I've been wiring stuff now for 10 years or so and have created lots and lots of circuits for all kinds of various stuff. Some digital some analog and most of them work as designed. However, I am totally and completely convinced that it is impossible to turn on and off a power MOSFET.
I am using a CMOS comparator to check two voltages which drives two CMOS logic gate. An AND gate and an OR gate. Everything works great. So far so good.
Now, with the output from the AND/OR gates, I want to drive the GATE on an N-CHANNEL MOSFET to act as a switch for a seperate 12Vdc power supply.
This has got to be the easiest circuit in the world and, according to Fairchild, the voltage from either logic gate "should" drive the gate on the MOSFET they recommended, with "no problem".
HOWEVER, the MOSFET's act like they are shorted (they are not)... With no voltage on the gate, there is voltage from DRAIN to SOURCE. But not all the voltage is there. I have tried this with numerous MOSFET's and can not get this thing to switch. I've tried different power supplies from 3.3V (which should work according to Fairchild) to 5 to 12. The logic and comparator circuits work fine with all voltages. With 5 volts on the same circuit, I can switch a DPDT latched relay with no problem. I have never given up easily on anything, but am just about ready to throw in the towel on the idea of using a power MOSFET as a switch.
So, can anyone recommend a power MOSFET that will act as a SWITCH with 3.3 low current volts on the gate. I'd like to switch 12-20Vdc at 20A. What am I misisng here?
If you have a circuit diagram could you post it to make it easier to visualize what you are describing? Also the part numbers for the devices in the circuit as well as the MOSFETs you have tried and are trying would help.
A common mistake with FETs is to try to use an N-channel device as a high-side switch. This usually won't work without adding special gate drive schemes to the device because it is the voltage difference between the gate and source that opens/closes the drain to source channel, not just the gate voltage relative to your circuit ground. Are you trying to use the N-channel device as a high-side (connect power to complete circuit) or as a low-side (connect ground to complete circuit) switch?
Other reasons for why it won't work correctly could involve, as you mentioned, the drive voltage, and also possibly the inability of the logic device you are using to drive the FET to properly source and sink the abrupt amounts of current needed to turn the FET on/off properly.
Is the gate you are trying to drive the MOSFET with, an open collector output? If so, you will need a pull up resistor. Also is the MOSFET a N-Channel Enhancement Mode Field Effect Transistor? Depletion mode transistors will not work as switches. I hope this has been some help.
Scidoc's suggestion about checking all the FETs you have tried for whether they are depletion mode or enhancement mode is also a very good thought. Depletion mode devices are less common, but if in your search you limited yourself to devices that could switch fully "ON" from a 3.3V drive you may have inadvertently limited yourself to a selection of only depletion mode FETs.
It's not so much that you can't use a depletion mode device as a switch, but the basic FET behavior and necessary gate drive scheme is different. With no gate to source voltage applied, depletion mode devices will inherently be in a "partially ON" transition state (called the linear region) between "OFF" (cutoff region) and "ON" (saturation region). To use them properly as a switch you must drive them with positive gate to source voltage to turn them fully "ON" and negative gate to source voltage to turn them fully "OFF." They are generally used in more specialized applications and not as basic power switches like you want.
Scidoc also mentioned the possible need for a pull-up resistor, which has to do with the previously mentioned current source/sink ability of the logic device you are driving the FET with. He is correct, if the intended gate drive device is open collector (or open drain) then it would only have the ability to sink current and you would need a pull-up resistor to facilitate the current source half. Conversely, if the intended gate drive device can only source current you would need a pull-down resistor to facilitate the current sink half.
Actually, now I think I may have been incorrect above about a depletion mode FET inherently sitting in the linear region. I found a couple of useful discussions of the topic that make me think they in fact are inherently in the saturation region with zero gate to source voltage (though neither explicitly states that fact, just seems to imply it). I'm not positive about this anymore, my uncertainty comes from the fact that depletion mode MOSFETs can be (and I have seen them) operated with gate to source voltage swinging not only negative to cutoff the drain to source channel but also positive to turn them more "ON."
This EETimes article authored by a technical writer from Advanced Linear Devices is well written and states the following within it.
Unlike enhancement-mode transistors, which are "normally-off" devices, depletion-mode MOSFETs are "normally-on".
A completely unique feature of depletion-mode MOSFETs is that they can also be made to work in the "enhancement-mode". This is achieved by making the gate-to-source voltage (VGS), slightly positive by a volt or two for N-channel, or slightly negative by a volt or two for P-channel devices. This allows increased current levels beyond the normal IDSS point.
This page on learnabout-electronics.org gives a less thorough but still good (and with pictures) explanation. It seems to imply that there are variations of N-channel depletion mode devices based on the physical geometry of the gate relative to the channel. One that can be turned more "ON" and one that is already fully "ON" at zero volts gate to source.
Depletion mode MOSFETS are also available in which the gate extends the full width of the channel (from source to drain). In this case it is also possible to operate the transistor in enhancement mode. This is done by making the gate positive instead of negative. The positive voltage on the gate attracts more free electrons into the conducing channel, while at the same time repelling holes down into the P type substrate. The more positive the gate potential, the deeper, and lower resistance is the channel. Increasing positive bias therefore increases current flow. This useful depletion/enhancement version has the disadvantage that, as the gate area is increased, the gate capacitance is also larger than true depletion types. This can present difficulties at higher frequencies.
Either way, it might be a little off topic from your original question (unless of course that was your issue after all), but I thought it would be a good idea to clarify nonetheless in case what I stated in my previous post is in fact in error.