Two parallel diodes in a circuit

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  • #1
Jahnavi
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The Attempt at a Solution



I am not quite sure how to deal with the two diodes in parallel . How do we know what is the potential applied across the two diodes so as to decide which one is conducting ?
 

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  • #2
.Scott
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As the voltage across the diodes rise, the Germanium will start to conduct first - essentially shorting out the Silicon.
So in the circuit shown, only the Germanium is conducting significant current.

With the Germanium reversed, it will not conduct at all. So the Silicon will conduct with a 0.7v voltage drop.
 
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  • #3
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Thanks :smile:

I have a very basic question .

Till the time the current is not flowing through any of the diodes , what is the potential difference across the two diodes ?

Is it 12 V ?
 
  • #4
.Scott
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Assuming ideal components, you will never have 12V across the diodes in the conducting direction.
You will have, at most, 0.3V across the Germanium and, at most, 0.7V across the Silicon.
So when the Ge and Si diodes are in parallel and in the same direction as the circuit flow, the voltage across the (ideal) diodes will never exceed 0.3V.
 
  • #5
Jahnavi
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Assuming ideal components, you will never have 12V across the diodes in the conducting direction.

Till the time none of the diodes is conducting , voltage across both of them will be 12 V but after germanium starts conducting voltage across it will be fixed at 0.3V .

After the current starts flowing , there will be a constant voltage drop of 11.7 V across the resistor and a constant voltage drop of 0.3 V across the germanium diode .

Is that correct ?
 
  • #6
cnh1995
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voltage across both of them will be 12 V
No, that violates circuit theory. That is not how voltages develop in a circuit.
Till the time the current is not flowing through any of the diodes , what is the potential difference across the two diodes ?
There is no specific answer to this question. How voltages develop across components in circuits is a very interesting topic, but it is out of the scope of circuit theory. In circuit theory, all that matters is Kirchhoff's laws, Ohm's law and v-i relationships of the components.
How do we know what is the potential applied across the two diodes so as to decide which one is conducting ?
In the original diagram where both the diodes are in forward connection, you have two probable scenarios:
1)Voltage across both the diodes is 0.7V, which means Vout=11.3V. So, I1=11.3/R
2)Voltage across both the diodes is 0.3V, which means Vout=11.7V. So, I2=11.7/R

In scenario 1), both the diodes conduct, so I1 should be more than I2 (almost double), but from scenario 2), it is clearly the opposite.

So which one do you think is the correct scenario?
 
  • #7
Jahnavi
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So which one do you think is the correct scenario?

2)Voltage across both the diodes is 0.3V, which means Vout=11.7V. So, I2=11.7/R
 
  • #8
cnh1995
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2)Voltage across both the diodes is 0.3V, which means Vout=11.7V. So, I2=11.7/R
Right.
That's what .Scott is saying in #4.
 
  • #9
Jahnavi
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No, that violates circuit theory. That is not how voltages develop in a circuit.

I don't understand . When none of the diodes is conducting , there is no current in the circuit which means no potential difference across the resistor . This means complete potential difference is applied across the two diodes .

You are agreeing that after the germanium diode starts conducting then potential difference across it is 0.3 V ?

Do you agree potential difference across the Silicon diode is also 0.3 V ( not conducting ) after the current starts flowing through the Ge diode ?

Isn't this contradictory to your objection that while the two diodes are not conducting , complete battery voltage 12 V is not applied across them ?
 
  • #10
cnh1995
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Isn't this contradictory to your objection that while the two diodes are not conducting , complete battery voltage 12 V is not applied across them ?
Not at all.
If the diodes have 12V across them, they should conduct, but the don't. That violates circuit theory.
In circuit theory, there is no instant when there is no current through this circuit. The moment you close the switch, 0.3V appear across the diodes and 11.7V appear across the resistor, and the current starts flowing instantaneously. One of the most important assumptions in circuit theory is that the components 'enforce' their v-i relationships across their terminals instantaneously.
 
  • #11
Jahnavi
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Not at all.
If the diodes have 12V across them, they should conduct, but the don't.

You are right :smile:

Would it be correct to say that a potential drop occurs across a diode after it starts conducting ? But if that is correct , then why do we say that potential drop across the Si diode is also 0.3 V ?

Another question -

Can we consider the potential drop across the Ge diode as a battery of voltage 0.3 V connected in series with opposite polarity ( such that it opposes the driving battery ) ?
 
  • #12
cnh1995
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Would it be correct to say that a potential drop occurs across a diode after it starts conducting ? But if that is correct , then why do we say that potential drop across the Si diode is also 0.3 V ?
No, again, there is no "before" and "after" in circuit theory. Everything is instantaneous.
If you want to study how voltages develop in circuits, you can study the role of surface charges in establishing electric fields in circuits. That would help you understand Kirchhoff's laws intuitively. Any deeper than that, you need Maxwell's equations and a very short time scale (of the order of a few pico/nanoseconds). Circuit theory assumes a larger time scale and there is absolutely no delay between input connection and output response.
Can we consider the potential drop across the Ge diode as a battery of voltage 0.3 V connected in series with opposite polarity ( such that it opposes the driving battery ) ?
Yes.

Check out this excellent insights article by @anorlunda, especially the last paragraph that talks about levels of studying electricity.
https://www.physicsforums.com/insights/ohms-law-mellow/
 
  • #13
Jahnavi
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But you do agree that once the Ge diode starts conducting , the potential drop across the Si diode is also 0.3 V ?

If you do agree , then how is it that despite not conducting current through it , the drop across Si diode is 0.3 V ?
 
  • #14
cnh1995
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then how is it that despite not conducting current through it , the drop across Si diode is 0.3 V ?
Because the Silicon diode is in parallel with the Germanium one. And silicon diode conducts when its voltage drop is 0.7V. Anything below that, it will act as an open switch.
 
  • #15
CWatters
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But you do agree that once the Ge diode starts conducting , the potential drop across the Si diode is also 0.3 V ?

If you do agree , then how is it that despite not conducting current through it , the drop across Si diode is 0.3 V ?
If two components are in parallel they are connected between the same two nodes so must have same voltage.
 
  • #16
CWatters
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One way to visualize how the circuit works is to replace the 12v with a 0-12v variable supply. Start it at 0V and gradually increase it.
 
  • #17
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If two components are in parallel they are connected between the same two nodes so must have same voltage.

I agree .

But by that logic , what is the voltage across the two diodes before any one allows current to flow ?
 
  • #18
cnh1995
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what is the voltage across the two diodes before any one allows current to flow ?
If the input voltage is 0.2V (less than 0.3V), the diode voltage is also 0.2V and no diode will conduct. So, if you vary the input voltage from 0V to 0.3V, all of that voltage will appear across the diodes. I don't know if this is what you meant by "before any diode allows current to flow".
 
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  • #19
CWatters
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I agree .

But by that logic , what is the voltage across the two diodes before any one allows current to flow ?
It's not clear what you mean by "before anyone allows current to flow"?

If you put a switch in series and open the switch the voltage across both diodes diodes will be zero.

If you replace the 12v source with a 0.2v source then that's not enough to turn on either diode so no current flows (using the model implied in the problem). If no current flows the voltage drop across the resistor is zero so the 0.2 v appears across the diodes.

At all times the drop across the ge is the same as the si. You can apply KVl to the little loop formed by the two diodes if you want.
 
  • #20
CWatters
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Aside: If you use a different model for the diode you will get a different answer. The problem statement assumes a crude model where diodes don't conduct until a certain voltage across them. Real diodes actually have a IV curve so at 0.2v a small current may flow. If you want to know exactly how much current in this circuit you can plot IV curves for the source (power supply and diode) and the load (resistor) and where they cross will be the operating point. This is an example of solving two simultaneous equations by the graphical method which you may have covered in a maths class.
 
  • #21
Jahnavi
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Why is the voltage drop across the diode a constant value , 0.3 V for Ge ? When a 12 V battery is applied across a Ge diode , the drop across it is a fixed value . Why ?
 
  • #22
cnh1995
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When a 12 V battery is applied across a Ge diode , the drop across it is a fixed value . Why ?
If you directly apply 12V across a Ge (or any) diode, the drop across it will be 12V and not fixed 0.3V.
I am not sure what exactly is confusing you.
 
  • #23
Jahnavi
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If you directly apply 12V across a Ge (or any) diode, the drop across it will be 12V and not fixed 0.3V.

Sorry .This is incorrect .

Voltage drop across a diode ( conducting ) is a constant value . 0.3 V for Ge .
 
  • #24
cnh1995
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Sorry .This is incorrect
No it's not (again, think of circuit theory). The Ge diode will draw an insanely high current at 12V and probably burn out.
Voltage drop across a diode ( conducting ) is a constant value .
That is just an approximation. It is called the constant voltage drop model.
Diode+Constant-Voltage-Drop+Model.jpg


Look up 'ideal diode equation' and you'll see that the v-i relationship of a real diode is exponential.

Anyways, I'd better be off to bed now. I am sure experts from other time zones will chime in to help. :smile: TTFN..
 

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  • #25
Jahnavi
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My earlier response was under the assumption that the 12 V battery placed across the diode is non ideal i.e it has some internal resistance .

Voltage drop across diaode will be 0.3 V , rest 11.7 V appears across the internal resistance .

I think you are assuming an ideal battery .

I might be wrong .
 
  • #26
cnh1995
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I think you are assuming an ideal battery
Yes, as they have used an ideal battery in your OP.
Voltage drop across diaode will be 0.3 V , rest 11.7 V appears across the internal resistance .
Using constant voltage drop model, yes.
 
  • #27
ehild
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I don't understand . When none of the diodes is conducting , there is no current in the circuit which means no potential difference across the resistor . This means complete potential difference is applied across the two diodes .
Yes, at the instant the switch gets closed the initial current is zero, and 12 V appears might appear across the diodes, but the diodes and all the circuit do not behave as ideal ones. The p-n junctions of the diodes behave as capacitors and all wires behave as resistors and inductors. There will be some transient current in the circuit (till the equilibrium charge distribution builds up at the p-n junctions) which ends in the stationary voltage drop across the diodes and stationary current in the circuit in a very short time.
 
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  • #28
Jahnavi
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Yes, at the instant the switch gets closed the initial current is zero, and 12 V appears across the diodes, but the diodes and all the circuit do not behave as ideal ones. The p-n junctions of the diodes behave as capacitors and all wires behave as resistors and inductors. There will be some transient current in the circuit which ends in the stationary voltage drop across the diodes and stationary current in the circuit in a very short time.

Thanks !
 
  • #29
Jahnavi
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Yes, as they have used an ideal battery in your OP.

Using constant voltage drop model, yes.

Thank you very much @cnh1995 :smile: . I really appreciate that you patiently answered all my queries .

Thanks @.Scott and @CWatters .
 
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  • #30
ehild
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Thanks !
That happens in all DC circuits when switching on or off.
 
  • #31
Jahnavi
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OK :smile:
 
  • #32
cnh1995
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Yes, at the instant the switch gets closed the initial current is zero, and 12 V appears across the diodes, but the diodes and all the circuit do not behave as ideal ones. The p-n junctions of the diodes behave as capacitors and all wires behave as resistors and inductors. There will be some transient current in the circuit (till the equilibrium charge distribution builds up at the p-n junctions) which ends in the stationary voltage drop across the diodes and stationary current in the circuit in a very short time.
Whie I agree with the rest of your explanation, I don't see how we can say 12V will appear across the diodes initially. When the equilibrium charge distribution builds up at the pn junction and surface charge distribution builds up on Ohmic components (resistors, wires), there is no way we can compute voltage across any component. It's all chaos for that very very short time. Even if we assume the diode junctions to be capacitances, I don't think we can say initially all the 12V appear across the junctions. I believe this process is very different, extremely short-lived and hence, can't be described by the steady state quantities like V and I.
 
  • #34
ehild
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Whie I agree with the rest of your explanation, I don't see how we can say 12V will appear across the diodes initially. When the equilibrium charge distribution builds up at the pn junction and surface charge distribution builds up on Ohmic components (resistors, wires), there is no way we can compute voltage across any component. It's all chaos for that very very short time. Even if we assume the diode junctions to be capacitances, I don't think we can say initially all the 12V appear across the junctions. I believe this process is very different, extremely short-lived and hence, can't be described by the steady state quantities like V and I.
Yes, you are right, we can not use the steady-state quantities in the circuit, but the 12 V emf is applied at the instant of switching up. Even the diode have inductance and resistance because of the leads and the bulk of the semiconductor, so not all the voltage drop appears across the pn-junction or even across the diode because of the inductance of all the wires and the resistor.
 
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  • #35
CWatters
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A lot comes down to which model of a diode you are using. The problem statement provides very little information so you have to assume a very crude model, almost an ideal diode but with a defined forward voltage.
 

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