# Can't understand how the Zener voltage reference works

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. 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.

• 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. Nov 9, 2021 at 18:10
• 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.
– user208862
Nov 9, 2021 at 18:18
• I've incrised the rail to -6, -7 but still dosen't work Nov 9, 2021 at 18:28
• 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
– G36
Nov 9, 2021 at 19:23
• As long as your zener is essentially shorted by the left and right branches in your circuit, this circuit will never function as intended. Nov 10, 2021 at 8:31