1
\$\begingroup\$

i want to test my output current of transistor 2N3906. I have two questions.

  1. Is my setup valid to measure the max load of the (2N3906) transistor? One plug of multimeter is on leftmost pin of PNP transistor, the other to rightmost ping of 47R (R2). I count only 13mA when transistor works and 6mA (!) when it is NOT working (extra question: why?). I expected more than 25mA during active transistor since I already operate a device at 25-35mA!

load resistor setup

  1. So the transistor is capable of 13mA only? (base with 9mA aka: 3.3V > 330R)

  2. Sorry, third question. According to spec (Hfe Ic=10mAdc Vce=1V =>> 100), I was expecting 100mA with 9mA to base.

Thank you, and excuse basic questions :)

UPDATE: it should be obvious I have installed transistor the wrong way (Collector on emitter - Thanks Passerby!)

\$\endgroup\$
4
  • \$\begingroup\$ Base isn't at 9mA. You forgot to add the base forward voltage to the equation. Base current is (V source - V be ) / Ohms. Based on the datasheet its between 0.65 and 0.95 volts. 3.3 - .95 / 330 ~= 0.007 A. Also is your gpio at 3.3V exactly or has it dropped down due to load? Wire this back up without the ammeter in the circuit and measure the voltage at each node, base, emitter, collector and both sides of each resistor and compare it to what you think you should see. \$\endgroup\$
    – Passerby
    Aug 9 '20 at 18:09
  • 1
    \$\begingroup\$ That said you shouldn't see any current when off and even at 7 mA base should see more when on. You may want to double check you have no shorts or leakage current. Take the microcontroller out of the equation and test the transistor just by itself with the resistors to confirm it is working right. Did you wire it up backwards? Collector and emitter on your specific transistor? \$\endgroup\$
    – Passerby
    Aug 9 '20 at 18:13
  • 1
    \$\begingroup\$ Sorry my bad. Went to measure and I inserted wrong way the transistor! Normally EBC, but I connected in reverse CBE! I am evaluating again the situation. \$\endgroup\$
    – krg
    Aug 9 '20 at 18:16
  • \$\begingroup\$ No need to be sorry. We all start out this way. \$\endgroup\$
    – Passerby
    Aug 9 '20 at 18:24
2
\$\begingroup\$

Two issues:

  1. Base isn't at 9mA. You forgot to add the base forward voltage to the equation. Base current is (V source - V be ) / Ohms. Based on the datasheet its between 0.65 and 0.95 volts. 3.3 - .95 / 330 ~= 0.007 A. Also is your gpio at 3.3V exactly or has it dropped down due to load? A typical microcontroller output will sag as the current increases. Wire this back up without the ammeter in the circuit and measure the voltage at each node, base, emitter, collector and both sides of each resistor and compare it to what you think you should see.

  2. That said you shouldn't see any current when off and even at 7 mA base should see more when on. You may want to double check you have no shorts or leakage current. Take the microcontroller out of the equation and test the transistor just by itself with the resistors to confirm it is working right. Did you wire it up backwards? Collector and emitter on your specific transistor may not be in the same configuration. Always compare yours to the exact datasheet by its manufacturer and not other with the same part number.

\$\endgroup\$
2
  • \$\begingroup\$ thanks. I had installed transistor the wrong way (collector on emitter). With correct connection 38.8mA on emitter. You have right: 8.35mA on base. My MCU can go up to 10mA so I changed the base resistor with 270R so now base current 9.60mA and Emitter 49mA with 10R resistor load. Is it ok with you if I change the question or better to make new? Now (with correct setup) my question is: I count 49mA on Emitter with 10R (load resistor) and 220R on base. I want 50mA on Emitter. Ideally 150mA to handle spikes. Should I go with p-channel MOSFET? Or a capacitor? If capacitor which? \$\endgroup\$
    – krg
    Aug 10 '20 at 8:11
  • \$\begingroup\$ @krg that would be a new question. \$\endgroup\$
    – Passerby
    Aug 10 '20 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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