What is the voltage drop across a diode when it's on?

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Homework Help Overview

The discussion revolves around understanding the behavior of diodes in a circuit, specifically focusing on the voltage drop across a diode when it is conducting. Participants are examining the conditions under which diode D2 is off and the implications of voltage levels at various points in the circuit.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the voltage levels at the anodes and cathodes of the diodes, questioning how these affect the current flow. There are discussions about the behavior of ideal diodes versus real diodes, and participants consider the implications of changing voltage levels in the circuit.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's assumptions. Some guidance has been offered regarding the behavior of diodes under certain conditions, but there is no explicit consensus on the voltage levels or the state of the diodes.

Contextual Notes

Participants note potential ambiguities in the circuit diagram and the assumptions made about the voltage levels at different points. There is an ongoing examination of the effects of these assumptions on the operation of the diodes.

theBEAST
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Homework Statement


Find V and I in the following diagram:
fl0SQ3Z.png


The Attempt at a Solution


I know what the answer is. Diode D2 is off but why?

Now if the voltage left of D2 was +4V then diode D2 is off because +4V dominates +3V so the current runs right and thus turns off D2. Is this correct? Is this how ideal diodes work?

Thanks!
 
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The symbol of a diode is an arrow, showing the possible flow of current. In the figure, the current can flow through both diodes from right to left - from the anode to the cathode, from the p-type part to the n type part. The current flows from higher potential to lower. The anode is at 0 V, the cathode of D1 is at -1 V. Does current flow though it?
The anode of D2 is at 0 V, the cathode is at 3 V. It is was a resistor the current would flow from the higher potential to the lower one, that is, to the right. But it is a diode, and the current can flow to the left only. Is D2 closed or open?


ehild
 
ehild said:
The symbol of a diode is an arrow, showing the possible flow of current. In the figure, the current can flow through both diodes from right to left - from the anode to the cathode, from the p-type part to the n type part. The current flows from higher potential to lower. The anode is at 0 V, the cathode of D1 is at -1 V. Does current flow though it?
The anode of D2 is at 0 V, the cathode is at 3 V. It is was a resistor the current would flow from the higher potential to the lower one, that is, to the right. But it is a diode, and the current can flow to the left only. Is D2 closed or open?


ehild


Isn't the cathode of D1 1V? The diagram is a bit blurry but there should be a + sign in front of the 1. Also, how do you know the anode is 0V?
 
theBEAST said:
Isn't the cathode of D1 1V? The diagram is a bit blurry but there should be a + sign in front of the 1. Also, how do you know the anode is 0V?

Well, I see the anode is not at 0 V, o is just a terminal, not voltage.

The cathode of both diodes is connected to 3 V through the resistor. Current can flow in the direction of decreasing potential, so it can flow through D1. If D1 is an ideal diode, you can take the potential difference zero across it. Then the whole potential difference from 3 V to -1 V falls across the resistor, and the cathode is at -1 V. Is D2 closed or open?
More common to assume about 0.7 V across a conducting diode. What is V then?
What would happen if the cathode of D1 was at +1 V instead of -1?

ehild
 
ehild said:
Well, I see the anode is not at 0 V, o is just a terminal, not voltage.

The cathode of both diodes is connected to 3 V through the resistor. Current can flow in the direction of decreasing potential, so it can flow through D1. If D1 is an ideal diode, you can take the potential difference zero across it. Then the whole potential difference from 3 V to -1 V falls across the resistor, and the cathode is at -1 V. Is D2 closed or open?
More common to assume about 0.7 V across a conducting diode. What is V then?
What would happen if the cathode of D1 was at +1 V instead of -1?

ehild

Isn't the cathode of D1 connected to 1V? The anode is connected to 3V through the resistor.

As for D2, would it be open because the potential difference from 3 V to 3 V is zero and thus there is no current conducting there?

If the cathode of D1 was 1 V instead of -1 it would not make a difference since the current would still want to flow from high to low potential?
 
Disregard this post, I accidentally double posted.
 
theBEAST said:
Isn't the cathode of D1 connected to 1V? The anode is connected to 3V through the resistor.

1 V or -1 V, anyway, it is open. In case it is an ideal diode, the potential difference is zero across it. So V is the same as the potential of the cathode, 1 V or -1 V.

theBEAST said:
As for D2, would it be open because the potential difference from 3 V to 3 V is zero and thus there is no current conducting there?

The anode is at potential V and it is 1 V or -1 V, as the potential drops on the resistor. So the diode is closed.

ehild
 
I think you probably understand the circuit by now but it might help to consider would happen if the output voltage on the right was somehow ramped up slowly from 0V toward 3V. Don't worry about how that might be done just go with it for now..

With the output at 0V both diodes would be reverse biased (eg OFF).

As the output voltage is ramped up there comes a point where it reaches 1V. Above that point D1 would be forward biased (eg ON). So at this point replace D1 with a wire.

Now you have the output connected to a 1V voltage source by a wire. The output cannot go any higher than 1V no matter how hard the resistor tries to pull it up. In effect Diode D1 "clamps" or limits the output voltage to 1V. (Aside: If it was a real world diode it would clamp the output to 1V plus the forward voltage drop, say about 1.7V instead of 1V).

With the output clamped at 1V by D1 the diode D2 must be reverse biased (OFF).
 
theBEAST said:

Homework Statement


Find V and I in the following diagram:
fl0SQ3Z.png


The Attempt at a Solution


I know what the answer is. Diode D2 is off but why?

Now if the voltage left of D2 was +4V then diode D2 is off because +4V dominates +3V so the current runs right and thus turns off D2. Is this correct? Is this how ideal diodes work?

Thanks!

The 3V on top is not the deciding voltage.

Look at the diagram. The voltage at D1 cathode is +1V. What then is the voltage at D1 anode? What then is the voltage across D2? Is it on or off?

(Since 3V > 1V D1 is on.)
 
  • #10
rude man said:
The 3V on top is not the deciding voltage.

Look at the diagram. The voltage at D1 cathode is +1V. What then is the voltage at D1 anode? What then is the voltage across D2? Is it on or off?

(Since 3V > 1V D1 is on.)

Hmm, I am actually not sure.. What is the voltage at the D1 anode? Is it 3? Is it because we first assume that D2 is off so the current flows to D1 and gives the cathode 3V?
 
  • #11
theBEAST said:
Hmm, I am actually not sure.. What is the voltage at the D1 anode? Is it 3? Is it because we first assume that D2 is off so the current flows to D1 and gives the cathode 3V?

No, it is not 3 V. The current flowing through the resistor causes potential drop IR across it. So the voltage on the anode of D1 is 3-IR.

ehild
 
  • #12
Have you never been told what the voltage drop across a diode is when it's on?
 

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