I'm toying with a simple temperature warning circuit on a 12V system where a buzzer would make a sound once a NTC type thermistor's resistance falls low enough. There is also a potentiometer to adjust the switching threshold, but this could probably be replaced with a fixed resistor since the thermistor threshold resistance for the buzzer to go on is 50 ohms. Here's what I have so far based on some light googling:

Transistor switch circuit

I built the circuit on a breadboard, but the buzzer seems to be on almost all the time. I would welcome any comments on the circuit design, selection/placement of components, how to calculate the values for the components and whatever else you might think of.

Thank you in advance!


1 Answer 1


the buzzer seems to be on almost all the time

Bipolar transistors do not have an exact threshold voltage. However, if \$V_{BE}\$ is 700 mV, they will typically be quite "on". Since your buzzer will require a significant current, I don't imagine it will be on if the \$V_{BE}\$ is much below 700 mV.

If the base of your bipolar transistor did not draw any current, the voltage applied to the base would be determined by the supply voltage and the voltage divider consisting of the nominally 50 \$\Omega\$ NTC, and the maximum 10 k\$\Omega\$ variable resistor. (Incidentally, the schematic you provide shows the 10 k\$\Omega\$ variable resistor always set at 10 k\$\Omega\$. In order for this resistor to have a variable resistance, you must connect part of your circuit to the wiper of the variable resistor. I don't know whether your schematic matches your breadboard, but that is definitely an issue. However, that is not the only problem you face.)

If the base of the transistor did not draw any current, the voltage between the NTC and the variable resistor would be

$$V_{B} = 12V \frac{R_{variable}}{R_{NTC}+R_{variable}}$$

Since \$R_{NTC}\$ is very much smaller than \$R_{variable}\$ unless the variable resistor is set to less than 50 \$\Omega\$ and is wired to actually provide a variable resistance,

$$V_{B} = 12V \frac{R_{variable}}{R_{NTC}+R_{variable}}\approx 12V$$

This is much more than the 700 mV needed to turn on the transistor, so the buzzer will always be on. Note that the actual \$V_{BE}\$ will probably be quite close to 700 mV, rather than 12 V, because the base-emitter connection acts like a diode, and the voltage cannot rise much above 700 mV without very large currents flowing through it.

So, in order to make your circuit work, you need to arrange things so that below a certain temperature, the voltage supplied to the base of the transistor is less than 700 mV (say 600 mV or less) and above a certain temperature, the voltage is 700 mV or above.

Note, that decreasing the resistance of the variable resistor might seem like an obvious solution to someone not very experienced in electronics. There is, however, a significant problem with that. If the resistance of the variable resistor is low enough for that solution to work, the current through the variable resistor and through the NTC would be very large. Depending on the NTC, it might survive. However, the variable resistor would get very hot (it would need to be very large) and the power supply / battery would need to supply a great deal of power which is wasteful and might possibly exceed their capability.

So, you will need to use another circuit. My suggestion is to swap the locations of the NTC and the variable resistor, and then add a logic inverter before the drive transistor and buzzer.


simulate this circuit – Schematic created using CircuitLab

In this case, Q1, R3 and R4 comprise the logic inverter. I will leave it to you to see if you can determine appropriate values for R3 and R4 so that if Q1 is off, Q2 is on, and if Q1 is on, Q2 is off. Remember what was said above about voltage dividers and voltage required at the base of a transistor to turn it on.

  • \$\begingroup\$ thank you so much, this was exactly the kind of help I was looking for. So let me see if I got this right looking at your circuit. Since the ratio between \$R_2\$ and \$R_{NTC}\$ determines the switch-on voltage of \$Q_1\$, and the threshold resistance of \$R_{NTC}\$ is ~50 \$\Omega\$, I can just replace \$R_2\$ with a 900 \$\Omega\$ resistor. Then, looking at the logic inverter part, I need to select \$R_3\$ and \$R_4\$ so that \$V_{B2} \geq 700mV\$, basically \$\frac{R_3}{R_4} < 17\$ or so. How accurate is all that? \$\endgroup\$
    – hezamu
    May 21, 2023 at 10:37
  • \$\begingroup\$ The follow-up question is that since transistors are amplifiers and I don’t want to run too much current through the buzzer, how do I control that? I assume with \$R_3\$ and \$R_4\$? \$\endgroup\$
    – hezamu
    May 21, 2023 at 10:38
  • \$\begingroup\$ Regarding 1st Q. Yes, you want R3/R4 < 17. Regarding 2nd Q. Is your desire to limit the current through the buzzer based upon the loudness of the buzzer? Or is it to conserve power overall? Do you care about the power when the buzzer is off? \$\endgroup\$ May 21, 2023 at 12:43
  • \$\begingroup\$ Maybe I misunderstood the current issue, I was thinking that I don't want to "force" too much current through the buzzer so that I don't damage it. But as I see it now, it will just act as a load and draw whatever it will if \$Q_2\$ is "on". \$\endgroup\$
    – hezamu
    May 21, 2023 at 13:07

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