Diode circuit- Vin Vs Vout

Thread Starter

gbox

Joined Dec 29, 2015
42
I am given the following circuit:

צילום מסך 2016‏.05‏.22 ב‏.21.09.46.png

I need to graph V_out (AB) as function of V_in.
How Should I start looking at this?
 

shteii01

Joined Feb 19, 2010
4,644
By defining Vin. All you show is that it is 1 kHz sine wave. If the sine wave never exceed 0.7 volts, then D3 is always off and Vout is always zero. See how simple this is?

Also D3 has 25 volt battery on the cathode end, which means you will need a lot more than 0.7 volts to turn it on.
 

Thread Starter

gbox

Joined Dec 29, 2015
42
By defining Vin. All you show is that it is 1 kHz sine wave. If the sine wave never exceed 0.7 volts, then D3 is always off and Vout is always zero. See how simple this is?

Also D3 has 25 volt battery on the cathode end, which means you will need a lot more than 0.7 volts to turn it on.
Yes I need to show what happens for different values of V_in
 

WBahn

Joined Mar 31, 2012
29,928
The answer is:

Vin<50V Vout=50V
50V<Vin<100V Vout=Vin
Vin>100V Vout=100V
Not if it's a sinusoidal source.

Is it supposed to be sinusoidal -- or is it supposed to be a DC source.

In either event, set Vin equal to zero. What will the output be? Remember, you've specified physical diodes, so you can't just assume that there is no voltage drop across them when they are conducting.
 

dannyf

Joined Sep 13, 2015
2,197
How Should I start looking at this?
Think of those diodes as perfect diodes (Vfwd=0v).

The current through R2+R3 is (100v-25v)/300k=.25ma. So the voltage drop over R2 is .25ma * 200k = 50v.

That means the voltage on top of R3, where the two diodes' cathodes meet, is 25v+25v=50v.

That means that when Vin < 50v, D3 is not forward-conducting and Vin is not in the circuit. so Vab=50v when Vin < 50v.

As Vin goes over 50v, D3 starts to conduct, and Vab is clamped to Vin. Vab = Vin when Vin > 50v.

That relationship breaks down when Vin goes over 100v. In that case, D2 is not conducting and no current flows through R2. So Vab=100v - 0ma * 200k = 100v.

To summarize:

Vin < 50v, Vab = 50v;
50v<Vin<100v, Vab = Vin;
100v < Vin, Vab = 100v
 

joeyd999

Joined Jun 6, 2011
5,220
Not meaning to pick nits:

Is the diode supposed to be ideal? Or, as shown, actually a 1n4148?

The Vr for 1n4148 is 75V. This will have an effect on the output for large input voltage swings...
 

Thread Starter

gbox

Joined Dec 29, 2015
42
Think of those diodes as perfect diodes (Vfwd=0v).

The current through R2+R3 is (100v-25v)/300k=.25ma. So the voltage drop over R2 is .25ma * 200k = 50v.

That means the voltage on top of R3, where the two diodes' cathodes meet, is 25v+25v=50v.

That means that when Vin < 50v, D3 is not forward-conducting and Vin is not in the circuit. so Vab=50v when Vin < 50v.

As Vin goes over 50v, D3 starts to conduct, and Vab is clamped to Vin. Vab = Vin when Vin > 50v.

That relationship breaks down when Vin goes over 100v. In that case, D2 is not conducting and no current flows through R2. So Vab=100v - 0ma * 200k = 100v.

To summarize:

Vin < 50v, Vab = 50v;
50v<Vin<100v, Vab = Vin;
100v < Vin, Vab = 100v
Thanks, how did you recognized that when Vin>50v "Vab is clamped to Vin"?
 

MrAl

Joined Jun 17, 2014
11,342
Hi,

Yes you should take this one step at a time, Starting with Vin=0.
Since you seem to be assuming that the forward voltage drop for any diode is zero, let's stick with that for now. You also seem to be assuming that there is no diode leakage current with reverse bias so let's stick with that for now too.

With Vin=0 the first diode (D3) is reverse biased because the voltage to the right of it will be higher than zero.
Next you must find the voltage at A with the first diode 'open'. Can you do that? This is a two source circuit so you have to know how to handle that situation. Can you find the voltage at A with the first diode open?
 

dannyf

Joined Sep 13, 2015
2,197
how did you recognized that when Vin>50v "Vab is clamped to Vin"?
In that case, both diodes are conducting and both Vin and the 100v source are powering the 25v source. However, Vin does so without any resistance soo it overwhelms.

You will arrive at the same conclusion if you were to short thee two diodes.
 
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