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Transistor gate driver circuit9/1/2023 ![]() R1 controls the current flow to the collector of NPN transistor Q1 so that, when a logic HIGH signal reaches the base of Q1, there is a significant voltage drop (usually 20+ volts) across R1.R0 should be sized according to the characteristics of Q1, to protect Q1's base from overcurrent damage.If you're handy with a soldering iron, it could be educational and rewarding to make your own gate driver circuit. The diode should be rated for whatever current the relay coil draws, and have a higher voltage rating than the power supply.Ī quick look at the datasheet shows that your irf540 has a gate-source threshold voltage of up to 4V, so you are definitely correct to assume that you cannot reliably operate it directly from your 2.4V TTL signal. Worse, sometimes it will take multiple operations to kill the transistor, so you think you've got a working circuit, but you can't understand why it's unreliable. If you don't, when you turn off the relay you'll get a big voltage spike across it, and you may well kill your transistor. Also, D1 is called a flyback diode, and you should always use one when switching anything with a coil. Without the combination, using only R2 would apply a maximum of 24 volts to the gate, and this exceeds the maximum specification. R2 and R3, when the transistor is off, set the gate drive at 12 volts. That is, a high output at the MCU will turn the relay off, not on. ![]() This will do, as long as you don't mind the fact that the signal is inverted. Simulate this circuit – Schematic created using CircuitLab However, since your MCU operates on 3.3 volts, you do need some sort of driver to guarantee that your MOSFET gets turned on strongly, and 3.3 volts on an output just won't do that.įor a simple case like this, all you need is a transistor and a few resistors, and Since you're just switching a magnet, you don't need high speed, or high gate current. ![]()
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