The relation between Vin and Vout (diodes circuit)

Thread Starter

mungs

Joined Mar 18, 2021
2
Hi, can anyone here tell me if the function I found that relates the Vin to Vout is right? So I made a mesh analysis to find the current in the first mesh (left to right) and in the second and finally got that the Vout = (2/11)*Vi +6.328 (is it right?)

threshold voltage = 0.7V

diode circuit
my shot
 

MrAl

Joined Jun 17, 2014
8,500
Hi, can anyone here tell me if the function I found that relates the Vin to Vout is right? So I made a mesh analysis to find the current in the first mesh (left to right) and in the second and finally got that the Vout = (2/11)*Vi +6.328 (is it right?)

threshold voltage = 0.7V

diode circuit
my shot
Hello,

Is your result Vout=2/11*Vi+6.328 ?

If that is so, then a cursory glance at this circuit tells me that can not be right.

The reason it can not be right is because first, usually diode circuits with a variable input will have an output expression that is conditional meaning there will be changes that occur more or less abruptly and that would be indicated by two or more expressions not just one. It's not always like that, but when you have an input of some sort that has no limits on it's value (as your Vi seems to have) and the other voltage(s) are fixed, then there will be some point where one or more diodes conduct for some input values and not for other input values which means the output can only be represented by some sort of conditional expression like:
2*Vin when Vin<2
3*Vin when Vin>2
0v when Vin=2

That is just a random example that has nothing to do with this circuit but you see the difference. The output may change to very different values depending on the input so you end up with two or more output expressions.

This circuit contains two bias sources V1 and V2 and they will establish the point(s) where the input voltage causes the first diode to conduct or not conduct.
Note that because the input can be virtually anything, the simplest starting point is to state that the first diode is not conducting which means it is open circuit and thus Vin has no effect. That means that the voltage at the junction of D1 and D2 is due only to the two bias sources V1 and V2 and the two associated resistors. The output is then simply that voltage plus the diode voltage drop of 0.7 volts.
Note that will mean there will be a constant voltage at the output for the entire time Vin is below some limit value, and thus the output expressions must at least be something like:
5v when Vin<VinA
2*Vin+4 when Vin>=VinA
although again these are random examples that are not actual results from this circuit.
There could be more conditionals required also it depends on when the two diodes conduct in relation to the input voltage Vin.

So check over your results again and see if this starts to make more sense.
Start with diode D1 open and what Vin has to be for that to happen, then go from there.
 
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