# full bridge rectifier equations

Discussion in 'The Projects Forum' started by suzuki, Apr 15, 2012.

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Hi, I am having some trouble deriving or finding some help on this.

I want to get an equation which describes the output of a full bridge rectifier.

I have found the following information thus far,

$V_o = V_{peak} - 0.5V_{ripple}$

$V_{ripple} = \frac{V_{peak}}{fCR}$

When you calculate this value, you will get a singular output dc value. However, I want to have an equation which gives more details, one that includes the ripple voltage. In my mind, this equation should be in a sinusoidal form, but I can't seem to think of have to arrive at that.

Does anybody know how i can do this analysis? or maybe have some references available for this?

tia

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3. ### #12AAC Fanatic!

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radical 2 C Er F = I
.707 Capacitance [Ripple voltage (p-p)] Frequency = current.

I don't know how to make that into a sinusoidal form.
4. ### MrChipsModeratorStaff Member

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After rectification it is no longer sinusoidal.

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i agree with this, but in reality, i don't think a pure dc output can be achieved. the output voltage should still exhibit some ripple, which (i believe) can be described by some sinusoidal equation. But this is the part where I cannot seem to find any information.

thanks again for all replies
6. ### mlogMember

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If you want a "sinusoidal" equation that describes diode conduction, then you will have to write an equation for each segment of time or alternatively as a segment of the angle (θ=ωt). For example, you might have one equation for the segment 0 < θ < ∏ and a second equation for the segment ∏ < θ < 2∏.

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i had thought about it in this way before as well. seems to be that the diode bridge itself can just be written as the abosolute value of the input ac voltage. But the tricky part is how to deal with the equations that involve the capacitor charging and discharging. Not quite sure how to approach this.
8. ### MrChipsModeratorStaff Member

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It is possible to work this out exactly but it is simpler if you make some approximations.
You need to include a load resistor as well as the filter capacitor.
Without a load the capacitor will reach full charge at the peak voltage (assuming the diode voltage drop is zero at zero current).
When the load is included you need to determine the phase angles when the diode is conducting and not conducting. From this you can determine the shape of the ripple voltage.

More importantly, you need to calculate the shape of the current pulse through the diodes.
9. ### waynehAAC Fanatic!

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Modeling the output, under load, of a full wave bridge + filter cap is not simple but can be accomplished by breaking the wave into segments (conduction, RC decay). Then you need iterative calculations to determine where the segments intersect to produce the continuous output.

I'm not sure I ever posted the spreadsheet I mentioned in that thread. I will if you're interested.
10. ### #12AAC Fanatic!

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For my own education, if you develop this equation, what purpose does it serve?

I've designed hundreds of these rectifier/capacitor/load circuits without the benefit of this equation and have very little idea what you are doing. Please educate me as to what the goal is.
11. ### waynehAAC Fanatic!

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No idea about the OP, but in my case I was interested in understanding data from the load side, and how it could tell what the AC supply was doing. I was studying an alternator but only had data from the DC side, eg. time-average DC voltage into a resistive load. So I calculated out what the relationship was that related DC voltage at a load level, to the peak voltage of a sine wave on the AC side. When you factor in the that diodes change with temperature, this becomes a pretty tough challenge. But I ultimately realized that small rpm changes driving my alternator were a bigger factor.
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12. ### #12AAC Fanatic!

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I see. It's a POV change. Obviously something I have never considered. Thank you.
13. ### waynehAAC Fanatic!

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One of the more interesting outcomes of the analysis was how the ripple filtering disappeared as the load increased. No surprise that it happens, but it was interesting to plot out just how the average voltage drops from the peak of the AC wave (less diode drop) at no load, down to the time-average of the AC wave at "big" current load. "Big" is where the filter cap no longer does any good and in fact just acts as additional load as it fully charges and discharges on every cycle.