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I've built a small amplifier circuit in which I've wanted to use a Zener diode as a reference but I can't figure out how it works.

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Knowing that the negative rail is -5V and the Zener breakdown voltage is 5.1V, I've done the static operating point for this current mirror, but the current that I've calculated differs from the one that Orcad did. I've tried to rotate the diode upside down, but I've got the same current as in the other way. I really can't figure it out why.

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    \$\begingroup\$ 5 V is not enough to turn on a 5.1 V zener significantly, hence the 185 pA through it. You need 1 mA or so for a zener to be functioning in specification. Raise the rail voltage, or reduce the zener voltage. Orcad and you might use different assumptions for when the applied voltage is below the expected zener voltage, hence your different results. \$\endgroup\$
    – Neil_UK
    Nov 9, 2021 at 18:10
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    \$\begingroup\$ It does not make much sense to me why you do the reference with left transistor (connected as diode with CB shorted) and zener also. \$\endgroup\$
    – user208862
    Nov 9, 2021 at 18:18
  • \$\begingroup\$ I've incrised the rail to -6, -7 but still dosen't work \$\endgroup\$
    – PowerTb321
    Nov 9, 2021 at 18:28
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    \$\begingroup\$ The Zener diode needs 5mA of current to work and the supply voltage larger than 5.V. So for example if Vsupp = -6V then R18 = (6V - 5.1V)/5mA = 180 ohms \$\endgroup\$
    – G36
    Nov 9, 2021 at 19:23
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    \$\begingroup\$ As long as your zener is essentially shorted by the left and right branches in your circuit, this circuit will never function as intended. \$\endgroup\$
    – tobalt
    Nov 10, 2021 at 8:31

1 Answer 1

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I see you want to make a current mirror. First things first.

To simulate this circuit, and to learn how the Zener works, you need a "complete" circuit. Don't leave the collector of Q6 or Q7 disconnected. Use a resistor of 40 ohms, then connected to GND. Similarly,on the collector of Q7, connect 40 ohms to GND=0.0 Vdc.

Just for a moment, keep your -5V negative rail, but choose a 4V Zener. You need to make sure the Zener "turns on". Keep your Zener = 5.1 volts, but for the simulation , use -8.1V rail voltage.

Suppose the Zener datasheet says the Zener needs 3mA to operate nominally; then the R18=10K resistor has to be 1K.

Now you are ready for the overview. The voltage from 0.0Vdc (across R18) to the -8.1 rail is 8.1 volts. 5.1 volts are across the Zener. The voltage at the node of R18, Q6base and Q7base is -3.0 volts. The voltage at the Q7 emitter is -3.7V, voltage at the Q6 emitter is -3.7V. The current thru R9 is (8.1 - 3.7) / 100 = 44mA, and Q7 base current is 44mA/100 = 0.44mA. Q7 has to operate linearly. So the voltage across Q7collEmmiter = 2V. Voltage at Q7 collector is -1.7V. The resistor from Q7 collector to 0.0V has to be 1.7/44mA = 38.6 ohms = 39 ohms, that is why I chose 40ohms when I started the discussion.

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