Friday, May 20, 2011

Push-pull output using complementary bipolar transistors

Over the past several years I've been using a transistor push-pull configuration as in Circuit B, with the NPN acting as a current sink for one of the loads and the PNP as current source for the other load. This to me is the "normative" way of using NPN and PNPs. And it's worked fine for me.

The transistors are driven to saturation and are operated such that when one is on the other is off. In Circuit B for instance net NR12 is either at VDD or at ground. When at VDD Q2 is switched on and Q1 off. With the NPN providing a path to ground LED1 then turns on. On the other hand because Q2 VCE is just a couple of hundred mV, current is shunted away from LED2 and thus cannot turn on.

When NR12 is pulled to ground, Q1 is turned on and Q2 off. As such, current is supplied to LED2 which then turns on. Q1 VCE is just several hundred mV, acting almost like a short circuit, and hence diverts current away from LED1 which then switches off.

There is also a push-pull configuration whereby the emitters, instead of the collectors, are tied together and become the output. This is illustrated in Circuit A. This configuration is  known as a Class B amplifier. The advantage of this setup is that only one base resistor is required. Two for each of the bases are needed for Circuit B. Using just one results in both transistors turning on and creating a short circuit path. Trust me. I've tried it and the transistors got searing hot.

To find out if the Class B kind would be a suitable alternative I breadboarded Circuits A and B and took various measurements which I've compiled in this Google spreadsheet.

The most important values of note are the comparative VCE figures. Even with very low base current limiting resistors (100 ohms) the VCE would not drop below ~0.7V for both the NPN and PNP. Circuit B was very different with VCE for the NPN ten times less at 70mV. Although the datasheet lists the S9015 PNP as having a current gain (hFE) of >100 I had to increase the base current by using a 1K resistor to bring down its VCE to a level comparable to the S9014 NPN.

My verdict? I'll stick to the configuration I've used all these years. Class B amplifiers are good for circuits that need the transistors working in the linear region. But for digital circuits the high VCE makes the transistors  inefficient switches which ideally should have zero voltage drop across the contacts (collector and emitter in this case) when they're closed.

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