Does the fact that both show ~ 0.6 v-drop rule out zeners?

Thanks.

"DaveC" wrote in message
...
Does the fact that both show ~ 0.6 v-drop rule out zeners?

Thanks.


No, only that they are most likely silicon. The zener effect is the reverse
breakdown
voltage. You need to dope out the circuit more to see if it could be a
zener.

DaveC wrote in
:

Does the fact that both show ~ 0.6 v-drop rule out zeners?

Thanks.



You sound like a man who imagines that by scrunchign his shoulders, no-one
will mind when he passes annopyingly several times through the same doorway.


And no. To test a zener you need to try a reverse voltage. Try a variable
voltage through a 10K resistor. A sine wave off a 30V transformer will do,
just look for the clipping voltages on a scope (or DC voltmeter if you add a
capacitor in parallel with the diode). Vz equals peak-peak voltage minus
about 0.58V. You may also have to remove one end of it from the circuit to be
certain.

Lostgallifreyan wrote in
:

Vz equals peak-peak voltage minus
about 0.58V.

Evidently my turn to pass annoying through the same doorway...

Scratch that, the rest stands though, you just have to do the test once for
each polarity of that diode. (You'll only see both peaks at once if you have
a zero-referenced AC voltage, and even then that capacitor idea wouldn't
apply, and I made it more complex than it needs to be, you only need to see
the zener peak to know...).