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Question about full wave bridge rectifier

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1

$begingroup$

Sorry if this is a stupid question, what I know is when a diode is forward biased, it conducts current, when it is reverse biased, it acts as an open branch if its $ PIV geq V_{in}$

But in this picture:

At the node between $D1$ and $D3$ ,$D1$ is forward biased and $D3$ is reverse biased, so that's fine, now at the node between $D2$ and $D3$ why does the current pass through $D2$ and not $D3$? Isn't $D3$ supposed to be forward biased at that junction too? Thanks.

electric-circuits voltage semiconductor-physics electronics

share|cite|improve this question

asked 4 hours ago

khaled014zkhaled014z

1137

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Would Electrical Engineering be a better home for this question?
  $endgroup$
  – Qmechanic♦
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know, I like to focus on the physics and theory so I'm not that fluent with electronics, but if you think they can help me out there I will ask there instead
  $endgroup$
  – khaled014z
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure what you mean when you write "forward biased at that junction". D3 can't be reverse biased and forward biased at the same time. Since you've already stated that D3 is reverse biased at the beginning of the previous sentence, why do you think it should be forward biased too?
  $endgroup$
  – Alfred Centauri
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doesn't the diode depend on the direction of the current? Or is it biased by default by the voltage source?
  $endgroup$
  – khaled014z
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @khaled014z The transformer creates an emf in the circuit. It is necessary to consider the current passing through the winding. This corresponds to the picture.
  $endgroup$
  – Alex Trounev
  3 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Sorry if this is a stupid question, what I know is when a diode is forward biased, it conducts current, when it is reverse biased, it acts as an open branch if its $ PIV geq V_{in}$

But in this picture:

At the node between $D1$ and $D3$ ,$D1$ is forward biased and $D3$ is reverse biased, so that's fine, now at the node between $D2$ and $D3$ why does the current pass through $D2$ and not $D3$? Isn't $D3$ supposed to be forward biased at that junction too? Thanks.

electric-circuits voltage semiconductor-physics electronics

share|cite|improve this question

asked 4 hours ago

khaled014zkhaled014z

1137

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Would Electrical Engineering be a better home for this question?
  $endgroup$
  – Qmechanic♦
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know, I like to focus on the physics and theory so I'm not that fluent with electronics, but if you think they can help me out there I will ask there instead
  $endgroup$
  – khaled014z
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure what you mean when you write "forward biased at that junction". D3 can't be reverse biased and forward biased at the same time. Since you've already stated that D3 is reverse biased at the beginning of the previous sentence, why do you think it should be forward biased too?
  $endgroup$
  – Alfred Centauri
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doesn't the diode depend on the direction of the current? Or is it biased by default by the voltage source?
  $endgroup$
  – khaled014z
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @khaled014z The transformer creates an emf in the circuit. It is necessary to consider the current passing through the winding. This corresponds to the picture.
  $endgroup$
  – Alex Trounev
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Sorry if this is a stupid question, what I know is when a diode is forward biased, it conducts current, when it is reverse biased, it acts as an open branch if its $ PIV geq V_{in}$

But in this picture:

At the node between $D1$ and $D3$ ,$D1$ is forward biased and $D3$ is reverse biased, so that's fine, now at the node between $D2$ and $D3$ why does the current pass through $D2$ and not $D3$? Isn't $D3$ supposed to be forward biased at that junction too? Thanks.

electric-circuits voltage semiconductor-physics electronics

share|cite|improve this question

asked 4 hours ago

khaled014zkhaled014z

1137

$endgroup$

share|cite|improve this question

asked 4 hours ago

khaled014zkhaled014z

1137

share|cite|improve this question

asked 4 hours ago

khaled014zkhaled014z

1137

asked 4 hours ago

khaled014zkhaled014z

1137

asked 4 hours ago

khaled014zkhaled014z

1137

3 Answers
3

active

oldest

votes

1

$begingroup$

what I know is when a diode is forward biased, it conducts current,
when it is reverse biased, it acts as an open branch

There's a simpler, clearer way to think about it - a diode (ideally) allows current through in just one direction.

So, it's easy to see that current entering the top node of the bridge can only exit through through D1 since there can be no current (ideally) 'backwards' through D3.

However, for the current entering the left node of the bridge, there could be a current exiting through D3 but only if the cathode is negative with respect to the anode. But we know this isn't the case, i.e., the cathode is positive with respect to the anode while, for D2, the cathode is negative with respect the anode.

Thus, the current is 'downhill' through D2 and not 'uphill' through D3.

share|cite|improve this answer

answered 3 hours ago

Alfred CentauriAlfred Centauri

48.3k350150

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you mean, but what made the cathode of D3 positive at the first place? This is weird to me because from my knowledge, The N region of the PN junction should be negative, but it is positive here?
  $endgroup$
  – khaled014z
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @khaled014z, note that the anode of D3 is connected to 'ground' (zero volt reference) and so a positive voltage applied to the junction of the D3 cathode and the D1 anode necessarily reverse biases D3. The positive voltage is applied by the transformer winding. Why should the N region be negative?
  $endgroup$
  – Alfred Centauri
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I get it now, thank you.
  $endgroup$
  – khaled014z
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

The node between D2 and D3 is grounded, so its potential is 0 V. The node between D3 and D1, as you stated, has positive potential, so the current can't go through D3.

On the other hand the node between D2 and D4 has a negative potential, so the current can go in that direction, and indeed it does.

share|cite|improve this answer

answered 4 hours ago

GRBGRB

9151722

$endgroup$

add a comment | 

1

$begingroup$

Assume that $V_{rm X}=0 ,rm V$ and noting the labels (signs) on the secondary of the transformer.

.

$V_{rm X}>V_{rm W}$ so diode D1 is forward biased and conducting.

$V_{rm Y}>V_{rm Z}$ so diode D2 is forward biased and conducting.

$V_{rm Y}>V_{rm X}$ so diode D3 is reverse biased and not conducting.

$V_{rm Z}>V_{rm W}$ so diode D4 is reverse biased and not conducting.

Assuming that the potential drop across the conducting diodes is very small nodes $X$ and $W$ are approximately at zero potential whereas nodes $Y$ and $Z$ are at a higher potential.

share|cite|improve this answer

answered 2 hours ago

FarcherFarcher

49.9k338104

$endgroup$

add a comment | 

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1

$begingroup$

what I know is when a diode is forward biased, it conducts current,
when it is reverse biased, it acts as an open branch

There's a simpler, clearer way to think about it - a diode (ideally) allows current through in just one direction.

So, it's easy to see that current entering the top node of the bridge can only exit through through D1 since there can be no current (ideally) 'backwards' through D3.

However, for the current entering the left node of the bridge, there could be a current exiting through D3 but only if the cathode is negative with respect to the anode. But we know this isn't the case, i.e., the cathode is positive with respect to the anode while, for D2, the cathode is negative with respect the anode.

Thus, the current is 'downhill' through D2 and not 'uphill' through D3.

share|cite|improve this answer

answered 3 hours ago

Alfred CentauriAlfred Centauri

48.3k350150

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you mean, but what made the cathode of D3 positive at the first place? This is weird to me because from my knowledge, The N region of the PN junction should be negative, but it is positive here?
  $endgroup$
  – khaled014z
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @khaled014z, note that the anode of D3 is connected to 'ground' (zero volt reference) and so a positive voltage applied to the junction of the D3 cathode and the D1 anode necessarily reverse biases D3. The positive voltage is applied by the transformer winding. Why should the N region be negative?
  $endgroup$
  – Alfred Centauri
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I get it now, thank you.
  $endgroup$
  – khaled014z
  3 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

what I know is when a diode is forward biased, it conducts current,
when it is reverse biased, it acts as an open branch

There's a simpler, clearer way to think about it - a diode (ideally) allows current through in just one direction.

So, it's easy to see that current entering the top node of the bridge can only exit through through D1 since there can be no current (ideally) 'backwards' through D3.

However, for the current entering the left node of the bridge, there could be a current exiting through D3 but only if the cathode is negative with respect to the anode. But we know this isn't the case, i.e., the cathode is positive with respect to the anode while, for D2, the cathode is negative with respect the anode.

Thus, the current is 'downhill' through D2 and not 'uphill' through D3.

share|cite|improve this answer

answered 3 hours ago

Alfred CentauriAlfred Centauri

48.3k350150

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you mean, but what made the cathode of D3 positive at the first place? This is weird to me because from my knowledge, The N region of the PN junction should be negative, but it is positive here?
  $endgroup$
  – khaled014z
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @khaled014z, note that the anode of D3 is connected to 'ground' (zero volt reference) and so a positive voltage applied to the junction of the D3 cathode and the D1 anode necessarily reverse biases D3. The positive voltage is applied by the transformer winding. Why should the N region be negative?
  $endgroup$
  – Alfred Centauri
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I get it now, thank you.
  $endgroup$
  – khaled014z
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

what I know is when a diode is forward biased, it conducts current,
when it is reverse biased, it acts as an open branch

There's a simpler, clearer way to think about it - a diode (ideally) allows current through in just one direction.

So, it's easy to see that current entering the top node of the bridge can only exit through through D1 since there can be no current (ideally) 'backwards' through D3.

However, for the current entering the left node of the bridge, there could be a current exiting through D3 but only if the cathode is negative with respect to the anode. But we know this isn't the case, i.e., the cathode is positive with respect to the anode while, for D2, the cathode is negative with respect the anode.

Thus, the current is 'downhill' through D2 and not 'uphill' through D3.

share|cite|improve this answer

answered 3 hours ago

Alfred CentauriAlfred Centauri

48.3k350150

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Alfred CentauriAlfred Centauri

48.3k350150

answered 3 hours ago

Alfred CentauriAlfred Centauri

48.3k350150

answered 3 hours ago

Alfred CentauriAlfred Centauri

48.3k350150

2

$begingroup$

The node between D2 and D3 is grounded, so its potential is 0 V. The node between D3 and D1, as you stated, has positive potential, so the current can't go through D3.

On the other hand the node between D2 and D4 has a negative potential, so the current can go in that direction, and indeed it does.

share|cite|improve this answer

answered 4 hours ago

GRBGRB

9151722

$endgroup$

add a comment | 

2

$begingroup$

The node between D2 and D3 is grounded, so its potential is 0 V. The node between D3 and D1, as you stated, has positive potential, so the current can't go through D3.

On the other hand the node between D2 and D4 has a negative potential, so the current can go in that direction, and indeed it does.

share|cite|improve this answer

answered 4 hours ago

GRBGRB

9151722

$endgroup$

add a comment | 

$begingroup$

The node between D2 and D3 is grounded, so its potential is 0 V. The node between D3 and D1, as you stated, has positive potential, so the current can't go through D3.

On the other hand the node between D2 and D4 has a negative potential, so the current can go in that direction, and indeed it does.

share|cite|improve this answer

answered 4 hours ago

GRBGRB

9151722

$endgroup$

share|cite|improve this answer

answered 4 hours ago

GRBGRB

9151722

answered 4 hours ago

GRBGRB

9151722

answered 4 hours ago

GRBGRB

9151722

1

$begingroup$

Assume that $V_{rm X}=0 ,rm V$ and noting the labels (signs) on the secondary of the transformer.

.

$V_{rm X}>V_{rm W}$ so diode D1 is forward biased and conducting.

$V_{rm Y}>V_{rm Z}$ so diode D2 is forward biased and conducting.

$V_{rm Y}>V_{rm X}$ so diode D3 is reverse biased and not conducting.

$V_{rm Z}>V_{rm W}$ so diode D4 is reverse biased and not conducting.

Assuming that the potential drop across the conducting diodes is very small nodes $X$ and $W$ are approximately at zero potential whereas nodes $Y$ and $Z$ are at a higher potential.

share|cite|improve this answer

answered 2 hours ago

FarcherFarcher

49.9k338104

$endgroup$

add a comment | 

1

$begingroup$

Assume that $V_{rm X}=0 ,rm V$ and noting the labels (signs) on the secondary of the transformer.

.

$V_{rm X}>V_{rm W}$ so diode D1 is forward biased and conducting.

$V_{rm Y}>V_{rm Z}$ so diode D2 is forward biased and conducting.

$V_{rm Y}>V_{rm X}$ so diode D3 is reverse biased and not conducting.

$V_{rm Z}>V_{rm W}$ so diode D4 is reverse biased and not conducting.

Assuming that the potential drop across the conducting diodes is very small nodes $X$ and $W$ are approximately at zero potential whereas nodes $Y$ and $Z$ are at a higher potential.

share|cite|improve this answer

answered 2 hours ago

FarcherFarcher

49.9k338104

$endgroup$

add a comment | 

$begingroup$

Assume that $V_{rm X}=0 ,rm V$ and noting the labels (signs) on the secondary of the transformer.

.

$V_{rm X}>V_{rm W}$ so diode D1 is forward biased and conducting.

$V_{rm Y}>V_{rm Z}$ so diode D2 is forward biased and conducting.

$V_{rm Y}>V_{rm X}$ so diode D3 is reverse biased and not conducting.

$V_{rm Z}>V_{rm W}$ so diode D4 is reverse biased and not conducting.

Assuming that the potential drop across the conducting diodes is very small nodes $X$ and $W$ are approximately at zero potential whereas nodes $Y$ and $Z$ are at a higher potential.

share|cite|improve this answer

answered 2 hours ago

FarcherFarcher

49.9k338104

$endgroup$

share|cite|improve this answer

answered 2 hours ago

FarcherFarcher

49.9k338104

answered 2 hours ago

FarcherFarcher

49.9k338104

answered 2 hours ago

FarcherFarcher

49.9k338104