i plan on doing a SC test on a power supply that the output is switched on with a p-mosfet i was just wondering if there is defined recovery time for the pmos to cool down before you can do another short circuit test.
Electrical – short circuit response P-MOSFET
pmos
Related Solutions
Wiki says...
In a depletion-mode MOSFET, the device is normally ON at zero gate–source voltage. Such devices are used as load "resistors" in logic circuits (in depletion-load NMOS logic, for example). For N-type depletion-load devices, the threshold voltage might be about –3 V, so it could be turned off by pulling the gate 3 V negative (the drain, by comparison, is more positive than the source in NMOS). In PMOS, the polarities are reversed.
So for a depletion-mode PMOS it is normally ON at Zero volts but you need 3V or more on the gate higher than the supply voltage to turn OFF. Where do you get that voltage? I think , that's why it is uncommon.
In practise now we call them High Side Switches or Low Side switches for power MOSFETs. They prefer not to combine enhancement and depletion mode in the same chip as the processing costs are almost double. This patent defines some innovation and better physical desc. than I can remember. http://www.google.com/patents/US20100044796
It is possible though what you are suggesting and performance are key issues. However when it comes down to low ESR, MOSFETS are like voltage controlled switches with ESR changing over a wide range of DC voltages unlike bipolar transistors which are 0.6 to < 2V for max peak in some case. Also for MOSFETs it is constructive to think of them as having an impedance gain of 50 to 100 when looking at loads and ESR of source. So consider you need a 100 ohm source to drive 1 ohm MOSFET and 10 ohm source to drive a 10mΩ MOSFET if you use 100:1, Conservative is 50:1. This is ONLY important during the transition period of the switch, not the steady state gate current.
Whereas bipolar hFE drops dramatically so you consider hFe of 10 to 20 good when saturated for a power switch.
Also consider that MOSFETS as charge-controlled switches during transition, so you want to have a big charge available to drive the gate capacitance and load reflected into gate with a low ESR gate drive, if you to make a fast transition and avoid commutation ringing or bridge cross-over shorts. But that depends on design needs.
Hope that isn't too much info and the patent explains how it works for all modes of P N type depletion and enhancement in terms of device physics.
Knowing the voltage being switched and max current would greatly improve available answer quality.
The MOSFETS below give examples of devices which would meet your need at low voltage (say 10-20V) at currents higher than you'd be switching in most cases.
The basic circuit does not need to be modified - use it as is with a suitable FET - as below.
In the steady state on mode the "problem" is easily addressed.
A given MOSFET will have a well defined on resistance at a given gate drive voltage. This resistance will change with temperature, but usually by less than 2:1.
For a given MOSFET you can usually decrease on resistance by increasing gate drive voltage, up to the maximum allowed for the MOSFET.
For a given load current and gate drive voltage you can choose the MOSFET with the lowest on state resistance that you can afford.
You can get MOSFETS with Rdson in the 5 to 50 milliohm range at currents of up to say 10A at reasonable cost. You can get similar at up to say 50A at increasing cost.
Examples:
In the absence of good information I'll make some assumptions. These can be improved by providing actual data.
Assume 12V to be switched at 10A. Power = V x I = 120 Watts.
With an Rdson hot of 50 milliohms the power dissipation in the MOSFET will be I^2 x R = 10^2 x 0.05 = 5 Watts = 5/120 or about 4% of the load power.
You would need a heatsink on almost any package.
At 5 milliohms Rdson hot dissipation would be 0.5 Watts. and 0.4% of load power.
A TO220 in still air would handle that OK.
A DPak / TO252 SMD with minimal PCB copper would handle that OK.
As an example of an SMD MOSFET that would work well.
2.6 milliohms Rdson best case. Say about 5 milliohms in practice.
30V, 60A rated. $1 in volume. Probably a few $ in 1's.
You would not ever use the 60A - that's a package limit.
At 10A that's 500 mW dissipation, as above.
Thermal data is a little uncertain but it sounds like 54 C/Watt junction to ambient on a 1" x 1" FR4 PCB steady state.
So about 0.5W x 54 C/W = 27C rise. Say 30C.
In an enclosure you'll get a junction temperature of maybe 70-80 degrees. Even in Death Valley in midsummer it should be OK. [Warning: DO NOT shut the door on the toilet at Zabriski Point in mid summer !!!!][Even if you are a woman and the Hell's Angels or similar have just arrived][My wife will tell you about it][But your MOSFET would be OK.]
Datasheet AN821 appended to datasheet - Excellent paper on SO8 thermal issues
For $1.77/1 you get a rather nice TO263 / DPak device.
Datasheet via here includes a mini NDA!
Limited by NDA - read it yourself.
30v, 90A, 62 K/W with minimal copper and 40 k/W with a whisper.
This is an awesome MOSFET in this type of application.
Under 5 milliohms achievable at many 10's of amps.
If you could access the actual die you could possibly start a small car with this as the starter motor switch (spec'd to 360A on graphs) BUT the bondwires are rated at 90A. ie the MOSFET inside greatly exceeds the package capability.
At say 30A power = I^2 x R = 30^2 x 0.003 = 2.7W.
0.003 ohms seems fair after looking at the data sheet.
Best Answer
If I understand your question correctly, you want to interrupt a voltage line with a series-connected P-channel transistor and turn it on by connecting its gate to GND. The bias in this case is the voltage \$V_1\$ connected to the source (see below sketch). Then you realize that the P-channel heats up seriously when you do a short circuit test. This is a classic in safety tests in which the quality department rejects the design because risks of fire exists when you short \$V_{out}\$ to ground. The reason is because you end-up in an equilibrium situation in which the bias collapses and ends up forcing the P-channel into a linear mode where it heats up big time. This is solution (a). A better solution is to use an N-channel (lower \$r_{DS(on)}\$ and cheaper than P-channel) as shown in solution (b):
In this application, the source \$V_2\$ is greater than \$V_1\$ by the amount of needed \$V_{GS}\$ to bias the MOSFET (10 V for a classical type). Then, when you short \$V_{out}\$, the MOSFET bias no longer disappears and it sustains the short circuit current with a low \$r_{DS(on)}\$ (low power dissipation). Of course the current is high and you may want to take additional precautions to harness it but the MOSFET is no longer in danger with risks of fire. Finally, if the body diode causes problems, you can connect two back-to-back MOSFETs as commonly done in battery protection circuits for instance (solution c).