# positive output terminal on full wave rectifier

Discussion in 'Homework Help' started by EL7819, Nov 12, 2011.

1. ### EL7819 Thread Starter New Member

Apr 15, 2011
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0
Im having trouble understanding this full-wave rectifier? I made a schematic on multisim to go along with my understanding of this circuit so far. Hopefully I am on the right track. I made lines on the schematic where there are nodes in the circuit for the question. The question is: In the circuit, which terminal will be the positive output terminal?

This is my attempt on explaining this circuit.
From the top-down.
1.)Node 3 positive but does not conduct because diode is reversed.
2.)center tap on same cycle is negative.
3.)center tap is positive, node 3 is negative and node 4 is positive.
4.)Node 2 negative.
Next cycle.
5.)node 3 is negative.
6.)center tap is positive, node 3 negative and node 4 is positive.
7.)center tap negative.
8.)node 2 is positive but does not conduct because diode is reversed.
*My answer is that node 4 is the positive output terminal.*
Please let me know If I need to clarify anything. I will try and explain better. Any help greatly appreciated. Thank you for your time.

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2. ### JoeJester AAC Fanatic!

Apr 26, 2005
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3. ### crutschow Expert

Mar 14, 2008
13,523
3,392
Where did you get that schematic? It's certainly not a standard full-wave rectifier. D3 is superfluous.

4. ### eng.mustafasalah Member

Nov 10, 2011
41
2
i think the C.T.R be [COLOR= ]constituted from 2 diode ! is not it ?????
[/COLOR]

5. ### EL7819 Thread Starter New Member

Apr 15, 2011
20
0
Thank you for your help. It seams pretty simple now. I kept looking at the circuit and trying to understand it from a DC current flow, when I look at it AC then the only way current would pass is when anode is positive as in a DC circuit. marking the terminals of the diode made me realize that when all the anode's are marked positive is when current would flow. Making me come to the conclusion that terminal 4 would be positive. Thank you for your help.

Your also right, the extra diode does seam superfluous.

6. ### crutschow Expert

Mar 14, 2008
13,523
3,392
The current flow is into terminal 1 and out of terminal 4 of the resistor when any of the diodes are forward biased. That would make terminal 1 positive with respect to terminal 4.

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7. ### EL7819 Thread Starter New Member

Apr 15, 2011
20
0
I'm not sure I understand anymore. The current goes the way the diodes face even If it is AC. Now Im looking at the output terminal as the 3rd output because it would need a positive charge to go through the diode and the resistor is the Load. I'm saying that since RL is the load wouldn't it be the output terminal. For the output as 1, I don't understand, does that mean the current starts from the top or is it relative on how you analyze it.

I thought the current would start at the center tap?
I understand the full wave rectifier as the current goes through the center tap, output 1 and 2 fluctuate between positive and negative, that eliminates them as the output terminals. Output 4 is after the load drop which makes it negative and output 3 is positive because it is positive before it goes over the resistor.

I appreciate all your help in helping me understand this circuit. I understand what it does but the way the current goes in the e-book confuses me and all the other books I have looked at. The arrows point going towards the diodes anode which makes the diodes open circuits.

Last edited: Nov 15, 2011
8. ### crutschow Expert

Mar 14, 2008
13,523
3,392
Current does not "start" anywhere. It's always a closed loop.

Positive and negative is always relative to some other point. So since the current, due to the diodes, is always flowing through the resistor from terminal 1 to terminal 4, that means terminal 1 is always more positive than terminal 4.

If you want to the voltage relative to some other point, than that may give a different relative polarity. You need to pick one point for you reference terminal (common) and then refer all the other polarities to that. Otherwise you will just confuse yourself (as you apparently already have done).

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9. ### EL7819 Thread Starter New Member

Apr 15, 2011
20
0
Yes. Thank you. I became confused over mixing the electron flow to the conventional current flow. Thank you for your patience and understanding.