I have no kind of knowledge about these things. I just copied some instructions from different projects and now my project died.

I have this 12V 0.8A solenoid that switched for about two weeks. Now the switch gives 2mV to the solenoid.

I measured a new transistor between base and emitter and I got 5.5kOhms. The one that is inside the circuit gives 12.5kOhms. The resistor is still 10k.

I am thinking maybe the 10k resistor was too much or the transistor had too much heat.

The thing is I have a headache from all the reading and would appreciate if someone would help. How do I solve it?

Edit1: If I measure the power supply alone I get 30V, but if I measure it while powering the solenoid directly it's just 14V and it works. Edit3: It turns out (after the discussion) that the signal should have been (and was and is again) 5V. I thought the problem was that I overloaded something but is not. Now that I know I need a higher current at the base I know what to measure in the future. So my problem was the 2mV current I measured when someone asked me.


simulate this circuit – Schematic created using CircuitLab

Thank you.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). \$\endgroup\$
    – Dave Tweed
    Jun 6, 2019 at 15:59

1 Answer 1


Since you confirmed in comments that the problem was the signal level, I'll explain that a little bit.

You are using an NPN transistor as what is usually called a "low side switch." Simply put, you are switching the connection between your load (the solenoid) and ground (usually the negative terminal of a powersupply or battery.)

When used this way, an NPN transistor operates as follows:

  1. When current flows through the base (B) terminal of the transistor to the emitter (E) terminal, a much larger current flows through the collector (C) terminal to the emitter. In this way, a low current signal can drive a high current device.

  2. The connection between the base and the emitter behaves very much like a simple diode. You have to have more than about 0.7V more on the base than on the emitter. Since the emitter is grounded, you need 0.7V on the base to make current flow.

Your circuit could never have operated if your control signal was always 2mV. You've confirmed that the circuit works with a higher voltage, but didn't mention the actual voltage used.

The TIP120 is in fact a darlington transistor:

enter image description here

As such, the required base voltage is much higher than for a simple transistor.

The datasheet says that you need 2.5V on the base to make the transistor switch on.

To reliably switch your solenoid, you should apply at least 2.5V to the base through a current limiting resistor.

It might be better to use a lower value resistor than the 10k you are using. The more current through the base, the more current through the collector.

If you get enough current through the collector, it will reach saturation. The voltage dropped across the collector to emitter will reduce, and you will waste less power in the transistor. You get more power to the solenoid, and the transistor won't get as hot.

Forgot to mention:

The base voltage value isn't set in stone. That 0.7V is a typical value that is close enough for most things. It varies according to how the transistor is made, the temperature, and the current.

The same applies to the 2.5V for your TIP120. The voltage is somewhere around 2.5V.

  • \$\begingroup\$ Will a lower value resistor increase the voltage on the solenoid? If so, using a 12V power supply what resistor should I use? Or what other parts are recomended? \$\endgroup\$
    – Joita Dan
    Jun 6, 2019 at 21:23
  • \$\begingroup\$ I got out the calculator, and checked what the gain does. It looks like your 10k is OK after all. \$\endgroup\$
    – JRE
    Jun 6, 2019 at 21:30
  • \$\begingroup\$ I found somewhere that a transistor requires to give their base 1/10th of the collector current. If that is correct then the signal should be 80mA. Could that be why the signal faded? But I am using it on the ground rail... \$\endgroup\$
    – Joita Dan
    Jun 14, 2019 at 8:58
  • \$\begingroup\$ To drive a transistor into "saturation" (so it acts like a switch,) the rule of thumb is that the base current needs to be 1/10 the collector current. That rule doesn't really apply to a darlington transistor like you have. \$\endgroup\$
    – JRE
    Jun 14, 2019 at 12:26

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