2
\$\begingroup\$

I have this circuit:

enter image description here

I am giving an input voltage from 0V to 40V shown at the input of the circuit.

When I give 1V at the input, I get the voltage values as such shown below (measured with multimeter and oscilloscope):

enter image description here

0.18V is dropped across D0002 and 0.8V is dropped across the 36V zener diode and the remaining 20mV is dropped across the remaining 2 resistors.

I tried the above circuit in simulator tool as well. I am getting similar results. The current through the components is only the reverse leakage current of the zener diode which is in the range of 10s of microamps.

My Questions :

  1. My understanding was that when I apply 1V at the input of the circuit, 0.7V would be dropped across D0002 and the remaining 0.3V will be dropped across the other components because diode requires a minimum of 0.7V to conduct. So, when voltage of above 0.7V is applied at the input, the diode conducts and 0.7V will be dropped across the D0002 diode right?

But in my above scenario, why is the voltage across the diode D00002 only 0.18V and D00003 is 0.8V?

  1. This also brings the question - What does it mean when one says, the diode is forward biased? Once forward biased, the diode should have 0.6V-0.7V voltage drop across it, right? Or what is it that I am misunderstanding?

Please help with the above questions.

\$\endgroup\$
2
  • \$\begingroup\$ The current is probably too small; see Why is the diode forward voltage constant? \$\endgroup\$
    – CL.
    Oct 6, 2020 at 11:08
  • 1
    \$\begingroup\$ In addition to what Andy aka has said... from a practical perspective, I only tend to bother with the behavior of circuits within their reasonable operating range :-) Supplying 1 Volt to a 36V zener in series with an Si diode, that to me is in the zone of pure parasitic behavior, and none that I care about... unless I'd have that circuit affect the base of a transistor in a threshold detector / linear regulator of some sort, and I'd expect the circuit to be perfectly dead in this range ;-) In that case, it's a matter of the impedance or pullup that I compare this "dead" circuit to. \$\endgroup\$
    – frr
    Oct 7, 2020 at 2:08

1 Answer 1

Reset to default
6
\$\begingroup\$

because diode requires a minimum of 0.7V to conduct

Not true. diodes conduct current all the way down to virtually zero volts applied across them: -

enter image description here

Picture from this very good article

I think that should explain all your questions.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.