1
\$\begingroup\$

The question I have is: how can I design a circuit so that I can determine the I-V curve of a diode.

I've found this one:

enter image description here

But is that a smart circuit or can I find a better one? Thanks for any advice.

\$\endgroup\$
5
  • \$\begingroup\$ It's a fine circuit for the general measuement, assuming you don't want to measure uA and 6 digits etc. \$\endgroup\$
    – Huisman
    Commented Jun 17, 2019 at 19:20
  • 1
    \$\begingroup\$ Don't ask "how can I design" because an answer would be: "By becoming a circuit designer" which is a "silly" answer. If you apply an AC voltage as Vs and use an oscilloscope in XY mode then you can plot the diode curve using this circuit. See w2aew's excellent video: youtube.com/watch?v=WWY-pakm_OM Professionals would use a semiconductor parameter analyzer: keysight.com/main/… but these are ludicrously expensive. \$\endgroup\$ Commented Jun 17, 2019 at 19:22
  • \$\begingroup\$ Hobbyists may want to google for semiconductor curve tracer kits or project articles. They're out there. \$\endgroup\$
    – TimWescott
    Commented Jun 17, 2019 at 19:25
  • 3
    \$\begingroup\$ What makes a particular circuit "smart" or "better" to you? The circuit you show meets the requirements you stated. \$\endgroup\$ Commented Jun 17, 2019 at 19:31
  • \$\begingroup\$ Over how many decades would you like to document the exponential-current or logarithmic-voltage behavior of the diode? In high school I tried to do this, using 1.5 volt battery, a number of resistors, and a 1,000 ohm/volt Air Force surplus volt/ohmmeter, and a struggling skill at error analysis because of the meter ----the lowest voltage range was 4 volts, the lowest current range was 4mA, and the shellac'd meter schematic was inside the case so I knew the series resistance during current measurements. You can do the same. \$\endgroup\$ Commented Jun 18, 2019 at 6:33

1 Answer 1

1
\$\begingroup\$

If you're not too concerned about the effects of self heating, then a circuit like the one shown in fig. 1 might be useful.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1.

Resistor R1 is the current sense resistor. If you know R1's resistance value (by measuring its resistance with an ohmmeter), and if you measure with a voltmeter the voltage drop across R1, then you know via Ohm's law how much current is flowing through R1 and through the diode (the "device under test" (DUT)).

Measure and record resistor R1's resistance value.

The desired current level is set via potentiometer R2.

Using Ohm's law, calculate the voltage drop across current sense resistor R1 that corresponds to the desired test current:

$$ V_{R1(TEST)} = I_{TEST} \times R1 $$

Connect your voltmeter across R1 (position VM1) and adjust R2 until the voltmeter indicates \$V_{R1(TEST)}\$ volts.

Disconnect your voltmeter from test position VM1 and reconnect it across the DUT (test position VM2). Measure and record the voltage across the DUT.

Repeat this procedure for additional test currents of interest.


A slightly more sophisticated test circuit is shown in fig. 2.

enter image description here

Figure 2.

Op amp U1A and its surrounding components comprise a transconductance amplifier circuit—i.e., voltage in, current out. The amplifier's input voltage range is 0V to 5V and output current range is ~0mA& to 100mA.

The desired input voltage value is set by potentiometer R2. Use a voltmeter connected between test point TP1 (+) and ground (-) to set the desired input voltage when adjusting R2. (Hint: If R2 is a 20-turn pot you can get good resolution when setting the input voltage.)

The transconductance amplifier circuit's output current is approximately

$$ I_{OUT} \approx \frac {V_{TP1}}{50\,\Omega} $$

The \$50\,\Omega\$ resistance is the equivalent resistance of the four parallel resistors R3 through R6. This \$50\,\Omega\$ resistance must dissipate about 0.5 Watts of power, so putting four \$200\,\Omega\$ 1\2 Watt resistors in parallel distributes this power dissipation across four resistors. The goal here is to reduce the level of self-heating in the \$50\,\Omega\$ load and thereby keep its resistance value more-or-less constant.

An op amp typically cannot source or sink much current (perhaps only a few milliamps); therefore, transistor Q1 is needed to amplify the op amp's output current to the desired test current levels (tens of milliamps or more).

To test the diode, start by choosing the desired test current value—e.g., 1 mA. Calculate the voltage at test point TP1 that corresponds to the chosen test current:

$$ V_{TP1} = I_{TEST} \times 50\,\Omega = (1\,mA)(50\,\Omega) = 50\,mV $$

Connect the voltmeter between test point TP1 (+) and ground (-) and adjust resistor R2 for an indication of 50 mV on the voltmeter.

Disconnect the voltmeter from test point TP1 and reconnect it across the diode at test points TP2 (+) and TP3 (-). Measure and record the voltage drop across the diode.

Repeat these steps for any additional test currents of interest.

For the schematic shown in fig. 2 I performed a DC sweep simulation that adjusts parameter POTSET from 0 to 1 (0% to 100%), which equates to adjusting potentiometer R2 from its minimum value (0 ohms) to its maximum value (5 kohms). Figure 3 shows the simulation results, plotted with the diode's forward current (y-axis) versus its forward voltage drop (x-axis).

enter image description here

Figure 3.

\$\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.