Modern microprocessors can deliver respectable currents from their I/O pins. Usually, they can source (i.e., deliver from the power supply) or sink (i.e., conduct to ground) up to 20 mA without any problems. This allows the direct drive of LEDs and even power FETs. It is sufficient to connect the gate to the output of the microprocessor (see Figure 1).
Driving a FET from a weaker driver (such as the standard 4000 series) is not recommended. The FET would switch very slowly. That is because power FETs have several nanofarads of input capacitance, and this input capacitance has to be charged or discharged by the microprocessor output. To get an idea of what we’re talking about: the charge or discharge time is roughly equal to V × C/I or 5 V × 2 × 10-9/(20 × 10-3) = 0.5 ms.
Not all that fast, but still an acceptable switching time for a FET. However, not every FET is suitable for this. Most FETs can switch only a few amps with a voltage of only 5 V at their gate. The so-called logic FETs do better. They operate well at lower gate voltages.
So take note of this when selecting a FET. To make matters worse, many modern microprocessor systems run at 3.3 V and even a logic FET doesn’t really work properly any more. The solution is obviously to apply a higher gate voltage.
This requires a little bit of external hardware, as is shown in Figure 2, for example. The microprocessor drives T1 via a resistor, which limits the base current. T1 will conduct and forms via D1 a very low impedance path to ground that quickly discharges the gate.
When T1 is off, the collector voltage will rise quickly to 12 V, because D1 is blocking and the capacitance of the gate does not affect this process. However, the gate is connected to this point via emitter follower T2. T2 ensures that the gate is connected quickly and through a low impedance to (nearly) 12 V.
In the example, a voltage of 12 V is used, but this could easily be different. Note that if you’re intending to use the circuit with 24 V, for example, most FETs can tolerate only 15 or 20 V of gate voltage at most. It is therefore better not to use the driver with voltages above 15 V. We briefly mentioned the 4000 series a little earlier on. There are two exceptions. The 4049 and 4050 from this series are so-called buffers, which are able to deliver a higher current (source about 4 mA and sink about 16 mA). In addition this series can operate from voltages up to 18 V. This is the reason that a few of these gates connected in parallel will also form an excellent FET drive (see Figure 3). When you connect all six gates (from the same IC!) in parallel, you can easily obtain 20 mA of driving current.
This looks like an ideal solution, but unfortunately there is a catch. Ideally, these gates require a voltage of two thirds of the power supply voltage at the input to recognize a logic one. In practice, it is not quite that bad. A 5-V microprocessor system will certainly be able to drive a 4049 at 9 V. But at 12 V, things become a bit marginal!
—Elektor, 060036-1, 6/2009