# Help with a PNP Switch

simulate this circuit – Schematic created using CircuitLab

I'm trying to design a simple switch for a electric valve. I'm using a 2N4033, and I'm providing 12.3V between the emitter and the collector. For the base, I have a voltage divider of a 100k and a 1k, which provides the base with about 12V. This should cause Vce to be 0, as the voltage difference between Vce and Vbe is under 0.7V. However, when everything is powered on, the Vce is 12V, when it should be zero. Can anyone help with what I'm doing wrong?

Edit: Added circuit diagram. The SPST is the control mechanism for turning the circuit on and off. The valve is represented by the inductor. The current draw from the valve is 540 mA at 12V. My power supply can supply 2A at 12V.

• You probably are not giving the base enough current, but we need to see a schematic and to know the current draw of the valve coil. Jun 2, 2020 at 13:56
• There is even a nice integrated tool to draw a simple schematic if you don't have any drawn already. Jun 2, 2020 at 13:56
• Where's the load? When you close the switch the transistor will short the rails. And finally the transistor will explode. Even before that, the LED will burn since you are loading the 12V battery with a single LED (or diode). Jun 2, 2020 at 14:30
• As I stated in the question, the LED was meant to represent the valve. I've since changed it to an inductive load since it was a source of confusion for a few people. Jun 2, 2020 at 14:49

Thanks for the question and for the schematic :-)

With the SPST switch open, the resistive divider settles the base at about 120 mV, which is below the opening voltage of a BJT B-E junction. So the transistor does not conduct from C to E. R2 is too high to have any meaningful effect, you could as well leave it out.

Note that you do not have a resistor in series with the SPST switch. If you close the SPST switch = make it conduct, the circuit between your 12V PSU, the switch and the BJT's B-E junction has no defined current limit. If your 12V PSU is strong enough, it will blow the base connection out of the transistor, or fry the chip, or both.

I suggest that you put a resistor in series with the SPST switch. Check how much current your transistor can sustain. In a randomly chosen datasheet for the commodity 2N4033, I cannot see Ibe(max) specified, but Ic(max) is 1A, so you can probably consider that. Actually much less should be sufficient. To allow for Ic=1A, considering a conservative Hfe=50, you could ballpark that you need to limit the base current to 20 mA. At about 11 V available differential voltage (the B-E junction will clamp at about 0.6 V) the resistor should be 11 / 0.02 = 510 Ohms (a particular standardized value in the series).

EDIT:

• You've probably blown the transistor and the LED by now, you may need new ones. Unless the LED has a current-limit resistor integrated, always use a resistor in series with an LED, to limit its current to about 20 mA (10 mA would be safer). I.e. for 12V supply, you need about 1k2 in series with the LED. And, the LED goes in series with the transistor's Collector electrode. Do not expect the circuit to work if your transistor just shorts the PSU :-)
– frr
Jun 2, 2020 at 14:41
• How much current does the valve need to "trigger" ? You could just measure it with an Amper-meter.
– frr
Jun 2, 2020 at 15:15
• Also, the one final touch to your circuit would be: put a diode in parallel with the valve (inductor) and place it such that it is normally reversed (not conducting). Its role is to absorb an inductive spike upon the transistor turning off.
– frr
Jun 2, 2020 at 15:17
• So if your valve does not "trigger", have you tried measuring current through the valve, while it should've triggered?
– frr
Jun 2, 2020 at 15:19
• After a bit of tinkering, the circuit works now! Thank you so much frr, I couldn't have done it without you. Jun 2, 2020 at 15:27

When the switch is closed, the transistor will be destroyed, with the circuit as shown, as the base current is not externally limited.

R1 and R2 do nothing of value. The 2N4403 is pretty marginal for switching 540mA and you should have a diode across the valve coil or the transistor will be destroyed when it switches off. Since the load is inductive, I think the 2N4403 will be outside the SOA limits and will fail either immediately or after a short time, even if properly driven with ~50mA base current.

This answers your immediate question, but there is not enough information to make helpful suggestions on how to do whatever it is you are trying to do. A MOSFET would be better in this voltage/current range.

I suggest looking at a MOSFET circuit And please add a diode in inverse parallel to your inductor to prevent voltage spikes from destroying your components, since inductors can create large spikes when switched off