Electronic – PNP Transistor Leak

You should put a resistor in series with each LED and eliminate R18. If you are just switching LEDs it's perhaps wasteful to construct a high-side switch- if you can switch the low side it only requires one NPN transistor per LED.

The high-side switch you show (top diagram) will work, however you can only switch a fairly small current due to the 10K base resistor. At 12V you'll get about 1mA base current so most transistor will be well saturated for up to ~20mA load current, but if you want to switch 100mA reliably you should reduce that resistor in most cases.

Also, it's good practice to add a resistor from base to emitter on the PNP. The reason is that the leakage in the NPN can be amplified by the PNP and result in excessive current at the output (particularly at high temperatures). Something like 20K to 100K will work fine (it's not critical). That said, you can guess that the gain of the PNP will be low at low base current and the circuit will typically work fine at moderate temperatures.

You are not using a PNP transistor the correct way. With the base raised to 12V and the emitter as the output, the PNP B-E junction is being reverse bias. The BC557 B-E junction breaks down and current is supplied to the load through the break down. The TIP32C displays a different behavior probably because it breaks down at a higher B-E voltage.

With the description of your circuit, you want to use a NPN transistor.

^{simulate this circuit – Schematic created using CircuitLab}

Better yet, use a N-MOSFET. Since the gate of a MOSFET needs essentially zero DC current, the 12V signal does not have to supply a big BJT base current (order of 100mA) for the 2A output case.