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enter image description here

Above on the left side is the original zener diode circuit.

I want to draw an approximation of this circuit, given the zener voltage Vz and zener resistance Rz (Rz is considered constant for simplicity). Vout should match the output of the original circuit.

Which figure (A or B) would be more correct?

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  • \$\begingroup\$ Which one models the change in output voltage with regulator input voltage? \$\endgroup\$ – Spehro Pefhany May 27 '16 at 12:11
  • \$\begingroup\$ What do you mean? \$\endgroup\$ – user16307 May 27 '16 at 12:46
  • \$\begingroup\$ I'm asking you to think about what happens when the input voltage changes. The output voltage should change to some degree or another, because the Zener diode is not perfect. It will be useful to you to be able to see this for yourself rather than just giving you the answer. \$\endgroup\$ – Spehro Pefhany May 27 '16 at 13:24
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Neither.

If Vcc falls below the zener voltage the battery will start to supply output current (which you will not get from a zener diode).

enter image description here

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  • \$\begingroup\$ Internal resistance can only be taken constant if Vcc is above breakdown voltage right? \$\endgroup\$ – user16307 May 27 '16 at 13:13
  • \$\begingroup\$ Yes and no. Its a reasonable approximation above the breakdown voltage but will vary with temperature (as any semiconductor will). For a simple model the answer would be a yes. The diode in this model can either be 'ideal' (no voltage drop, no forward resistance) or 'real' with a forward drop of about 0.6V and forward resistance which varies with current. Rz is simply a 'lumped' resistance that you have no direct access to. Below the breakdown voltage the 'zener' will appear as an open circuit. \$\endgroup\$ – JIm Dearden May 27 '16 at 13:24
  • \$\begingroup\$ Well, obviously, but in this case, it is also mising another diode in parallel to the whole thing, in the opposite direction. You can refine the approximation forever... \$\endgroup\$ – dim May 27 '16 at 14:03
  • \$\begingroup\$ @dim Quite correct. If Vcc goes negative then the zener would then turn into a forward biaseddiode. I shall add that to the diagram, thank you \$\endgroup\$ – JIm Dearden May 27 '16 at 14:42
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In schematic A, the Rz element increases the zener voltage when the current through the zener increases. Like a real zener, because their I/V relation in the breakdown range is never exactly vertical.

In schematic B, the Rz element serves no real purpose. And how would that be even physically possible ? A zener has two leads. In schematics B, you have three leads.

So, answer is A.

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  • \$\begingroup\$ I was a bit confused even though seems like a cheesy question \$\endgroup\$ – user16307 May 27 '16 at 12:13
  • \$\begingroup\$ I was trying to write an equation where I can see the change in Vout wrt change in Vin. How would you do ? thanks \$\endgroup\$ – user16307 May 27 '16 at 12:15
  • \$\begingroup\$ It is a cheesy question, indeed... Where does it come from, by the way ? It looks like homework, no ? I can't really say when I look at your profile... Now, regarding the equation, you need to involve Iout to write the relation between Vcc an Vout. \$\endgroup\$ – dim May 27 '16 at 12:20
  • \$\begingroup\$ alright not an homework \$\endgroup\$ – user16307 May 27 '16 at 12:27
  • \$\begingroup\$ Well, actually, I don't really care if it is homework or not. I'm not a teacher, and nobody can force a student to study. Anway, just write all the equations this circuit gives: Iin = Iout + Izener and Vout = Vcc - R*Iin = Vz + Rz * Izener. Then solve. \$\endgroup\$ – dim May 27 '16 at 12:33

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