# Designing a full wave bridge rectifier

Discussion in 'Homework Help' started by sunshine27, Feb 24, 2011.

1. ### sunshine27 Thread Starter New Member

Feb 24, 2011
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Design a full-wave bridge rectifier circuit to deliver 10 volts dc with less than 0.1 volt (pp) ripple into a load drawing up to 10 mA. Choose the appropriate ac input voltage (by means of a transformer) assuming 0.6 volt diode drops. Be sure to use the correct ripple frequency in your calculation.

**See my attached diagram**

So far, i determined that my f=120 Hz, so I used the equation

dV=I/2Cf to solve for C=4.17E-4F

Next, I used Ohm's law to find the resistance: R=V/I=.6V/10E-3A=60 Ohms.

Since to get to the load we have 2 voltage drops of .6V, our DC from the transformer has to be 10+1.2=11.2V

I don't know how to solve for the input voltage without the number of turns . Any help would be great!

• ###### circuit.png
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Dec 26, 2010
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I think that you were asked to state the input voltage for the rectifier - this is actually the transformer output voltage. The transformer input voltage might be your local mains supply voltage.

More importantly, you don't seem to have allowed for the peak of the AC voltage being the √2 times its RMS value. The AC output from the transformer would normally be qouted in terms of its RMS voltage.

As for your load resistance, you seem to have divided the diode forward drop by the output current. Why??? The output load would be connected across the output, which is not 0.6 Volts!

3. ### sunshine27 Thread Starter New Member

Feb 24, 2011
7
0
We haven't learned about RMS voltage yet, so I don't think that I have to do that....I accidentally used the voltage drop across the diode as the voltage drop across the resistor. Do they mean 10V drop across the resistor? So that would mean 11.2V input? Also that would make my resistor 880 ohms....

Not really sure if that is what they meant by delivering the DC current to the resistor....

Apr 20, 2004
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5. ### sunshine27 Thread Starter New Member

Feb 24, 2011
7
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I still am confused .

Now I'm assuming that we have 10 V going into the load, meaning we would have an input of 10.6V (but that is still DC so I don't know...). So that means we need to drop 10V through the load and through the 2nd diode. Thus, we would need to have a voltage drop of 9.4V through the resistor.

So, R=9.4/1E-2=940 Ohms.

Don't know what they want me to do with AC....
any help would be great!!!

Dec 26, 2010
2,147
300
You seem to be getting things completely muddled up. The load voltage is required to be 10V, and the load current is 10mA. The load resistance is... not 940Ω

When you began, you correctly added on two diode drops to get the the peak input required from the transformer. Why have you now forgotten one of the diodes?

7. ### sunshine27 Thread Starter New Member

Feb 24, 2011
7
0
ok so the R would be 1000 ohms, and the Vin would be 11.2 ohms. How am i supposed to predict the AC voltage though? Could I do it as Vpeak-to-peak possibly?

Dec 26, 2010
2,147
300
Vin is a voltage. Its units are volts, not ohms. If you want the equivalent RMS sine wave value, divide the voltage by √2

9. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
You just said it! Since you know that the rectifier raises any negative voltage on the positive half-plane, your peak DC voltage will be equal to your peak AC voltage (after the diodes). Hence, you need 11.2Volts (not Ω as you wrote) in the secondary end of the transformer.

Now only the filtering capacitor remains to be calculated. It seems you have the wrong formula for this one. The correct one is:
$V_r=\frac{V_p}{2\cdot f \cdot C \cdot R_{load}}$

10. ### sunshine27 Thread Starter New Member

Feb 24, 2011
7
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sorry.. i know about the ohms i think i'm just a bit tired

11. ### sunshine27 Thread Starter New Member

Feb 24, 2011
7
0
thanks so much for the help everyone!