# NPN driving PNP, Diode from NPN Base to PNP Collector?

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? • 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. – jonk Jul 27 '17 at 6:32
• 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. – jonk Jul 27 '17 at 7:08