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vermin

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## Homework Statement

The diagram:

http://images.fr1ckfr4ck.fastmail.fm/probB.jpg

These zener diodes;

- Have a test current of 5mA when 6.8V is applied (V_z=6.8V, I_z=5mA)
- Have an incremental resistance (r_z) of 20 ohms
- Knee current (I_zk) is 0.2mA

- Calculate the output voltage of the voltage regulator, assuming no load (R_L = infinite) and V
^{+}is at nominal value. - Calculate the line regulation of the voltage regulator, assuming no load.
- Calculate load regulation of the voltage regulator.

## Homework Equations

see below

## The Attempt at a Solution

To get the output voltage at nominal voltage and no load, I first went to solve for V_zo using V_z=V_zo+r_z*I_z and the test current I_z=5mA. That told me V_zo = 6.7V. Then I solved for I_z using I_z=(V+ - V_zo)/(R+r_z). I decided to go ahead and find I_z for one diode, then solve for half of V_o later. I_z turned out to be, for one diode, I_z = (15 - 6.7)/(500+20) = 16mA. So then to solve for output voltage V_o, I used

(1/2)V_o=V_zo+r_z*I_z

I_z being not the test current but the one I just solved for. So;

(1/2)V_o=6.7+20(.016)

V_o=14.04 V

It makes sense to me that the voltage should be ~1V less under a 500 ohm resistor, so I think that's a good answer.

As for line regulation - I'm looking for the change in V_o resulting from the +/- 1 volt change in V

^{+}(14 to 16 V).

If there were only one zener diode it would be as simple as ΔV_o=ΔV

^{+}*(r_z/(R+r_z))

But since there's two in series, I have to decide what to double and when.

Since that equation is asking what the change in V_o is when you consider the change contributed by the zener diode, I decided that the equation should be done twice and involve both diodes. It looks like voltage division to me, so if there's three resistors it should involve all three.

So, I changed r_z to 2*r_z:

ΔV_o=ΔV

^{+}*(r_z/(R+2*r_z)) + ΔV

^{+}*(r_z/(R+2*r_z))

Since ΔV

^{+}=2V, it all comes out to ΔV_o=2(20/(500+40))+2(20/(500+40))=8/54=0.148

expressed in mV/V --> 148 mV/V

and since it's a deviation they want I'd say the answer is (+/-) 74 mV/V.

Load Regulation:

Looking for the ratio of the difference in V_o vs. the difference in I_L, the current through the load.

A load resistance that draws a current of 1mA will cause the current through the zener diodes branch to decrease by 1mA. Normally if there's one diode that change would be;

ΔV_o=r_z*ΔI_z

But in this case there's two. Still, though - since the total current change will remain 1mA, it should be divided evenly between both diodes, and the end result stays the same. The load regulation should be proportional to r_z; -20 mV/mA

I have a sinking feeling I messed this all up. Any tips on where I went wrong or what I did right, would help. Thanks in advance.

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