simulate this circuit – Schematic created using CircuitLab

I currently have a circuit with a 2N3904 BJT that acts as an on/off switch. I want to add a second BJT in parallel that toggles an LED on/off to indicate the on/off status of the first BJT.

I've tried simply connecting the bases of the two transistors together, but this has not worked, I think due to insufficient base current now that the current i is being split between 2 transistors.

If I halve the biasing resistor values to double the amount of current, but maintain the R1/R2 ratio, should this theoretically work?


Edit: Apologies for not adding a schematic, I wasn't aware there was a built in feature. This is the schematic I tried, which obviously did not work. Upon reading the comments, it seems that adding a resistor in series with the base of each transistor could help solve the biasing issue?

  • 6
    \$\begingroup\$ Hit the edit link under your question and then the CircuitLab button on the editor toolbar and draw a quick schematic of what you've tried. Double-click to set or delete component properties. \$\endgroup\$ – Transistor Dec 26 '18 at 14:48
  • \$\begingroup\$ Can you use a MOSFET instead of a BJT? \$\endgroup\$ – evildemonic Dec 26 '18 at 15:46
  • 1
    \$\begingroup\$ Due to no schematic provided, I am voting to close this question as "unclear what you're asking". \$\endgroup\$ – The Photon Dec 26 '18 at 17:02
  • \$\begingroup\$ There are some details missing: what is the dc value of the input sine wave, and what is the amplitude of this sine wave? \$\endgroup\$ – Lazarus Dec 26 '18 at 20:10
  • \$\begingroup\$ You could also consider using the collector of your NPN switch, which will be value close to ground when ON, to pull down on the base of a PNP (via a resistor) to turn on the PNP. Hang an LED from the PNP collector (with current limiting resistor to ground.) This directly measures your switch function rather than reaching back to the controlling signal, so it has its advantages that way. \$\endgroup\$ – jonk Dec 26 '18 at 21:20

You cannot simply connect the bases of two NPN transistors like you describe if the emmiters of both the transistors are connected to GND. Instead you have two alternative choices:

  1. Add a resistor in series with each transistor from the bias source. This will allow each transistor to operate independently without interaction problems with the differing VBE of the two transistors.
  2. Add a resistor on each transistor from the emitter to GND. This will introduce some voltage drop across each resistor that will help to offset the VBEdifferences. This is a less optimum solution because it will limit the total IC and may lead to less optimum drive of the two loads.

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.