# Two parallel diodes in a circuit

## Homework Statement ## 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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.Scott
Homework Helper
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.

• CWatters and Jahnavi
Thanks 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 ?

.Scott
Homework Helper
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.

• Jahnavi
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 ?

cnh1995
Homework Helper
Gold Member
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?

• Jahnavi
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

cnh1995
Homework Helper
Gold Member
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.

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 ?

cnh1995
Homework Helper
Gold Member
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.

• Jahnavi
Not at all.
If the diodes have 12V across them, they should conduct, but the don't.
You are right 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 ) ?

cnh1995
Homework Helper
Gold Member
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/

• Jahnavi
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 ?

cnh1995
Homework Helper
Gold Member
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.

CWatters
Homework Helper
Gold Member
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.

CWatters
Homework Helper
Gold Member
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.

• NTL2009
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 ?

cnh1995
Homework Helper
Gold Member
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".

• Jahnavi and CWatters
CWatters
Homework Helper
Gold Member
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.

• Jahnavi
CWatters
Homework Helper
Gold Member
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.

• cnh1995
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 ?

cnh1995
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Gold Member
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.

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 .

cnh1995
Homework Helper
Gold Member
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. 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. TTFN..

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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 .