Why Do Different Paths in a Diode Circuit Show Different Voltages?

[Saitama]

Homework Statement

Determine $V_0$ and $I$ for the network shown in attachment 1.

Homework Equations

The Attempt at a Solution

I can replace Si diode with a battery of emf of 0.7 V and Ge diode with that of 0.3 V.

If I take Path-1 (orange) in attachment 2, I get:
$$V_0=10-0.3=9.7 V$$
which is correct but if I take Path-2 (blue), I get:
$$V_0=10-0.7=9.3 V$$
Why do I get two different answers?

Any help is appreciated. Thanks!

 

[mr_pavlo]

Voltage across Ge diode will be always smaller than Si threshold so Si diode will never open so you can omit it from circuit and calculate series combination of Ge diode and resistor. There is small current through Si diode but it is negligibly smaller than current through Ge diode.

 

[NoPoke]

You can model a silicon diode as a 0.7V battery in series with a "perfect" diode. A little more care is required when modelling a germanium diode as a 0.3V battery and a "perfect" diode as germanium diodes leak a lot more than silicon under reverse bias. Leakage won't be an issue for this circuit.

 

[BvU]

I can replace Si diode with a battery of emf of 0.7 V and Ge diode with that of 0.3 V.

Only approximately, and only when they are in a "conducting state". Diodes have a I-V relationship that is a bit more continuous than this step function approximation. (V_d is your 0.3 or 0.7).

Don't worry about the precise exponential shape for the answer to this exercise (unless you were given a lot more details on these two particular diodes). Electronics is a coarse science compared to real physics .

 

[Saitama]

Hello everyone! :)

Welcome to PF, mr_pavlo!

Voltage across Ge diode will be always smaller than Si threshold

Yes, agreed.

so Si diode will never open so you can omit it from circuit

Why?

Only approximately, and only when they are in a "conducting state". Diodes have a I-V relationship that is a bit more continuous than this step function approximation. (V_d is your 0.3 or 0.7).

Don't worry about the precise exponential shape for the answer to this exercise (unless you were given a lot more details on these two particular diodes). Electronics is a coarse science compared to real physics .

You can model a silicon diode as a 0.7V battery in series with a "perfect" diode. A little more care is required when modelling a germanium diode as a 0.3V battery and a "perfect" diode as germanium diodes leak a lot more than silicon under reverse bias. Leakage won't be an issue for this circuit.

@BvU: Yes, I am aware of the I-V relationship but since most of the solved examples in the book replaced the diode with an "equivalent" battery of emf I have shown above, I thought I should also solve the problem the same way.

...but how does that answer my question?

 

[NoPoke]

You don't get two answers if you use a better model. In particular the model that you are using of a diode equals a battery is insufficient for this problem.

Batteries can both source and sink current. Batteries have finite internal resistance. Adding a perfect diode in series with the battery can prevent it from sourcing current. Adding a finite internal resistance will avoid the infinite loop current which you currently have.

 

[mr_pavlo]

Thanks for welcome,

so Si diode will never open so you can omit it from circuit

because Ge diode will not allow voltage across Si diode to rise to more than 0.3V and Si diode starts conducting when voltage across it is around 0.7V. It is like when you connect Blue LED to circuit and it shines and then you connect Red LED parallel to Blue one and the Blue goes off. It's because blue LED has higher forvard voltage drop than red so red takes over all current.
Of course in real, there would be small current through Si diode compared to that through Ge diode but you would have to know exactly the type of diodes (1N4148, ... ) for precise calculations but it would have no practical relevance as the result would be almost the same with or without Si diode. That is just my practical experience and I can be wrong .

 

[Saitama]

Hi again! Sorry for the delay in reply.

because Ge diode will not allow voltage across Si diode to rise to more than 0.3V and Si diode starts conducting when voltage across it is around 0.7V.

I think this clears my doubt, thanks a lot mr_pavlo!

It is like when you connect Blue LED to circuit and it shines and then you connect Red LED parallel to Blue one and the Blue goes off. It's because blue LED has higher forvard voltage drop than red so red takes over all current.

Yep, I have seen this when I made a circuit involving differently coloured LEDs.

I hope you spend a great time at PF. :)