What is the normal convention(safe and reliable), should one connect NPN transistor base to ground or should a resistor be used?

My logic tells me that in common emitter mode, shorting base to ground turns off transistor so no current flows between emitter and collector and also base current must be zero. If current is zero there is no need for resistor?

At the same time I think there should be some minor base capacitance that will discharge when base is shorted and thus could exceed rated base current? (However I cant see base capacitance mentioned in BC817 datasheet)


simulate this circuit – Schematic created using CircuitLab

  • 2
    \$\begingroup\$ some minor base capacitance that will discharge when base is shorted and thus could exceed rated base current? Indeed there will a small current peak, but these capacitances are generally small enough not to contain any amount of energy that can do actual harm. Also these are depletion layer capacitances formed by the PN junctions, these always have some series resistance that limits the current. I dare to say that is impossible to damage a transistor with your circuit on the left. \$\endgroup\$ Commented Feb 1, 2021 at 15:26
  • \$\begingroup\$ Either one should work fine. If ESD at the button is a possibility, maybe it would be nice to have R1 there to provide a tiny bit of protection for the transistor. \$\endgroup\$
    – user57037
    Commented Feb 1, 2021 at 19:09

2 Answers 2


The base of a CE (common emitter) NPN transistor circuit can be safely shorted to ground and the effect will be to turn the current in the collector OFF.

Some designers prefer to use a resistor, often because the base is also driven to a + voltage level and they want to limit the base current. But in your circuits above, you can leave the resistor in the path to ground out if desired.


Basically, the second circuit is more reliable because the base-emitter junction has low resistance and it is controlled by low voltage (several hundred millivolts). So, if the switch is connected directly in parallel, it must have a much lower resistance and a low voltage drop when closed. This is easily done for ordinary electromechanical switches but if it is a transistor switch or other device (e.g., a photoresistor) there can be problems.

The combination of the switch in parallel to the base-emitter junction can be considered as a current divider where the base current must be diverted through the switch.

Splitting the base resistor into two resistors helps to solve this problem because the second resistor increases the input resistance and the voltage threshold of the transistor.

These considerations applied more to transistors in the past which were of lower quality.


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