I want to build a simple V-I tester and based on this article (or this) I did this easy circuit:

enter image description here

R1 = 560Ω
R2 = 100Ω
R3 = 1kΩ

where R1 and R2 is a voltage divider to set the \$V_{(DUT+V_{R_3})}=1VAC\$ and R3 limit the current in the Device Under Test.

I used this transformer, with 6.3V, 10 VA output (is dual, but I use only one output).

My question:

  1. If I change the voltage divider resistors to \$R_1=56Ω\$ and \$R_2= 10Ω\$ (especially), isn't better? With 10Ω I will reduce the output impedance of the circuit that feed the DUT, thus I don't load the source because it is in parallel to the \$R_3+R_{DUT}\ge1000Ω\$ and then \$R_3+R_{DUT}\gg R_1\$, isn't?
    BTW the current I draw is \$I =\frac{6.3}{66} \approx 100mA\$ and the output power is at the "safe" value of: \$6.3V\cdot100mA = 0.63VA\$
  2. With this voltage divider I have a voltage less than 1VAC across the DUT. The article says that I can test also Diode, Zener, etc. How is possible with this small voltage?
  • \$\begingroup\$ Just to say that R3 "limit" the current through device, but is also the "measure" sensor of this current. Point 2: diodes need only 0.7V ... current is always "exp", and the scale doesn't matter. You can't determine the Zener point if the voltage is too low. \$\endgroup\$
    – Antonio51
    Apr 21, 2022 at 17:15
  • \$\begingroup\$ Point 1: changing values is not really useful, because R3 limits the current. NB: adding a battery, one pot, and one resistor to control current, should also be used for testing and tracing curves of quasi any type of "transistor" ... \$\endgroup\$
    – Antonio51
    Apr 22, 2022 at 7:09
  • 2
    \$\begingroup\$ @Circuitfantasist - Divider R1- "R2 variable" only to limit the excursion of max voltage on DUT (some kind of current protection). For low current, it is enough and easy through R2. For high currents, a variable transformer (or a "fast" triangular voltage pulse with a low duty cycle) should be needed to adjust the voltage applied on DUT (as on a professional Curves-tracer), but with a scheme of "power" protection (so the calculation of v*i on the DUT is necessary to remain under Pmax) ... \$\endgroup\$
    – Antonio51
    Apr 22, 2022 at 18:19
  • 1
    \$\begingroup\$ @Circuitfantasist It is a very good exercise in a lab and helps to understand some drawing functions (mathematical with scope) and electronic characteristics of DUT. This simple circuit is very useful. I used this simple curve tracer about 55 years ago with my "own scope made" ... and a professional tool Philips (radiomuseum.org/r/philips_transistor_curve_tracer_pm_6507.html) 5 years later at university. \$\endgroup\$
    – Antonio51
    Apr 22, 2022 at 19:56
  • 1
    \$\begingroup\$ Right. I forget that the axes for the "tunnel" curve are reversed! I used only the load line for the curves of UJT (V vs i) ... :-( \$\endgroup\$
    – Antonio51
    Apr 23, 2022 at 8:46

1 Answer 1


Here is what you can do, with a little effort ...
This is for BJT-type NPN. Here, R4 is stepped 100 k -> 1000 k.
See the "visual" change of "beta" ...

enter image description here

@Circuit fantasist ... and all those interested :-) ...
Here is a simulation to explain the choice of "voltage" and "resistor" to see the full curve of a DUT with "negative" resistance (NB: Voltage vs current, as with UJT device) ...
NB: the part of the "unstable zone" can't be seen if the 2 conditions (voltage, resistance) of "loading" are not met.

enter image description here


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