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I am high side switching a 5v collector with a 12v base. On the emitter side is just an led and 10k ohm resistor to ground. The base resistor was 1k ohm.

This circuit works just as I expected it with a BC557 PNP transistor, (when 12v base is removed led goes off, with 12v base applied led is on @ 5v) but as my intended load will have a current of 2A or so I need to use the tip32c instead. Problem is, I have no voltage at all on the emitter pin when I try it. I've checked the pinout multiple times (EBC on bc557 and BCE on TIP32C) but it seems that not the issue. I've also experiment with every combination of pinout. I guess what's left is the base resistor, but does that need to be varied with the transistor?

Thanks

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    \$\begingroup\$ Schematic ... ? \$\endgroup\$ – brhans Aug 28 '16 at 0:49
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    \$\begingroup\$ Also, did you notice the different packaging??? The BC557 likes Ic at or less than 100mA. The TIP32C likes Ic that is 10 times higher and even more. It does tend to be the case that some design decisions are different when wrapping a circuit around one or the other. \$\endgroup\$ – jonk Aug 28 '16 at 1:00
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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.

schematic

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.

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  • \$\begingroup\$ It seems like everything I read says PNP should be used for high side switching (learn.sparkfun.com/tutorials/transistors). If that's not true, what actually drives the need for a PNP vs npn? \$\endgroup\$ – DrTarr Aug 28 '16 at 12:57
  • \$\begingroup\$ Take the schematic I put up and change the 0-12V signal to 0-5V signal. Now the design parameters have changed. With an input signal maxing out at 5V, the emitter output cannot reach higher than 5V minus the BE drop of the transistor. If that is not acceptable, then a different circuit would be needed, using a PNP would be a potential way. \$\endgroup\$ – rioraxe Aug 28 '16 at 19:40
  • \$\begingroup\$ By the way, go to your link and to the Application I: Switches section. There is a motor switch schematic. The PNP emitter is connected to 12V and the control to the base is 0-5V. That circuit is bad. At 0V, the circuit is ON as it says. At 5V to base, with 12V at emitter, a current still flows from emitter to base turning the transistor on. Therefore the circuit is also ON. That circuit never turns off. That circuit would function as intended if the off signal voltage is same as the supply voltage. \$\endgroup\$ – rioraxe Aug 28 '16 at 19:50
  • \$\begingroup\$ I tested this circuit with an irf740 n-channel MOSFET and found that the gate floats and I need a pull down resistor. Is this correct? \$\endgroup\$ – DrTarr Sep 15 '16 at 0:54
  • \$\begingroup\$ If the input signal always goes either around 12V or 0V, then you don't need a pull down. If the input signal can go high impedance or disconnected, then you should put in a high value resistor pull down. \$\endgroup\$ – rioraxe Sep 15 '16 at 17:31

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