1
\$\begingroup\$

I have built the "circuit" illustrated below. Basically its a 5V supply with positive terminal connected to a schottky diode's cathode, multimeter in voltage measurement mode across the same diode's anode and supply's negative terminal.

enter image description here

My first instinct would be that the multimeter would read a floating value, but after trying its the supply voltage. Simulation also indicates its 5V.

Is it because the PN junction is still in equilibrium, and P and N halves are part of the same "conductor"?

\$\endgroup\$
1
  • 1
    \$\begingroup\$ The Shockley diode equation says that \$V_{_\text{D}}= V_T\cdot \eta\cdot \ln\left(1+\frac{I_{_\text{D}}}{I_{_\text{SAT}}}\right)\$. Since \$I_{_\text{D}}=0\$ it follows that the voltage drop across the diode is also zero volts. That said, if the voltmeter draws a tiny current (and it will) then there will be a very slight drop. A simulator's voltmeter doesn't draw current, though. So it's perfect. \$\endgroup\$ Commented Oct 15, 2023 at 6:18

1 Answer 1

1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Here I have simulated a real Schottky diode (1N5819) and a realistic digital voltmeter (10MΩ input resistance).

Your simulation may also be accurate for the models you are using, however your voltmeter model (at least) is suspect.

Schottky diodes can leak relatively high currents, especially at high temperatures so you should be surprised to see the above 390nA leakage with 1.1V across the diode (reverse bias). Review the Shockley diode equation and a typical diode model (below is Diodes Inc. 1N5819 model, where the saturation current is of particular interest here):

*SRC=1N5819;DI_1N5819;Diodes;Si;  40.0V  1.00A  3.00us   Diodes Inc. Schottky Barrier Rectifier
.MODEL DI_1N5819 D  ( IS=390n RS=0.115 BV=40.0 IBV=1.00m
+ CJO=203p  M=0.333 N=1.70 TT=4.32u )

Spice uses a somewhat more complex set of equations than the Shockley equation to model a diode (there are three regions of operation), but the latter should be adequate for now. enter image description here

\$\endgroup\$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.