FET Switching Question

[tommy.genes]

This is not technically a stompbox application, but I figure I'm more likely to get a technical answer here than in the Lounge.

I have to switch a resistive but fairly high-current load with a Hall effect sensor as shown below. IC1 is a Hall effect device that outputs a constant current in the presence of a magnet. That current passes through R1 to create the gate voltage to turn on Q1. With the fan and LED only, which only draw about 1/8th of an amp, the drain-source voltage drop across Q1 is minimal, i.e. almost the full voltage from J1 is across the load.

When the full load is attached, however, it draws about 1.25 amps and the drain-source voltage of Q1 goes up to almost 1.5V. I would expect some increase in Vds at higher current due to the on resistance of the transistor, but Q1 is an FDP8880 and it has an on resistance of less than 20 milliohms. Also, at 1.25 amps and 1.5V, the transistor should be dissipating almost 2W, yet it remains cool to the touch.

Is there some other source of voltage drop that I'm missing here? Any other critique of the circuit would also be appreciated.

Thanks,
-- T. G. --

[Seljer]

Whats the voltage on R1?

[R.G.]

Yes, that's the right question. MOSFETs above the threshold voltage can best be thought of as a transconductance - channel amperes per volt of change on the gate-source.

If you don't have enough Vgs enhancement, the MOSFET pulls out of what passes for saturation to a low resistance and acts like a current source, which lets the load voltage vary. "Enough" means "Vgs bigger than about 1V per ampere". The 1V/A varies from MOSFET to MOSFET, of course, but you may need lots of volts if you have big currents, just like with a bipolar you may need lots of base current for big collector currents to saturate it.

That means that if your voltage sags at that jack because of loading from the external load, the enhancement voltage sags too because the Hall effect device may act like a current source, but there's probably some minimum voltage it takes. So if the supply voltage sags, so does the enhancement voltage on the MOSFET gate, so the MOSFET drain-source voltage rises.

[Gurner]

Or put another way (because you can never have enough "ways" put), this is possibly due to the load sagging the power rail & therefore, in turn, altering the fet gate voltage - so possibly RDS ain't 20 milliohms (but much higher) & therefore there ain't 1.25 amps running through the fet (as you think there is)  - this is bourne out by the cool to touch observation.

One solution would possibly be to regulate the VCC supply voltage to the hall device (zener etc)

[earthtonesaudio]

1k posts for Gurner!  Woo!

...and now back to your regularly scheduled program.

[Gurner]

1k posts for Gurner!  Woo!

...and now back to your regularly scheduled program.

Good spot there earthtonesaudio!

Tell you the truth I never cast my eye that way ....but I'm surprised I've contributed that many posts ...I guess time & posts fly when you're having fun 

[tommy.genes]

Thanks for the input so far. I will try some more testing, but that might have to wait until the weekend. But from what I've done so far...

The voltage at R1 is about 5.6V, which is as expected with the "on" current from IC1 rated from 12 to 17mA and R1 having a value of 430 ohms. I specifically chose the FDP8880 since it is designed to turn on at TTL levels, presumably so you can switch loads directly from the output of a microprocessor. To be sure that Vgs wasn't a problem, however, I actually removed the Hall sensor from the circuit and fed the gate of Q1 directly from a second bench supply. Varying Vgs from 5V to 10V had no effect on the rest of the circuit. It was only below about 3V that the FET started to shut off.

I didn't actually measure Vds directly. Instead I measured voltage across the load, and to get the 9V across the load that I wanted, I had to turn the bench supply up to 10.5V (that's how I came up with 1.5V for Vds). At that voltage, the bench supply reported delivering about 1.25A, which is what the combined loads draw if they are connected directly to the supply at 9V.

Perhaps I should also note that the "resistive" load is actually a Peltier cooler. They seem to have a linear V/I curve so I thought of it as a resistive load, but maybe there's something else I'm not considering. If we have to spec a 10.5VDC supply to insure we get the desired 9VDC across the Peltier, that's certainly not the end of the world. I was just concerned that there are 1.5V unaccounted for, and I don't want that to come back and bite me in the butt.

Thanks,
-- T. G. --

[tommy.genes]

Oh, and the Hall sensor is an Allegro A1158LUA-T, which can operate from 3 to 24V and includes its own on-board voltage regulator and transient protection. It's amazing what they can pack into that 3-pin SIP.

So I'm not sure that sagging voltage to the Hall device is the problem.

-- T. G. --

[R.G.]

So I'm not sure that sagging voltage to the Hall device is the problem.

Me neither - but it's the first place I'd look.

[tommy.genes]

Finally got around to digging into this a bit deeper. When measured directly (duh...), Vds was less than a tenth of a volt! The other 1.5 volts that went missing were being lost in the breadboard and jumper wires, as verified by taking readings at various points along the path from the + terminal to the - terminal of the bench supply.

All voltages shown are referenced to the bench supply ground except as noted. "Off" measurements marked with * were tricky - meaning they would vary significantly from reading to reading - because the Peltier junction (LOAD) actually becomes a variable voltage source as it heats up back towards room temperature when it is shut off.

I've had PCBs made for this circuit, so I will have to populate one and see how much that improves the voltage loss.

-- T. G. --