# bridge rectifier

Discussion in 'Electronics Homework Help' started by bennethos, Jan 6, 2013.

1. ### bennethos

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Jan 6, 2013
Hi guys,

I'm using multisim to simulate a simple bridge rectifier.
One thing I'm not understanding is "how to measure the rectified DC voltage correctly with a multimeter".

I'm getting values way too high, I would expect it to be lower.
the XMM3 Multimeter -> way too high of what I'm expecting 24.635V . (or i'm not understanding something)

This lead me to think the multimeter it was showing Peak voltage instead of RMS.
So I calculated : RMS= pkV/sqrt2 --> 17,419V but it doesnt seem to be correct if I compare it with the XMM2 RMS voltage.

Hopefully you guys can give me some advice.

Last edited: Jan 6, 2013
2. ### bennethos

15
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Jan 6, 2013
Some of the things I'm definately not understanding :

- How I calculate a resistive load and the correct capacitance & voltage for the filter.

- where do I ground ?

I messed around a bit with the resistor value, DC measurement of XMM3 changes drastically. So my error lies there somewhere I suspect

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,449
2,809
Jan 21, 2010
I would be intuitively expecting 24.45V - 1.2V. However the figure you get is slightly higher than this. I suspect you'd find the reason if you looks at the V vs I graph for the diodes.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,449
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Jan 21, 2010
When you say "drastically", what do you mean?

5. ### bennethos

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Jan 6, 2013
Well as an AC power source I use 18V. After rectification I have 26.5V ?
This can't be true right. I was expecting 18V - diode drops ... Or is it normal my voltage increases by 7 volts after rectifying ?
Can't be can it ? I'm lost here

Last edited: Jan 6, 2013
6. ### pwdixon

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Oct 14, 2012
Sounds like the 18V is rms and the 26.5V is no-load dc which is about right.

7. ### bennethos

15
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Jan 6, 2013
I've encountered the "no load" measurement issues a couple of times. But I don't understand why it is giving these results.

Would it be possible to elaborate a bit ? Or point me to a good link ?

Last I had to calibrate the output of a negative voltage regulator to -12v . I kept measuring -24v... Until I calculated a decent power resistor to go between the Vout and the common ground and suddenly I measured -12.3V. I was happy of course, but... I don't get it entirely...

8. ### pwdixon

52
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Oct 14, 2012
you will most probably measure rms when using ac on a dvm.

Rectified the no=load dc value would be root2 times the rms value hence 18Vrms=1.414*18dc ie 25.45Vdc.

9. ### bennethos

15
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Jan 6, 2013
Exactly it works out !

Now I figured out the AC value @ my meter was RMS due to the fact that I had put 18V as an AC PK source and a lot less came out of it. Hence I gave it 25.45PK-PK (18V*sqrt2)

What I still don't understand is... Why does the multimeter @ AC side shows the value in RMS or effective value vs DC showing it in PK Value ?

Thx a lot for your help, it motivates me a lot when you're investing hours on reading and trying out.

10. ### bennethos

15
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Jan 6, 2013

there is no effective or RMS value for DC , there's only 1 value and the oscilloscope shows this perfectly clear right ? Namely peak value.

11. ### davennModerator

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Sep 5, 2009
thats correct

Dave

12. ### Harald KappModeratorModerator

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Nov 17, 2011
RMS is the effective value of an AC waveform. The definition of "effective" being that this waveform produces the same amount of power as an equivalent DC signal.
When talking about DC it makes no sense to use RMS or "peak" values, as the voltage (ideally) is the same at all times. However, mathematically the RMS value of a DC signal is the same as the steady state value of that signal, because RMS is defined that way.

When you measure AC with a multimeter, the meter usually will give you an RMS value of the AC waveform. For a sinusoidal signal the relation is Vpeak=Vrms*sqrt(2).
Therefore after rectification of an 18Vrms by a bridge rectifier signal the voltage is 18V*sqrt(2)-2*Vdiode. The terms in this equation are:
18V*sqrt(2) = peak voltage of the sinusoidal signal
-2*Vdiode = voltage drop across two diodes of the bridge rectifier.

Note that above musings are valid for sinusoidal signals. If the waveform is other than sinusoidal, the equations change accordingly with another factor between RMS and peak.

Harald

13. ### bennethos

15
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Jan 6, 2013
thanks so much for taking the time to explain all of you. It's clear as crystal now !