Recent content by d3thkn1ght

Great to know. Thanks again for all the help.

The question wanted to know the contribution of VI which it called Vout and then the contribution from vi which was called vo. For RL of 2K ohms: E of 9.5 V gave a value of 5.001997004. I put in 5.1 since from my understanding the Zener diode keeps a constant voltage when open and it worked...

I was able to get the right answer using the following formula: $$\frac{E-V_{out}}{R_{in}}=\frac{V_{out}-5}{rz} + \frac{V_{out}}{R_L}$$

Is this equation any better: $$\frac{E-V_z}{R_{in}} = \frac{V_z}{rz} - \frac{V_L} {R_L}$$

Okay, thanks for clarifying.

Is this formula correct: $$\frac{ (E - Vz)} {Rin} = Iz + \frac {VL} {RL}$$ If so, is Vout = VL?

Thanks for clarifying. I was confused a little on the Vout part of it all. I'll try this. Thanks again.

So far, I came up with the following equation for KCL: $$\frac{ (E - Vz)} {Rin} = Iz + \frac {Vz} {RL}$$

Great, I will try the equations and post my results. Thanks again for all your patience.

For the zener model, I would guess a 5 V source and a 1 ohm resistor? With the + end of the source pointing up

Yes, I see my error on the voltage. I did it incorrectly in my head. So, the operating point is 5V from the graphical method? In the two sentences, would I use a single source of 9.55 Volts (9.5 from VI and .05 from Vi) and then use KVL at the top node?

And the units I used were milliAmps

I divided the Voc by the Rth and figured that would be the Isc. I believe I can get the operating point graphically, and would be interested in how to do it mathematically. Also, now that I have the operating point, how do I found Vo and vo?

Okay, so the load line appears to be from (0,10) to (6.33,0). Thus it crosses the iv line of xener diode slightly past 5 Volts. I don't have any easy way to draw the line to scale. Looked online but didn't find any apps that let me do it.

Thanks again. You are making this very understandable. I will draw the load line and try to see if I can find the operating point.