With help of this site, I have redeveloped my voltage alarm circuit (see my previous question for the history). I started off with an 741 Opamp and it did not work as I wanted it to work (Thanks for your input here). So I did some research and found a LT1017 comparator.

Based on the data sheet, I assumed that I can drive a LED with the output (it says in the datasheet that it can source 70 mA). But it does not work. The output voltage collapses under load.

Anyway, I switched it around and used a PNP transistor (2N2907) to drive a relay, LED and buzzer (See schematics).

So I still have issues and I don't understand why. The 2N2907 has a hfe of around 250 (measured with my multimeter). The buzzer, relay and LED use about 70 mA. Let's keep it simple and use a hfe of 100. This means I need 0.7mA as ib and with 14V Vcc we are at 14-0.7 at 13.3V/0.7mA, that would be 20k.

My problem: With this calculated Rb nothing happens. Even with Rb = 1k only the relay and LED works, the buzzer does not produce sound.

If I play with Rb and measure ib and ic, I get ib of around 10mA and a ic of 50mA. So the transistor has a hfe of 5 according to this.

I know that I must be making a massive error, but where is it? I learned that stuff 30 years ago and I am going back into it now a little. Of course I forgot many many things from back then, but what am I doing wrong?

Thanks in advance for your input.


  • \$\begingroup\$ I made a new discovery. the transistor does not seem to be problem. The comparator does not switch between Vcc (14V) and 0 (which is my Vss) but stays on about 3.5V when the two input voltages are very close together. So this influences the transistor basis current and therefore the Ic. I thought that there will be a "clean switch" between Vcc and 0 when the one input becomes higher than the other. Is this not the point of a comparator? As always, looking forward to your thoughts. \$\endgroup\$ – Dsch Nov 20 '18 at 0:04

An hFE of only 5 implies you have swapped emitter and collector on the 2N2907. But at 14V, that would usually result in breakdown of E-B junction and can damage the transistor.

  • \$\begingroup\$ I double checked the polarity, so that should be fine. Is there anything wrong in your opinion with the schematics? \$\endgroup\$ – Dsch Nov 19 '18 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.