Hi, i am posting the fullwave center tapped rectifier circuit diagram can any one tell me that how can I calculate Dc voltage which is observed by the RL or load? is there any formula to calculate DC voltage in this type of circuit? secondary voltage here is 6 plus 6 rms and each diode is working on the 6v rms in each half cycle.

 

It looks to me like you have approximately 6 volts d.c. If you add a capacitor across the output, it will rise to 1.414 X 6 = 8.48 volts no load. The voltage will be less when you put a load on the output; best to use a reliable d.c. voltmeter.

 

It looks to me like you have approximately 6 volts d.c. If you add a capacitor across the output, it will rise to 1.414 X 6 = 8.48 volts no load. The voltage will be less when you put a load on the output; best to use a reliable d.c. voltmeter.

Thanks for your kind reply but sir u have only calculate the peak value of the 6v rms.I think there is no relationship b/w capacitor and peak value b/c 6v rms is always equal to 8.48v peak.

 

And you would be wrong. The action of the capacitor is to be charged up to the peak value of the AC waveform, and in the absence of a load to very slowly leak off while the AC input goes to 0 VAC and then rises to the next peak. The actual droop in voltage from one peak of the AC waveform to the next is quite small relative to the AC voltage itself and the short time interval between peaks which is on the order 8.333... milliseconds for a 60 Hz. input or 10 milliseconds for a 50 Hz. input.

It is quite easy to see this ripple on an oscilloscope. You should try it sometime.

 

This is probably for a homework problem.

If the theoretical transformer would put out 6v+6v RMS no matter what the load, or if the transformer rating matched the load, the peak voltage across the load would be equal to the RMS value multiplied by 1.414, less the Vf (forward voltage) of the diode at the peak load current.

I don't know what your instructor told you to use for the diode Vf; 0.7v would be likely. Real-world diodes' Vf varies significantly depending on If (forward current), type of diode (silicon, Schottky) maximum If, and Vr (reverse voltage) ratings.

But, using 0.7v for Vf, you would get about 6v X 1.414 - 0.7v = 8.48-0.7v = 7.88v peak.

 

Right -- I should have said:

"charged up to the peak value of the rectified AC waveform"

I was thinking that since I knew that the frequency of the rectified AC waveform is twice the frequency of the mains.

Sorry.

 

Hello,

Take a look at page 13 and on from this PDF: Unit12.pdf

Bertus

 

The peak current is high in a rectifier that very quickly charges the capacitor. So the rectifier forward voltage drop is close to 1V. It is only 0.4V or 0.5V if the DC load current is small.

 

The peak current is high in a rectifier that very quickly charges the capacitor. So the rectifier forward voltage drop is close to 1V. It is only 0.4V or 0.5V if the DC load current is small.

If the diode is a 1N4000 series and the load at the peak voltage is 1A, and the diode junction temp is 25°C, then the diodes' Vf at 1A would be nominally 1.1v.

A 1N5400 series would have a Vf of around 0.76v with the other parameters being the same.

A 1N5820 Schottky diode would have a Vf of around 0.37v, other parameters being the same.

I'm not being argumentative here; just trying to illustrate that our original poster will need to use the Vf @ current specification that their instructor told them to use; if a particular model/series/part# diode was specified, they will need to find the datasheet for it.

 

@Wookie,

An friend gave me this to keep for a reference. Please look at it and tell me if it's ok to use

 

=soda;326320A friend gave me this to keep for a reference. Please look at it and tell me if it's ok to use

It wrongly ignors the forward voltage drops of the rectifier diodes.

 

I don´t see how the numbers work out. For example what is supposed to be the difference between the first two pictures?

 

@soda,
As already mentioned, there are shortcomings in those schematics; they are simplified to the point where critical information is missing.

While you'll be able to get within 20% or so of actual, that's not saying much.

Even what I posted is pretty basic; but I covered most of the bases for a reasonably accurate answer.

 

Ok, thanks for your reply, Wookie

 

I don´t see how the numbers work out. For example what is supposed to be the difference between the first two pictures?

Center tap is grounded in 2nd one.

 

I don´t see how the numbers work out. For example what is supposed to be the difference between the first two pictures?

Sorry for me answer you so late. The first pic is a normal single taped transformer like 12v-0v where the second pic is a center taped transformer like a 12v-0v-12v. What i experienced in the past is that you get a difference in the max Amp out between the 2 of them. Normally the 12v-0v will give you a higher amp out.

 

That is weird, and really doesn´t explain the 0.82 vs. 0.62 difference. The current should be the same or a half, depending on what exactly those equations actually mean.

 

Center tap is grounded in 2nd one.

No, the center tap is not grounded. The arrow is only to show you it's a center tapped transformer. Sorry for my misleading pic.