# How to measure the inverse saturation current of a diode?

I'm trying to measure the inverse saturation current of a diode, for using it in the Schockley equation. I would like to know what is the best circuit for doing this. I already tried the following circuit, but I can't detect anything.

simulate this circuit – Schematic created using CircuitLab

I'm using a 0-30V direct current power supply, a resistance of 230 ohms and a IN4001 diode. Do you have any suggestions?

• What ammeter are you using? And what approximately do you expect the I_s to be? Sep 12, 2017 at 1:08
• I'm using a steren mul-600 RS232C. I expect a I_s in the order of 10^-6 amperes. Sep 12, 2017 at 1:21
• If you actually need to measure the 65 pA, you will need a pretty fancy meter, like the Keithley 6400 series. Alternately, you can try sweeping the I-V curve and fitting the $n$ and $I_s$ parameters...but you might not get very accurate results if you don't have accurate measurements of the reverse currents. Sep 12, 2017 at 1:27
• Or replace R1 with 10 - 100 megohms and measure the voltage across it (possibly requiring an electrometer rather than a typical DMM). Sep 12, 2017 at 1:33
• make a plot of log(If) vs Vf of diode, extrapolate for Is Sep 12, 2017 at 3:34

The last time I made some diode leakage measurements, I used a cheap (distributor's own brand, <£10) DMM, with 1999 full count, 10M input impedance, and a 200mV range, as the current meter.

On that range, the full scale is 200mV/10M = 20nA, with the nominal current resolution 100uV/10M = 10pA.

You can use an external shunt to get a higher current range for big diodes, use an external 1.1M resistor for 200nA full scale, and 100k for 2uA.

It's worth using a variable power supply, and inching it up from zero, so as not to embarrass your meter's 200mV range with a leaky or failed diode. I found (and the Schockly equation will tell you) that once above a couple of volts, the reverse leakage current is more or less constant.

FWIW, I measured about 35nA for a BAT42 (schottky), 4nA for a 1N4148, and failed to measure (so <10pA) the current for a BAS116 which is advertised as a 'low leakage' diode.

If you want to measure lower currents, then you need to build a pico-ammeter round a low bias op-amp, next-hack's answer shows you how.

Two cautions when working like this. (1) You might want to verify that your meter is actually 10M input impedance on the 200mV range, put an external 10M in series with it, and check that halves the reading when supplied with a voltage. (2) When using non-native ranges like this, the decimal point will usually be in the wrong place. I am skilled at moving it the wrong way in my head, so I simply enter the voltage I read into a spreadsheet, and the relevant resistance in another column, and let it do the sums.

A "simple" way to measure small currents, is by using a low input bias op-amp.

The output voltage is proportional to the current. To get a 1V/nA you should still use a 1GOhm resistor. With a 10MOhm resistor, you have only 10mV/nA. However any 3.5 digit DMM has the 200mV scale, i.e. with 100uV resolution. I.e. 10pA resolution @10MOhm, 100fA @1GOhm.

You must apply externally "Vtest", while the other terminal will be forced to ground, due to the negative feedback through R1.

Notes:

• The circuit is dual supply (not shown). That particular OP amp has a Vcc-Vee max of only 15V. Still I expect that the reverse current does not vary a lot with the reverse voltage.
• You might need a suitable compensation capacitor in parallel to R1.
• Also the decoupling capacitors are not shown.
• Be aware that, since the LMC6482 has 20fA of leakage current, the overall leakage will be likely determined by the rest of the system. If possible use the dead-bug style mounting.

simulate this circuit – Schematic created using CircuitLab

To be able to find the diode saturation current (scale current) you need to take two measurements in the forward biased region.

simulate this circuit – Schematic created using CircuitLab

Notice that $V_{F2} > V_{F1}$ and $I_{D2} > I_{D1}$

And the diode equation:

$$\Large I_D = I_S \left(e^{\frac{V_F}{n*V_T}} -1\right)$$

And from the Shockley diode equation we have:

$$\Large nV_T = \frac{V_{F2} - V_{F1}}{\ln\frac{I_{D2}}{I_{D1}}}$$

And finally the $I_S$ current:

$$\Large Is = I_{D2}*e^{\left (\frac{-V_{F2}}{nV_T}\right)}$$

And we done.