# Understanding mosfets

Discussion in 'Homework Help' started by greek, Sep 22, 2012.

1. ### greek Thread Starter New Member

Sep 22, 2012
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This is a very basic question. As I understand it, in a MOSFET the conductivity from source to drain depends on the voltage from gate to source.

I have a a K2718 , the datasheet tells me this:

So I set up the transistor like this:

With a voltmeter I checked Vgs = 3.8V. From the graph I expect the drain current to be around 150mA. However if I try to short R2 with an ammeter I am effectively shorting my power supply (which starts making funny noises). Why?

2. ### t06afre AAC Fanatic!

May 11, 2009
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Because your ammeter function as a short. And short out R1. If you are going to measure current. You must use it in series not in parallel. As you do with a voltmeter. The strange noise is probably the short circuit protection in your supply. A tip. In your circuit you can perfectly well use a voltmeter and measure the voltage over R1. And then use Ohms law to calculate the current

3. ### greek Thread Starter New Member

Sep 22, 2012
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OK now I'm even more confused.
The current through R1 should simply be 3.8V/1Kohm.
If I short R2, the reading should be the transistor's drain current.

What am I getting so terribly wrong?

4. ### t06afre AAC Fanatic!

May 11, 2009
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oh sorry I meant of course R2

5. ### Ron H AAC Fanatic!

Apr 14, 2005
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No, the current through R1 is (5V-3.8V)/1k=1.2mA, assuming the voltage across the diodes is 3.8V. Are your diodes LEDs?
The curves in the datasheet are typical. The drain current may be a lot higher than 150mA @ Vgs=3.8V.
If you want to measure drain current, connect R2 from drain to +5V, connect the source to ground, and then short out R2 with your ammeter. There is a subtle difference when you do it this way vs the way you drew it. Ammeters have resistance. The voltage drop across that resistance will reduce Vgs slightly, so Ids will be less than if you do it the way I said.
You can burn out your MOSFET if your power supply can deliver the current drawn by the transistor. For example, if Ids is 500mA, your FET will be dissipating 2.5W. It will quickly get too hot, and be damaged.

EDIT: High drain currents are typically measured using short pulses. See the schematic on p.2 of the datasheet.

Last edited: Sep 22, 2012
6. ### greek Thread Starter New Member

Sep 22, 2012
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I chose to start experimenting with that transistor precisely because it comes attached to a very large heat sink so I figured it would be less likely to blow up

As expected I was making a trivial mistake: confusing source and drain. I don't care if the whole world thinks they are appropriately named, they are not.

My original goal was to provide a current limiter (at about 1A) for a motor driver circuit. It seems to work now - is there a reason I shouldn't do it this way? All current limiter examples I have found use more than one transistor.

7. ### panic mode Senior Member

Oct 10, 2011
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if your circuit is current limiting circuit, you don't have load connected. for example it should be inserted between positive terminal of the power supply and your circuit. Also at only 5V this barely has chance to work.

next, your circuit shows J-Fet instead of Mosfet (blasphemy in my book )

also swapping drain and source is quite serious mistake. datasheet shows clearly pinout and names of all terminals (check package info on the very first page):

http://www.datasheetcatalog.org/datasheet/toshiba/3014.pdf

Datasheet also shows that there is internal diode across drain and source (quite common in power transistors). So when you had D and S swapped and R2 shorted by ampmeter, you had only that diode across 5V supply (load would be handy). This diode was clamping power supply output to about 0.8V and the noise you heard was power supply complaining that something is not right.

the diodes D1 and D2 are shown as rectifier diodes so I would expect Vg=1.4V (gate voltage relative to ground). Higher gate voltage is possible with different diodes (LEDs for example) but this is not indicated (Ron already asked this but you still didn't answer).
Note, actual Vgs is going to be smaller than Vg due voltage drop across R2 (which is negative feedback in this case). I wonder how did you measure Vgs=3.8V!? If you are certina this was Vgs then something else must be going on (hopefully this was measurement of voltage across R1).

8. ### Ron H AAC Fanatic!

Apr 14, 2005
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I think it was because he had mentally swapped the surce and drain. He has 1.2V across the diodes, but actually measured from gate to drain, which would be 3.8V.

9. ### Ron H AAC Fanatic!

Apr 14, 2005
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The only stable (with temperature, voltage, etc.) way to make a constant current source (or sink) is to use a current sense resistor and negative feedback. Then the current can be insensitive to voltage, temperature, and individual device selection (MOSFET parameters vary widely from device to device, even among those with the same part number).

10. ### greek Thread Starter New Member

Sep 22, 2012
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If I got it right, I could provide feedback to the mosfet like this:

How should I choose R2 and R3? I think as long as R2 >> Rs and Q2 can handle the power they should be irrelevant. Am I right?

11. ### Ron H AAC Fanatic!

Apr 14, 2005
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R3 just determines the collector current of Q2. Its value is relatively unimportant. R2 is not needed unless there is a fault somewhere in the circuit. 1k should be OK.

12. ### greek Thread Starter New Member

Sep 22, 2012
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without R2, wouldn't the load current be shorted through Q2?

edit: I'll try to answer myself. Q2's base should act as an open circuit until the drop across Rs goes above Q2's threshold voltage, at which point the whole load current tries to go through Q2, but in doing so the feedback on Q1 brings the drop across Rs below threshold again.

Last edited: Sep 23, 2012
13. ### Ron H AAC Fanatic!

Apr 14, 2005
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I think you've got it. The load current will be ≈0.7V/Rs. If it tries to exceed that, Q2 turns on and holds the gate at the voltage required to make Ids≈0.7V/Rs.