Voltage from high impedence? (Newbie)

Don't forget the capacitances involved. If the gate was previously set to 5V, and you change the pin to an input, the 5V will stay there, at least for a while, making the FET conduct. Similarly, a previous 0V output will keep the gate low, at least for a while until it gets charged/discharged by leakage currents.

It is best not to count on any useful behavior from leakage currents. If you want a well defined behavior, use a pull-up/pull-down resistor, or only use the uC pin as an output.

The leakage currents are not specified by the uC supplier. Also, the high impedance of the circuit makes it very susceptible to noise. Just holding your hand over the circuit may turn the FET on.

I prefer BJTs for different reasons. I understand both FETs and BJTs, but BJTs have a very well defined turn on voltage of 0.7 volts +- 0.1 volts. FETs have a very poorly defined and variable (on process parameters) threshold voltage which makes it hard to do tricky things using the threshold voltage.

One of my favorite configurations for this sort of circuit is the emitter follower. It has unity voltage gain and a current gain of beta (or is it beta + 1 or maybe beta - 1, I always forget those details). So with a base current of a fraction of a mA, you can turn on and off a 20 mA LED. Since the voltage across the BE junction is well defined, you can do a lot of interesting things, like use two BJTs of opposite polarity to control two different LEDs from the same I/O pin. Isn't that what you are trying to do?

The threshold of a FET is not a universal constant. There are FETs that don't turn on with 5 V on the gate and there are FETs that do turn on with only 2.5 V on the gate. It depends on the application they are designed for. Even if you find one that turns on at say 3.5 V, it is not a sharp transition and it will vary between different FETs with that same part number. Look at the data sheet. There should be a spec of a maximum voltage that will cause a minimum current to flow between drain and source (turned on). There should also be a spec on the minimum voltage that will allow some maximum current to flow (turned off). The difference in voltage is typically a volt or more.

That is pretty accurate, but I don't know how high a value resistance. The data sheet should spec this as a leakage current when the output is high Z. It may be to 5 V or it may be to GND. But there will be a small leakage current (very small, microamps max).

No, the DC resistance of a MOSFET gate is a **very** high value. I don't remember how many zeros you need to tack on, but it is *way* up there. Even a few Mohms resistance will drive it very easily. But since you can't depend on the current polarity or level of an output in high Z mode, there is not much you can do with that. Like you said, your 1 Mohm meter swamps the reading. You need to bias the gate with resistors when the MCU output is high Z... or use BJTs which are current controlled and so are easy to define when the drive is high Z.

Rick

That a simplification. What matters is the voltage difference (Gate to source for NFets, Gate to Drain for PFets). That voltage has to reach a certain level to turn on and if it's too high the FET will fail. For some FETs a 5V difference will not be enough. With a high enough voltage on the drain it's easy to have a P-Fet fail by driving the gate to zero, a voltage as low as 20V might do it.

Generally you slow down the turn-on and turn-off times. Another oversimplification, Think of the gate as a capacitor. The drive is charging the cap so adding a resistor introduces an RC.

Simplifications are fine, you just have to realize that is what they are and when to avoid them. (Just like knowing when you can substitute theta for sin(theta))

Maybe. You won't see many data sheets specify minimum leakage which you would need to make this work reliably.

See leakage current. You could specify a resistance large enough that you could not be assured of turning the FET on.

Robert

Oh how they forget. Germanium transistors (they still exist) generally have a forward diode drop of about 0.3 V. Transistors amplify input current, FETS and vacuum tubes amplify input voltages.

It's very unlikely, the voltmeter showing low reading because of its' internal impedance. why don't you try with a different meter? to make sure it's not the case. I suspect, the internal latch may have blown off. Accidently, you may have applied high voltage to the input pin. You can check it by using another uc, but I would re-check the circuit before put a new chip in. you can send your circuit diagram to me, if you wan't me to have a look.

Thanushan

The emitter follower is best used with a touch of base resistance, which makes it the same number of components as saturating switch, but significantly more loss.

Or nominally zero. Or even that are guaranteed on with 0V and you have to put a voltage on the gate to turn them off.

Best regards, Spehro Pefhany

I've fallen into the reverse of that trap before now. Spent a little time head scratching wondering why sin(theta) was returning theta without doing anything. Would have been there all day until I half remembered something from my A-level maths. ;-)