This question already has an answer here:


simulate this circuit – Schematic created using CircuitLab

Hi, usual apologies here, total enthusiastic amateur noob, etc.

The situation is: a Common Collector NPN transistor.

Question: Is it okay to omit the resistor R2? (that is, R2 = 0 Ohms)?

The way I understand it, when the switch is closed, the voltage at the Emitter (as indicated) would be 5V – 0.7V = 4.3V. The reason for that is that both the Collector and the Base are connected directly to 5V, and that VBE must be 0.7 V for silicon-based transistors. Therefore the voltage drop from the Collector to the Emitter will 0.7 V rather than the normal 0.05 to 0.10 V.

But if that is the case, if I want to calculate the value of R2, the total voltage drop is the same, irrespective of whether I calculate it through the loop containing the switch, or going through the transistor form the Collector to the Emitter. Therefore IR = 0 for the Base resistor so that R2 must be 0 Ohms.

Elsewhere I came across the injunction: “Do NOT omit the Base Resistor!” so now I’m uncertain.

(I’m really trying to design a circuit to drive an PWM input to an H-bridge from an Intel Edison, so the above is not my complete circuit – it just crystallises one of the many current conceptual difficulties I have. This particular part has to do with an AND gate, so there will at least be another transistor and a pull-down resistor, and so on.)


marked as duplicate by CL., pipe, Community Oct 14 '16 at 15:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ You can find your answer here (from a similar question): electronics.stackexchange.com/a/134064/103420 \$\endgroup\$ – Rohat Kılıç Oct 14 '16 at 14:20
  • \$\begingroup\$ You're welcome. But you'd be quicker if you made a search ;) \$\endgroup\$ – Rohat Kılıç Oct 14 '16 at 14:30
  • \$\begingroup\$ I did, I did. But I am not yet at the stage where I know what to search for. "Do I need a base resistor" is still a bit generic for me. \$\endgroup\$ – Pieter Beneke Oct 14 '16 at 14:33
  • \$\begingroup\$ @RohatKılıç that is for linear load R's in series.... \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 14 '16 at 14:37
  • \$\begingroup\$ @pipe not same load \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 14 '16 at 14:39

Yes you can drive it direct in most DC cases, but without a small Series R Emitter followers can become oscillatory with capacitive loads and step pulses.

The value needs to degenerate positive feedback and can be as small as the output impedance, but usually more. Series R's are often placed between bridge drivers to MOSFET Gates to tradeoff between instability and slew rate.

I don't have a short answer, but I references. The Rs loop including (Rb/hFE + Re ) must have a Q<1 thus total gain <1 with positive feedback with ESL and ESR of load, otherwise you have an emitter Follower "Hartley Oscillator". When you add a series R, you reduce the Q of parasitic ESL and load Ciss, it is ok to reduce Rb to 0.


  • \$\begingroup\$ I’ve built 2 rudimentary circuits in my entire life (as in the past week), so I’m still trying to do everything with NPN transistors – I’ve not progressed to PNP transistors yet. But picking up from a reply mentioned by Rohat Kılıç above, it occurs to me that maybe I should do this with PNP transistors instead: I won’t get the voltage drop, and I would HAVE to use base resistors anyway, so maybe that would address the oscillating issues you mention as well? Would that be a better design? \$\endgroup\$ – Pieter Beneke Oct 14 '16 at 15:20
  • \$\begingroup\$ insufficient details on layout and components. best to follow expert designs that work , if you can find. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 14 '16 at 15:29

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