NPN Base Resistor when only switching ground? - Page 1
Forum section: electronics / beginners

author: mike519
author_rank: Contributor
timestamp: April 19, 2025, 06:03:22 pm
content: I am pretty sure I still need a base resistor for the NPN I will be using, even though all it will be doing is using a 2.5V trigger pulse on the base to switch a ground signal. I need a low level relay trigger but it needs to be fed by a 2.5V positive signal.How do I calculate the base resistor needed when there is no collector current? Or is there and I'm just not understanding?

author: Zero999
author_rank: Super Contributor
timestamp: April 19, 2025, 06:42:21 pm
content: Sorry, I don't know what you're talking about.How is the transistor configured? Is it an emitter follower? If it's common emitter, then it needs a base resistor, unless the base is being driven by a constant current source.Please post a schematic.

author: pqass
author_rank: Super Contributor
timestamp: April 19, 2025, 06:49:01 pm
content: See "NPN Relay Switch Circuit"herewhich is what I think you're describing.Think of the NPN base-emitter being forward-connected diode of about 0.7V voltage drop. Your drive voltage of 2.5V minus this drop of 0.7 equals 1.8V. When the NPN is fully on, the collector-emitter path will have a low voltage drop; connecting your relay coil do ground.What's the coil current rating of the relay? Lets say it's 50mA. And if you're using a small signal transistor like a 2N3904, the the HFEis around 100. That means that to switch a 50mA relay coil on you need 50mA/100HFE= 0.5mA base current at a minimum.How do you achieve at least 0.5mA base current using the 1.8V above? Ohms law... 1.8V/0.0005A=3600ohms.So you can make the base resistor 1K and you're good.Having more base current compensates for HFEdifferences (you can't count on it in a batch of transistors). If you don't provide enough base drive current then the collector-emitter voltage starts creeping up; behaves more like a resistor than a switch. This causes it to heat up and let out the magic smoke.QuoteThere is collector current when the NPN is on because you are connecting the load (relay coil) to ground (also known as 0V).

author: pcprogrammer
author_rank: Super Contributor
timestamp: April 19, 2025, 06:55:11 pm
content: When the NPN transistor has it's emitter connected to ground and the voltage to switch on the transistor is 2.5V, then yes it does need a resistor in series with input and the base of the transistor.To calculate it one gets the datasheet of the transistor chosen for the job and checks out the specifications of the transistor. What is needed is the hfe and the expected collector current when the transistor is turned on. The needed base current is obtained by dividing the collector current by the hfe. Now that you have the base current you can calculate the needed resistor. For this take the 2.5V and subtract the needed base emitter junction voltage, which most likely will be about 0.7V. (Again, check the datasheet for this) Divide the calculated voltage by the base current and you have yourself a resistor value that is needed to do the job.

author: TimFox
author_rank: Super Contributor
timestamp: April 19, 2025, 06:59:01 pm
content: If you connect a "stiff" voltage source between base and emitter, with no series base resistor, and zero (or very low) collector current, you are driving a PN diode from a voltage.This results in a base-emitter current that is a very strong (exponential) function ofVbe, limited only by the capability of the voltage source.With +2.5 V, the resulting base current would be huge and probably burn out the diode.With only +0.3 V, the base current would be very small, and nothing would happen.

author: golden_labels
author_rank: Super Contributor
timestamp: April 19, 2025, 07:18:49 pm
content: Note that the diagram pqass linked has a 10 kΩ resistor in parallel with the B-E junction. If you’re just starting to learn about transistors, you may skip it. You may ignore the rest of this post too.The purpose of this resistor is give a path for charges to bleed from the transistor. This makes transistor turn off faster. You may not need it at all in your circuit. In particular if the input is actively driven low. A secondary function is it acting as a pull-down resistor to ensure that with no input signal the transistor stays off. Again, it’s up to you to decide the need for that.If that resistor is present in your circuit, it forms a voltage divider with the base resistor. The B-E junction can’t get voltage higher than the output of this voltage divider. Usually this isn’t a problem. But given you work with 2.5 V only and it is a pulse,(1)the final voltage may be not enough to fully turn on the transistor, if the base resistor is too high. Have that in mind.(1)With very short pulses it may happen that the signal hardly reaches that 2.5 V before the pulse ends.