# How to amplify 3.3V to 10V by using an NPN transistor? How to select resistor based on current, how does the amplification happen?

I have basic knowledge about NPN transistors and I know how they work: for current gain based on the small input current in the base we are able to drive large amount of current in between collector and emitter.

In my circuit I need to amplify 3.3V of my stm32 micro controller input to 10 V output to drive a Mosfet switch.

I want to give 10V input to the gate of the MOSFET to get very low Rds. I am using STM32 micro controller for controlling. Its maximum output voltage is 3.3V but I need 10V so I planned to use an NPN transistor in common emitter configuration so that I can amplify 3.3V to 10V and then I can give that to the gate of the MOSFET.

I understood that the amount of voltage we give in input to the base we will get as large voltage between collector and emitter based on the resistors. I also understood the voltage gain based on Rc and RE for example if Rc = 1k and Re = 100 ohm the gain is 10. But I cant understand how to choose a resistor and how much current should I use how much current to amplify to amplify the voltage, and how much voltage to give to get 10V as output. How this entire thing works based on voltage amplification based on my circuit.

I have searched on google but I can't 100% understand it.

• How accurately do you need to make the 10 V? For example, would it be okay to get somewhere between 8 and 12 V? Between 9 and 11 V? Apr 2, 2020 at 16:14
• between 9 t 11 v would be nice Apr 2, 2020 at 16:25
• Also your diagram shows the FET in a high-side switching configuration. To make this switch well, you actually need the gate voltage to be ~10 V above the load's power supply voltage, not just 10 V above ground. In my answer, I assumed you can move the FET to a low-side switching position. Apr 2, 2020 at 16:34

As another answer says, you can't get it with your proposed design. You need a source of 10 V. But you can get that by adding a single additional resistor:

simulate this circuit – Schematic created using CircuitLab

Now when IN is high, you'll get about 0.2 V at the MOSFET gate. And when IN is low, you'll get about 10.4 V at the gate.

There is a trade-off here. You can adjust the values of R1, R2, and R3 up to reduce the quiescent power consumption of the circuit. Or you can adjust them down to reduce the transition time of the FET between its on and off states. The FET likely consumes much more power when transitioning between states than when fully on or fully off, so if the transition time is too long it could burn itself up.

• The input is 3.3 v DC .i cant understand Vin is high and low what does that mean csn you explain that. and can you kindly explain the circuit please. Apr 2, 2020 at 18:29
• @Muthu, if the input is always 3.3 V (and never 0 V) how will the output know when to change from one state to the other? Why bother connecting it to a uC (as shown in your diagram) and not just to a 3.3 V supply? Apr 2, 2020 at 18:47
• yeah brother which means micro controller PWM signal from 0.2 v to 10.2 volt .Is that what you mean Apr 2, 2020 at 18:58
• @Muthu, please edit your question to make it clear what kind of signal is present at the uC output. If it's a DC 3.3 V, say that. If it's PWM, say that. They are not the same thing, so now you have me confused about what your signal really is. Apr 2, 2020 at 19:01

Is this the sort of thing you are after?

simulate this circuit – Schematic created using CircuitLab

You can't. There's no source of 10 V that your transistor could switch.

So, get a different MOSFET. Also, don't forget that for MOSFET switching, the Gate-Source voltage needs to be defined – so there has to be some common notion of potential (e.g. a common ground); your schematic doesn't have that.

• If OP isn't too worried about power consumption, they can add another resistor from the FET gate to ground. Apr 2, 2020 at 16:20
• @ThePhoton I didn't want to recommend that: 10V Vgsthreshold MOSFETs are likely meant to switch something relatively lethal... Apr 2, 2020 at 16:23
• It is possible i think because already a person have done that . iam using 48 v as a source voltage from that can we do Apr 2, 2020 at 16:29