Troubleshooting Bridge Rectifier in LT Spice Simulation

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Homework Statement


I am to build a linear power supply in four stages. However, at stage two, i have to add a bridge rectifier to the ouput of my transformer. My lecturer at university says i should get a graph that is a sine wave, with the negative values reflected into the positive x-axis (Volatage). I was wondering were am i going wrong?


2. The attempt at a solution
http://img399.imageshack.us/img399/1367/ltspice1gm7.jpg


Any help would be greatly appreciated.

Thanks.
 
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You need to plot the difference between rectifidp and rectifidn. Or, alternatively, plot the voltage across your load resistor, R5.
 
Yeah, had a suspicion it might have been that, but it took me a while to work out how to enter an expression. Only started using the program a few days ago, so I'm still getting used to it.

http://img218.imageshack.us/img218/138/ltspice2rc0.jpg

Many thanks for your help.
 
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I've got another problem. For a full wave bridge, with a capacitive filter, we are told the following relationship between Vdc [V(Regout)-V(Rectifidn)] and Vac [V(Secondary)]:

Vdc = 1.41 x Vac

However, I'm obtaining the following results:

http://img386.imageshack.us/img386/5310/lt4vr1.jpg

I'm just wondering if i have the wrong formula, or taking incorrect readings?

Thanks.
 
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Redbelly98 said:
A couple of things to keep in mind here:

That formula uses the rms voltage, not the amplitude, for Vac.

Also, it does not account for the 2 diode drops in the rectifier voltage.
Ah, forgot about rms. Thanks.

That equation was given a bridge rectifier (4 diodes) with a capacitor and load resistor, but exluding the regulator. So i assume to calculate Vac, if i want Vdc of 5, i simply calculate 5+the voltage acrross the regultor, all divided by 1.41.
 
Yeah, thanks for your help.