calculations above.

Thank you in advance.

#### AlbertHall

Joined Jun 4, 2014
11,108
For the peak inverse voltage, is the multimeter set to record the peak voltage?

Thread Starter

#### Kuturan

Joined Jan 11, 2020
11
For the peak inverse voltage, is the multimeter set to record the peak voltage?
I am not sure what you mean. If you mean the sinusoidal wave, I'd get 10.774V which if you multiplied by sqrt(2), would give you around 15.237V, which is still way off from my calculations.

#### AlbertHall

Joined Jun 4, 2014
11,108
The voltage across the diode varies between the peak inverse voltage in the reverse direction and 0.7V in the forward direction. You want to know the peak of that voltage, not the average or RMS value.

Also consider what shape the output voltage waveform is.

Thread Starter

#### Kuturan

Joined Jan 11, 2020
11
The voltage across the diode varies between the peak inverse voltage in the reverse direction and 0.7V in the forward direction. You want to know the peak of that voltage, not the average or RMS value.

Also consider what shape the output voltage waveform is.
Do you mean something like this? If so, then the results should make sense.

Thread Starter

#### Kuturan

Joined Jan 11, 2020
11
If this is the case though, I am still a bit confused over my V(DC) results compared to the simulation, as I do not believe that the margin of error between my results (17.11V) and the simulation (16.693V) is small enough.

Do you see any mistake/missing calculations on my part?

#### AlbertHall

Joined Jun 4, 2014
11,108
Have a look at the output waveform on the scope.

#### magicChristian

Joined Nov 20, 2016
24
My problem involves a full wave bridge rectifier with no capacitor or transformer attached to a 20Vrms 60Hz AC source.

I am trying to find Peak Inverse Voltage (PIV) across the diodes alongside DC voltage across the load. In this case, the problem asks that I assume that the diodes are not ideal and in forward bias, thus resulting in a 0.7V voltage drop across each one.

Here is my work so far.

Vout = (20 * 1.414) - 1.4 = 28.28 - 1.4 = 26.88V

V(DC) across the load = (2*(Vout)) / pi = (2*26.88) / pi = 53.76 / pi = 17.11V

PIV = Vin - 0.7 = Vout + 0.7 = 26.88V + 0.7V = 27.58V

However, these results do not match up with the simulations that I've done with an equivalent circuit, with virtual diodes at 25 degrees celsius and all other configurations set to their defaults.

I would like to know what I am missing in regards to my calculations above.

Thank you in advance.
Hi, You set the Multimeter for DC measuring and there is no capacitor. Therefor it gives you the averag over time. Switch to Ac and the values should change. But this circuit needs a capacitor if used as normal!