Didoe limiter-finding the maximum current

[intelwanderer]

Homework Statement

https://www.edx.org/static/content-mit-6002x/images/circuits/diode-limiter.013c535f04ae.gif [Broken]

The AC voltage source puts out a symmetrical triangle wave that looks like:
https://www.edx.org/static/content-mit-6002x/images/circuits/triangle-wave.52377c602e8e.gif [Broken]

The peak voltage of this waveform is 9.5V.

The resistors both have the resistance R=4.7kΩ. The voltage of the DC sources is V1=3.0V and V2=2.0V.

They ask for the maximum current that can go through both diodes(both through D1 and D2). They also ask for the maximum positive and negative voltages across R, but I got that already.

Homework Equations

Standard: KCL, KVL, blah blah blah...

The Attempt at a Solution

OK. So if the maximum current is going through a diode, it acts like it's a short. I assume that the other diode would act like it's an open branch, so I can disregard that. From there, I use KCL to solve. I get a ridiculously small amount of current(in Amperes), that is wrong. What am I missing?

 

[gneill]

Hi intelwanderer.

Are the diodes to be treated as ideal (no forward resistance or junction voltage --- e.g. "real" silicon diodes have an approx. 0.5V to 0.7V forward conduction potential)?

Anyways, what do you consider to be a ridiculously small amount of current? Can you show your calculation so that we can see how to help?

 

[intelwanderer]

Yes, it's ideal.

Something like .0001 Amps or something...
The equation is:

Calling the node voltage A and for D1, V1 = 3, Vs = 9.5

(A-Vs)/R = (A-V1) + (A)/R. Neglecting V2, as that is a open.

 

[gneill]

On the positive half-cycle of the input, the D1 branch will clamp the node voltage at V1=3V. So you don't need to solve for the node voltage after the input exceeds a threshold that allows D1 to conduct. Since you know that voltage, you should be able to calculate the currents in the two resistors...

 

[intelwanderer]

It asks for the maximum current through the diode. I calculate the current from the AC, subtract that of the R, and the difference between 0 and that value is the current through the diode. But that's apparently wrong.

Assuming that the node voltage is 3 here.

 

[gneill]

It asks for the maximum current through the diode. I calculate the current from the AC, subtract that of the R, and the difference between 0 and that value is the current through the diode. But that's apparently wrong.

Assuming that the node voltage is 3 here.

I don't understand the meaning behind your phrase, "the difference between 0 and that value"? How does 0 enter into the picture?

On the positive half-cycle you have a maximum input voltage and the fixed maximum node voltage. So what is the maximum current that can flow through the first (left) resistor?

 

[intelwanderer]

-.00139 or thereabouts. 3-9.5/4700

 

[gneill]

-.00139 or thereabouts. 3-9.5/4700

Okay, although you would be better to think of the current as flowing from the higher voltage source and into the lower voltage node. So about 0.00138A, or 1.38 mA.

Now, given once again the fixed node voltage, what's the current through the second resistor? How does it compare to the previous current?

 

[intelwanderer]

3/4700. 6.38e-4


I subtract the above value(which I made positive like you suggested), with this one, and it's something like 7.4468e-4. That's a wrong value, apparently.

 

[intelwanderer]

Ah. Now I got it.

Thank you so much gneill! I appreciate it.

 

[gneill]

3/4700. 6.38e-4


I subtract the above value(which I made positive like you suggested), with this one, and it's something like 7.4468e-4. That's a wrong value, apparently.

Right. So in other words, a certain maximum current flows into the node from the source and a certain fixed current flows out of the node through the load resistor. Whatever current remains takes the only other path available, through the diode branch.

Now apply the same logic for when the supply reaches its minimum (negative) peak.

 

[intelwanderer]

Gotcha. Thanks again, gneill. Darn, I made that more complicated than it needed to be.

I got one last problem to do, if I need help on it, I'll make a separate thread.