Basic diode analysis problem

In summary,The homework statement is that you need to determine the voltage across two resistors for the network shown in the attachment.voltage across Ge diode will be always smaller than Si treshold so Si diode will never open so you can omit it from the circuit.
  • #1
Saitama
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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? :confused:

Any help is appreciated. Thanks!
 

Attachments

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  • #2
Voltage across Ge diode will be always smaller than Si treshold 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.
 
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  • #3
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.
 
  • #4
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. (Vd 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 :smile:.
 
  • #5
Hello everyone! :)

Welcome to PF, mr_pavlo! :smile:

mr_pavlo said:
Voltage across Ge diode will be always smaller than Si treshold
Yes, agreed.
so Si diode will never open so you can omit it from circuit
Why? :confused:

BvU said:
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. (Vd 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 :smile:.

NoPoke said:
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? :confused:
 
  • #6
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.
 
  • #7
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 :smile:.
 
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  • #8
Hi again! Sorry for the delay in reply. :redface:

mr_pavlo said:
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! :smile:
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. :)
 

What is a basic diode analysis problem?

A basic diode analysis problem involves using circuit analysis techniques to determine the behavior and characteristics of a diode in a given circuit.

What types of circuits typically involve diode analysis problems?

Diode analysis problems are commonly found in rectifier circuits, voltage regulators, and signal processing circuits.

What are the key parameters to consider in a diode analysis problem?

The key parameters to consider in a diode analysis problem include the diode's forward voltage drop, reverse breakdown voltage, and current-voltage relationship.

What are some common techniques used to solve diode analysis problems?

Some common techniques used to solve diode analysis problems include the diode equation, Kirchhoff's laws, and load line analysis.

What are some potential challenges in solving a diode analysis problem?

Solving diode analysis problems can be challenging due to the non-linear behavior of diodes, and the need to consider both forward and reverse biasing conditions. Additionally, temperature effects and parasitic components can complicate the analysis.

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