Adafruit Trinket 5v https://learn.adafruit.com/introducing-trinket/pinouts

I'm really sorry to bother you all with my ignorance but I completely suck at math and couldn't figure out the equation for Ohm's law. I'm at a crossroads and don't know where to turn. This Adafruit Trinket will only produce 20 milliamps from GPIO pins. I need to add a transistor to one of the pins I've programmed to push 20mA to 7.5v.

Per Adafruit's description:

All the GPIO pins can be used as digital inputs, digital outputs, for LEDs, buttons and switches etc. They can provide up to 20mA of current. Don't connect a motor or other high-power component directly to the pins! Instead, use a transistor to power the DC motor on/off

I'm not trying power a motor. I'm trying to power a self-made cup warmer made from kanthol wire and polymer clay. Can anybody advise what transistor will boost 20 milliamps to 7.5v? Your assistance will be greatly appreciated.

Thanks Again!

Image of Adafruit 5V Trinket

  • \$\begingroup\$ Need more information. Is the cup warmer powered from a 7.5V battery? How much current/power will it need. Just to point you in the right direction. A power MOSFET is a good choice in these situations. Even better a 5V relay will do the job as well. \$\endgroup\$
    – user183368
    Commented Mar 26, 2018 at 5:56
  • 1
    \$\begingroup\$ Kevin, you haven't been responding. If possible, you should check in on your questions frequently for at least a half hour after posting and respond per questions seen. Revisit periodically as time allows afterwards. It's your question, of course. For example, how will you control the resulting temperature or will you instead simply supply a certain wattage (unstated) and run it open-loop (no temperature feedback to control it?) \$\endgroup\$
    – jonk
    Commented Mar 26, 2018 at 18:08
  • \$\begingroup\$ Sorry for the delay and thank you for the response back jonk 20,. I will check for responses more frequently. The Chip is going to be powered by USB 5 volt. The chip in his nature is a 5 volt chip. Any current that comes into it is automatically converted down to 5 volts. Thank you user 183368, because I was looking into the MOSFET transistor. If iuse that transistor, I know have to figure out what kind of resistor I need to bring the current down to a steady 7.5 volts. \$\endgroup\$ Commented Mar 27, 2018 at 14:28
  • 1
    \$\begingroup\$ @KevinBoston The power being produced by your heater is going to be related to the current. WHAT IS THE CURRENT YOU NEED FOR YOUR HEATER? Transistor selection depends almost entirely on that. \$\endgroup\$
    – Daniel
    Commented Mar 28, 2018 at 14:34
  • 1
    \$\begingroup\$ The information you've provided is grossly incomplete, and if you expect an answer, you need to take the time to completely explain your problem. \$\endgroup\$
    – Daniel
    Commented Mar 28, 2018 at 14:36

1 Answer 1


I need to add a transistor to one of the pins I've programmed to push 20mA to 7.5v.


what transistor will boost 20 milliamps to 7.5v?

Ampers can't be boosted to Volts!!!

--You must know the current consumed by the kanthol-wire by:

--first: measuring it's Resistance (R) using a MultiMeter...

--after, you have to use the Ohm's law:

I = V/R

--with V=7.5V , and R measured in first step; let's say it's 30Ω, then:

I=V/R = 7.5 / 30 = 0.25 A

wich is 250mA

--the GPIO pin can only deliver 20mA, and you need to boost those 20mA to 250mA , and here is the function of the Transistor...

there are millions of transistors with millions of characteristics... it's up to you to choose...


simulate this circuit – Schematic created using CircuitLab


  • 2
    \$\begingroup\$ "Ampers can't be boosted to Volts!!!" probably true for this case, but technically that is a "transimpedance amplifier" \$\endgroup\$
    – Daniel
    Commented Mar 28, 2018 at 14:31
  • \$\begingroup\$ Man that was awesome Daniel!! Thanks! I will apply!! \$\endgroup\$ Commented Mar 29, 2018 at 19:28
  • \$\begingroup\$ @kevin-boston: you will apply what??? \$\endgroup\$
    – moyoumos
    Commented Mar 30, 2018 at 8:13

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