Power supply question

Thread Starter

Mothrog

Joined Mar 21, 2006
6
I'm having a problem visualizing what this circuit does. When the transformer has the indicated polarity, and the bottom diode is forward biased, won't the potential from the transformer essentially be passing through a loop with no load, and so be short circuited? If not, what happens?

 

windoze killa

Joined Feb 23, 2006
605
Originally posted by Mothrog@Mar 22 2006, 03:19 PM
I'm having a problem visualizing what this circuit does. When the transformer has the indicated polarity, and the bottom diode is forward biased, won't the potential from the transformer essentially be passing through a loop with no load, and so be short circuited? If not, what happens?


[post=15278]Quoted post[/post]​
I don't think the circuit has been drawn correctly. Without trying to analize it totally I think C1 and C2 should be connected to the centre of the transformer. The catchode of the diode on the bottom should be connected to the cathode of the top diode.

See if that makes sense.
 

alim

Joined Dec 27, 2005
113
Originally posted by windoze killa@Mar 22 2006, 07:28 AM
I don't think the circuit has been drawn correctly. Without trying to analize it totally I think C1 and C2 should be connected to the centre of the transformer. The catchode of the diode on the bottom should be connected to the cathode of the top diode.

See if that makes sense.
[post=15285]Quoted post[/post]​
Hi you are essentially correct. He needs to remove the bottom cathode from the centre tap and take it to the top diode cathode as you stated.
 

Thread Starter

Mothrog

Joined Mar 21, 2006
6
Originally posted by alim@Mar 22 2006, 08:23 AM
Hi you are essentially correct. He needs to remove the bottom cathode from the centre tap and take it to the top diode cathode as you stated.
[post=15287]Quoted post[/post]​
Unfortunately this is the circuit as drawn in a review set given to me by my instructor. Is it safe to say that this circuit cannot function as drawn? Could you guys provide me a drawing of what you think the circuit should look like? I'm very new to circuitry and am having difficulty understanding what you mean when you say the bottom cathode should be removed from the center tap.
 

Thread Starter

Mothrog

Joined Mar 21, 2006
6
Originally posted by JoeJester@Mar 22 2006, 09:20 AM
So the instructor drew it and you copied his drawing?

If your having a hard time visualizing how a full wave rectifier circuit works, review this webpage.
[post=15290]Quoted post[/post]​
It's a scan of his actual drawing. I understand the full wave rectifier, but I'm not quite seeing what the others mean when they describe how to redraw the circuit.
 

JoeJester

Joined Apr 26, 2005
4,259
If you understand the full wave rectifier circuit, you should be able to see what is wrong with the drawing you posted.

It's times like this that you need to clear up with the instructor any misunderstandings.

If it's a full wave rectifier circuit, you should know how to draw one to show the instructor why their drawing is not correct.
 

pebe

Joined Oct 11, 2004
626
Originally posted by JoeJester@Mar 22 2006, 04:55 PM
If you understand the full wave rectifier circuit, you should be able to see what is wrong with the drawing you posted.

It's times like this that you need to clear up with the instructor any misunderstandings.

If it's a full wave rectifier circuit, you should know how to draw one to show the instructor why their drawing is not correct.
[post=15293]Quoted post[/post]​
I am amazed that an instructor can draw such a simple circuit - and get it wrong!
 

JoeJester

Joined Apr 26, 2005
4,259
As drawn, the bottom diode is useless and the circuit will function as a half-wave rectifier once that bottom diode blows up [which will be in about 1/120th of a second].

If it is meant to be a full wave rectifier circuit, the instructor needs to be more careful in the future with their drawings.
 

Thread Starter

Mothrog

Joined Mar 21, 2006
6
Originally posted by JoeJester@Mar 22 2006, 03:53 PM
As drawn, the bottom diode is useless and the circuit will function as a half-wave rectifier once that bottom diode blows up [which will be in about 1/120th of a second].

If it is meant to be a full wave rectifier circuit, the instructor needs to be more careful in the future with their drawings.
[post=15311]Quoted post[/post]​
I was thinking of a full wave bridge rectifier when I replied to your post. I looked at my reference and saw the full wave rectifier you were talking about. I have it figured out now. Thanks! One more question - will a common emitter transistor circuit function properly if it has zero input impedance? This guy also has a question based around finding gain and specifically tells you to set input impedance to zero. But, if you do that, wouldn't you essential get infinite base current and fry the transistor?
 

JoeJester

Joined Apr 26, 2005
4,259
Start a new post and include whatever drawings you may need. Include your analysis and questions in that post.

That would keep things easier.
 

Thread Starter

Mothrog

Joined Mar 21, 2006
6
I've got the transistor stuff figured out, but I have another question concerning the power supply problem initially discussed. I went to office hours yesterday and my instructor provided the corrected circuit depicted below



However, making that one connection into a jump would mean that C_1 would only play a role when the bottom diode was forward biased, while leaving it unconnected when the top diode was forward biased. I would think that would interfere with full wave rectification, as the capacitor would not discharge when the upper diode was forward biased, but would charge when the lower diode was forward biased and would provide equal but opposite voltage provided by the transformer. I would think that would lead to the bottom diode not contributing at all to the rectification, essentially turning what should be a full wave rectifier into a half wave rectifier. Should C_1 really be there? I think the circuit should look like this:

 

JoeJester

Joined Apr 26, 2005
4,259
I think you misunderstood the previous posters.

When they said to disconnect the cathode of the bottom diode from the bottom of C1 and reconnect the cathode of the bottom diode to the cathode of the top diode.

That makes the full wave rectifier circuit.

In the link I provided earlier, you should remember



The picture above doesn't include the pi filter network [C1 - resistor - C2].

Your bottom drawing is correct if you add C1 between the common [transformer centertap] and the line between the cathodes and the resistor.

The attachment illustrates the connection change.
 

Thread Starter

Mothrog

Joined Mar 21, 2006
6
Originally posted by JoeJester@Mar 23 2006, 10:42 AM
I think you misunderstood the previous posters.

When they said to disconnect the cathode of the bottom diode from the bottom of C1 and reconnect the cathode of the bottom diode to the cathode of the top diode.

That makes the full wave rectifier circuit.

In the link I provided earlier, you should remember



The picture above doesn't include the pi filter network [C1 - resistor - C2].

The bottom drawing is correct if you add C1 between the common [transformer centertap] and the line between the cathodes and the resistor.

The attachment illustrates the connection change.
[post=15341]Quoted post[/post]​
I understand what the others are saying, but what is it that C_1 as drawn in the circuit does? Wouldn't the circuit work perfectly fine and filter the output without C_1 in place as I've drawn in my circuit? Am I correct in my critique of my instructor's redraw of the circuit?
 

JoeJester

Joined Apr 26, 2005
4,259
If your single capacitor is sufficient enough to meet whatever design specifications you have for ripple voltage, then one is enough.

Your top diagram, a few posts back is not a good design. It's only full wave rectification when the cathodes are both connected to a common point, as in your bottom diagram.
 
Originally posted by Mothrog@Mar 23 2006, 09:52 AM
I've got the transistor stuff figured out, but I have another question concerning the power supply problem initially discussed. I went to office hours yesterday and my instructor provided the corrected circuit depicted below



However, making that one connection into a jump would mean that C_1 would only play a role when the bottom diode was forward biased, while leaving it unconnected when the top diode was forward biased. I would think that would interfere with full wave rectification, as the capacitor would not discharge when the upper diode was forward biased, but would charge when the lower diode was forward biased and would provide equal but opposite voltage provided by the transformer. I would think that would lead to the bottom diode not contributing at all to the rectification, essentially turning what should be a full wave rectifier into a half wave rectifier. Should C_1 really be there? I think the circuit should look like this:


[post=15339]Quoted post[/post]​

First of all dont use 2 diodes use a full wave bridge recitifier. It way better. Anyways this wesbsite provides free electronics stuff, but you have to have a bussines email address, ( you cant have yahoo mail or aol mail). I posted a pic of what it should look like. If you need any help with getting the free full wave recitifier or any electronic stuff, caontact me on aim at

AIM: lildeezul ( l i l d e e z u l ) ( no spaces)

free electronic stuff

PS. SRYY about the pic, i had to contruct it on paint, but hit me on aim, if u need help reading the schematic ok.



and add more eltrolyte capictor to filter the power supply.
 

thingmaker3

Joined May 16, 2005
5,084
Mothrog's second drawing (not the instructor's drawing) is indeed a full wave rectifier.

Capacitor value should be equal to or greater than I / (2 *f * Vr)

where I = load current
f = line frequency
and Vr = maximum acceptable ripple voltage
 
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