Zener Diode Problem: Solutions and Guidance

[Weightofananvil]

Hey,
I'm studying zener diodes and the various ways they are used. This question is throwing me for a loop however. I can't get my simulator to display the circuit like it is in this question because it only sets reverse breakdown voltages @ 5mA.

I'm just wondering if I'm on the right track.If I am wrong please give me a slight nudge in the right direction. This is a homework problem though, so for the sake of learning let me struggle a bit.. :) Thanks

http://s704.photobucket.com/user/nrozka/media/File%202016-01-16%208%2010%2031%20PM.jpeg

 

[Svein]

I'm just wondering if I'm on the right track.If I am wrong please give me a slight nudge in the right direction. This is a homework problem though, so for the sake of learning let me struggle a bit.. :) Thanks

For what values of R would the Zener not regulate?

 

[Tom.G]

I wouldn't worry about it. Simulators are, after all, just an approximation. You may be able to edit the definition of a diode so the knee is at 1mA instead of 5mA but it's probably not worth the climb up the learning curve. So you've already learned that simulators have some limitations.

Your pencil&paper solution is fine, you obviously understand the concepts and used the right approach.

P.S. - I was recently simulating a five stage amplifier and the rather high-end simulator reported an output voltage of 300V. The thing is that the supply voltage was only 80V (must have had some magic transistors in there.) Not a problem because all I needed at that point was the openloop gain (60 in this case). Normal operation was at a gain around 5.

 

[LvW]

Weightofananvil - may I ask you: For what purpose you have computet the voltage in step1 ?
This is the voltage across Rs - but you didn`t use it in step3. Why not?

A short comment to simulation programs:
According to my experience, in 99.9% of all cases it is the USER who makes errors -and not the program.
In many cases, there is a confusion between simulations in the time and frequency domain, respectively.
For example, the ac analysis is a linear small-signal analysis. As a consequence, the simulation does not know any supply voltage limitations - and the output signal can be much larger than the suplly voltage.
The ac analysis serves only one purpose: It shows the frequency-dependent properties of a circuit and does not take into account any non-linearities.
In your case, I would use only DC and/or TRAN analyses.

 

[Tom.G]

LrW - Oops. You are correct in asking about calculating the voltage in step 1. I fell into the same trap. So I hereby retract the second paragraph in my earlier post. (Sorry Weightofananvil, close but no cigar.) You are also right about AC vs. DC analysis. My purpose in mentioning it, however, was to point out that there are quirks in simulators, and a human brain is needed to throw out the occasional absurdities, or to decide to accept the "good enough" results.

 

[LvW]

... a human brain is needed to throw out the occasional absurdities, or to decide to accept the "good enough" results.

Yes - one should never blindly trust the results obtained by simulation.
As a first step, it is important to select the proper analysis with respect to the information needed.
And, secondly, one must be able to interpret the results properly.

Example: Opamp with resistive POSITIVE feedback. We all know that this circuit will not find a stable bias point.
However, simulation of the circuit will show that (a) bias point calculation, (b) a DC analysis and (c) AC analyses will result in a stable system with gain.
Did the simulator fail? No - it was not wrong. All calculations are correct because of IDEAL constraints.
The user should know that all three analyses assume (a) a power supply which is available for t<0 (no switch-on effect) and (b) no external disturbances.
A mechanical analogon is a small ball riding upon a larger ball. Under IDEAL conditions, this scheme is stable.
For the opamp example, we have to perform a simulation in the time domain (TRAN analysis) with a realistic supply voltage switch-on transient to show that the circuit willl be unstable.

 

[Weightofananvil]

Thanks for all your help everyone, Some very helpful comments.