# Which transistor should be used as a switch for a load of 5 V, 100 mA

I'm new to electronics. I want to use a transistor as a switch for a load of 5 V and around 100 mA. And what should be the base current or voltage to activate the transistor. Please help me.

• Almost any MOSFET you are able to find can do it. Do a search on DigiKey, Mouser or similar. I would recommend a minimum of 2x the voltage rating and 3x the current rating unless you have knowledge about how and why. May 29, 2021 at 17:21
• A guess that you're switching power on/off? That's often done with a high-side switch. @winny suggests a MOSfet...a high-side MOSfet switch would be a p-channel MOSfet to switch +5V DC power on/off. May 29, 2021 at 17:29
• What switching frequency are you interested in? For BJTs I think pretty much any "general purpose" transistor will work, eg, 2n3904, 2n2222, 2n4401, bc547 for NPNs or their PNP counterparts. For base current use the "1/10th forcing beta rule" described here: electronics.stackexchange.com/a/369591/95488 May 29, 2021 at 17:35
• any, and the current to activate should be enough to sature the base of your transisor. May 29, 2021 at 18:25
• I'd use whatever random transistor I had to hand, this is not a demanding task. May 29, 2021 at 18:45

You need to read the device datasheets to determine which transistors may be suitable. But what parameters do you look for and how should you interpret them?

There are two main things to consider:-

1. Voltage drop when the transistor is switched on.

If 5 V is going into the transistor 'switch' then less than 5 V will come out. You must decide how much voltage loss you can tolerate. This also determines the transistor's power dissipation when passing maximum load current, which affects how much heat sinking it may need.

For a bipolar transistor the parameter to look for is the Collector to Emitter 'Saturation Voltage', typically measured with Base current 10 times lower than Collector current (a smaller Base current may still work but the voltage drop will be higher). For a MOSFET it is RDSON, which you must multiply by Drain current to get the voltage drop.

1. For FETS only, the voltage required between Gate and Source to fully turn it on.

This is not the 'threshold voltage', which is the voltage where it is just on the threshold of turning on. Usually the minimum recommended Gate drive voltage is either stated in the device summary or implied in parameters that specify a particular Gate drive voltage.

You don't need to worry about turn-on voltage for bipolar transistors because it is always around 0.7 V to 0.9 V in saturation. This won't be a problem unless you are working with very low drive voltages.

The advantage of MOSFETs over bipolar transistors is that they don't need any current to keep them turned on. This makes switching high currents easier and improves circuit efficiency, especially if the typical load current is much less than its maximum value.

Notes:-

• For lowest voltage drop the transistor needs to be configured in 'Common Emitter' or 'Common Source' configuration. That means to switch the 'high' side of the power supply (which is generally required when the load shares a common ground with anything else) you should use a PNP bipolar transistor or P-channel MOSFET. If the driving circuit runs at a lower voltage then you may need a level shifter or 'high side driver' to raise the drive voltage up to the power supply rail where the transistor is situated, like this:-

simulate this circuit – Schematic created using CircuitLab

• Transistors have 'typical' and minimum/maximum specifications due to process variation. If you design a circuit based on 'typical' characteristics then it may or may not work depending the individual transistor you have. To guarantee that it works you must use the 'worst case' parameters, eg. minimum current gain, maximum saturation voltage, maximum Gate threshold voltage (but go with the recommended Gate drive voltage and you won't have to worry about this).

• Never run any semiconductor device at or near its 'Absolute Maximum' ratings. If a transistor's maximum Collector or Drain current rating is 100 mA then it is not good enough for your application. Always calculate power dissipation using the worst case saturation voltage or RDSON, and check that it won't get too hot with practical heat sinking - not the 'infinite' heat sink (TCASE = 25 °C) and extreme junction temperature specified in the datasheet. Better to use a transistor with much higher current rating than you need, both for reliability and to simplify calculations.

I'd use a BC337 for that which can easily handle the required 100mA collector current.

There are 3 variants of the BC337 : BC337-16, BC337-25 and BC337-40.

The 3 variants differ in their DC current gain (hFE).

Which ever variant you purchase, look up its data sheet specified minimum DC current gain (hFE) and use this value to calculate the required base current using the equation below:-

The multiplier of 5 is included to ensure that there is sufficient base current to fully saturate the transistor, some people use a higher safety factor of say 10 but I've always found 5 to be high enough.

Next calculate the base resistor.

With Vcontrol asserted the base of the transistor will be at about 0.7V and so:-