# Computing stuff tied to the physical world

## Switching with a PNP transistor

In Hardware on Sep 7, 2012 at 00:01

This is an exploration to find out what circuit can be used to switch small electronic loads using a digital I/O pin.

The simplest solution by far, is to use the I/O pin to directly drive the circuit. This works great for low currents, such as an LED with series resistor. But there are limits to how much current you can draw:

As you can see, from the datasheet, drawing 5 mA will already cause a voltage drop of about 0.2V, so the load will not get the full supply voltage. With larger loads, the drop becomes even more pronounced.

Another approach is to use a PNP transistor, as follows:

Here’s how it works, in case you’re not familiar with them PNP transistor thingies:

• pulling the input pin down towards 0V, will turn the PNP transistor on
• when on, the transistor will conduct and supply power to the load
• leaving the pin floating or pulling it up, will cause the PNP transistor to turn off

In my setup, I’m using a very light load for now. The following measurements will be different as the load increases, but not as substantially as with the raw I/O pin drive. A circuit like this could easily drive a 100..250 mA load, if the base resistor has the right value – as we’ll see.

Time to try this out. I am quite interested in how the voltage drop over the transistor depends on the exact voltage placed on the base junction. Or rather: how much current I’ll need (since the resistor passes a known current once we know the voltage drop.

So as input, I’m using a 10 Hz sawtooth signal, which varies from 0 to 3V. Due to the way things are hooked up, it starts at 3V below the 3.3V power rail, i.e. 0.3V, and then rises to the level of the power rail.

Here are the results with a 10 kΩ resistor – both signals have been inverted, as I’m measuring relative to +3.3V:

(oops, ignore the yellow trigger point baseline)

At the start, the transistor is turned on strongly, and then the input voltage falls to zero (yellow trace). As you can see, the voltage drop over the transistor increases non-linearly as it gradually turns off. With only 1V driving it, the voltage drop increases from about 40 mV to 100 mV, i.e. 0.1V. With even less, it starts to switch off and gets the full power supply voltage (so nothing reaches the load).

Now let’s do some math. I used my transistor tester to determine that this particular BC557 PNP transistor had 0.8V drop over its base-to-emitter junction, and that its current amplification factor (hFE) is about 220x.

In other words: with 1.0V on the input, there is 1.0-0.8 = 0.2V on the 10 kΩ resistor. Ohm’s law (E = I x R, or I = E / R) implies that the current through the resistor will be 0.2V / 10kΩ = 20 µA. This current is essentially “wasted”, but given the 220 factor, it will allow the transistor to drive up to 20 µA x 220 = 4.4 mA. Not so great…

Update – not sure what I was smoking at the time I wrote the above paragraph. That’s 1.0V below VCC, roughly the switching threshold of this transistor when used with a 1 kΩ load. I’m not sure what point I was trying to make here, other than that a 20 µA base current is not enough to switch an RFM12B.

But if we pull the input to almost ground level, as would be the case with a digital I/O pin, then the base current increases to (3.3 – 0.8) / 10,000 = 250 µA, supporting a load of up to 250 µA x 220 = 55 mA.

If power waste is not an issue, we could reduce the base resistor to 1 kΩ, and get over 500 mA load switching capability (assuming the transistor is powerful enough). The base current will then be around 2.5 mA, a value which an ATmega I/O pin can still easily supply.

But what if power use is important, i.e. if we can’t afford to waste those 20 µA or more?

Here’s the same circuit, with a 100 kΩ resistor instead:

Note how the emitter-to-collector voltage drop (blue line) rises. With 3V input differential, i.e. almost pulled to ground, the drop over the transistor is still under 0.15V, but we’ll need to keep the input voltage at least 1.5V below the power supply voltage to make this work. So in very low-voltage / low-power scenario’s (i.e. running off two almost-depleted AA cells), this might become tricky.

The current losses are now only 1/10th, i.e. 25 µA when the power supply is 3.3V and the input is tied to ground. Then again, in the ultra-low power world that wasted 25 µA is not really such an impressive figure.

Still, for switching loads which draw up to a few hundred milliamps, this circuit using just a PNP transistor and a resistor is really quite practical. If you use a 1 kΩ resistor, the base current will be well with the I/O pin’s capabilities, and the transistor will usually have enough drive to switch its load.

The only other drawback is that the transistor will always add a small voltage drop of perhaps 50 .. 200 mV.

There is a third solution using MOSFETs, with its own set of trade-offs. To be continued…

1. When the base drive power ‘lost’ is a concern, using a Darlington configuration (TR1 driving TR2 as the actual switcher) helps. The gain is higher, so the base current can be truly miserly and the heavier base drive current generated for TR2 flows through the load.

The penalty is a higher Vce drop of ~0.6v and that the Darlington input pin needs > 2*Vbe to turn on, but at ~ 1.2v, this is still compatible with driving from an I/O pin.

2. The first part of your math doesn’t seem right. If the input is at 1V then there is 3.3 – 1.0 – 0.8 = 1.5V across the resistor, not 0.2V. Which makes the resistor current 150uA, not 20uA. And that would support a 33mA load, with your 220X gain. Did I miss something?

• Whoops – you’re absolutely right, that’s not a very meaningful calculation. But the logic remains that we’re going to have to “waste” over 100 µA to control power on an RFM12B – even when the RFM12B is asleep, since that base current is independent of the actual load. I’ve updated the post, thank you.