0
\$\begingroup\$

I am reviewing some old 12V alarm designs that use the circuit below to activate a siren. The design has been built on PCB and I have confirmed proper operation, proving the schematic is not flawed (not that I can see anyway). Simply put, when the MCU pin goes HI (5V), it switches the NPN transistor, which in turn switches 12V to the siren via high-power PNP transistor. That is easy to understand, but what is the purpose of Diode "D1"? It's Anode attaches to the Base of the NPN and its Cathode attaches to the Collector of the PNP. But what does it do?

I am also curious if Diode "D2" is necessary. Why not tie the NPN's Emitter directly to Ground?

enter image description here

\$\endgroup\$

1 Answer 1

Reset to default
2
\$\begingroup\$

I think it's short-circuit protection. If the output is accidentally shorted to ground, D1 pulls the drive to the NPN down shutting the output off. D2 is necessary since D1 can only pull the base down to a diode drop above ground, so D2 compensates the drop making Vbe (NPN) approximately zero.

\$\endgroup\$
8
  • \$\begingroup\$ Yup. Though I'm not convinced that the PNP can consistently deliver the indicated current with such a low base current. It's only guaranteed \$\beta\ge 50\$. Oh, well. Probably doesn't matter for this. \$\endgroup\$
    – jonk
    Jul 27, 2017 at 6:32
  • \$\begingroup\$ @jonk good catch, you're right. I guess the siren probably makes some noise at lower current anyway, though no excuse for poor design. \$\endgroup\$
    – John D
    Jul 27, 2017 at 6:55
  • \$\begingroup\$ Thanks. Working backward, \$I_{C_2}\le 1.5\:\textrm{A}\$ and let's say the high power PNP (\$Q_2\$) works as a switch with \$\beta_2=15\$, then \$I_{C_1}\le 100\:\textrm{mA}\$. And let's say the lower power NPN (\$Q_1\$) works as a switch with \$\beta_1=20\$, then \$I_{B_1}\le 5\:\textrm{mA}\$. The I/O pin resistor should be \$680\:\Omega\$, I think, and the NPN collector resistor to the PNP base should be \$100\:\Omega\$ and rated for \$2\:\textrm{W}\$ (twice what it will dissipate.) If they want to support all the way up to \$1.5\:\textrm{A}\$, anyway. \$\endgroup\$
    – jonk
    Jul 27, 2017 at 7:08
  • \$\begingroup\$ John D, thank you for the complete answer to my question! @jonk, I appreciate your comments as well, but what datasheet tells you the minimum β is only 50? The guaranteed Hfe varies from manufacturer to manufacturer and varies by p/n too (e.g., the NEC version 2SB772 offers 4 minimum β "ranks"). I will say this though, sirens connected to the above circuit draw about 1.0A (AVG) when blasting, and I have confirmed this via Fluke 8845A with multiple boards and multiple sirens. Even when the sirens are direct-connected to 12V (not the above circuit), they still draw 1.0A. \$\endgroup\$
    – JDW
    Jul 28, 2017 at 2:08
  • \$\begingroup\$ @JDW I checked the first link I got: svntc.com/TPDF/2775.pdf (as my history shows me.) But ST.COM's version shows 30 at 3 A and 80 and 1 A. So 1.5 A with 50 wouldn't be a shock. That's two sheets. Others may have different figures, of course. None of that really matters, though, since when in saturation the beta you use is probably less than the active mode value, anyway, because of ambient temperature variation, part variation, manufacturer variation, etc. So a distinctly saturated estimate should probably be considered. Which is what I used. The ST.COM sheet uses 20. That's fine. \$\endgroup\$
    – jonk
    Jul 28, 2017 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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