Simple MOSFET question

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rackley

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I've done a bunch of reading and some testing with MOSFETs but there's something going on that I don't see in the datasheets and I don't see much mention of when I search..

I've been playing with 3 different MOSFETs here, all three are logic level power MOSFETs. I'm driving the gate of all three with +5v, which is far above their minimum "on" threshold levels. Source is connected to GND and Drain is connected to the load (LEDs mostly at this point.) Simple, right?

According to the datasheets I should be seeing resistances far under 1 ohm so I should be seeing virtually zero voltage drop, right? Well, that's not the case...

STS8DNH3LL - 0.018 Ohm 8A, SOIC-8. I'm seeing a 1.2-1.5 volt drop from S to G. (https://www.st.com/stonline/products/literature/ds/10590/sts8dnh3ll.pdf)

VNS3NV04D-E - 0.012 Ohm 3.5A, SOIC-8. I'm seeing a 0.8 volt drop from S to G. (https://www.st.com/stonline/books/pdf/docs/7396.pdf)

STD16NF06L - 0.060 Ohm, 24A, DPAK. I'm also seeing a 0.5 volt drop from S to G. (https://www.st.com/stonline/books/pdf/docs/11188.pdf)

Why the heck am I getting this voltage drop? The closest thing I can find in the datasheets is mention of the Vsd for the source-drain diode voltage - is this the drop I'm seeing?

Thanks,
Ray
 
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The gate to source voltage turns on the Mosfet and makes its drain go down to ground which is the source voltage. The very low resistance is between the drain and the source.

If you don't have a current-limiting resistor in series with the LED which is between the positive supply and the drain then the LED will burn out.
 
The closest thing I can find in the datasheets is mention of the Vsd for the source-drain diode voltage - is this the drop I'm seeing?
Click to expand...
You'll only see this if the source and drain are swapped. ie; the drain going to ground and source to Vdd via the LEDs. But then the FET would never seem to turn off either. Does it turn off with 0V on the gate?
 
You say the source is grounded, but you list voltages from source to gate. Is that a typo?
 
kchriste said:
You'll only see this if the source and drain are swapped. ie; the drain going to ground and source to Vdd via the LEDs. But then the FET would never seem to turn off either. Does it turn off with 0V on the gate?
Click to expand...

They aren't swapped, I've checked and re-done the circuits many times They do indeed turn off with 0V on the gate - I get 0.00v on the drain.
 
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kchriste said:
You'll only see this if the source and drain are swapped. ie; the drain going to ground and source to Vdd via the LEDs. But then the FET would never seem to turn off either. Does it turn off with 0V on the gate?
Click to expand...
Source and drain would be interchangeable except for the S-D diode. With source and drain swapped, the transistor should still turn on with 5V from gate to GND. With the gate at zero volts, the transistor should turn off, but the diode would still conduct.
 
Ron H said:
You say the source is grounded, but you list voltages from source to gate. Is that a typo?
Click to expand...

Typo alert. Yes, source is grounded, gate is going to +5v and drain is going to the load/LED.

I also measured the source terminal voltages with just a voltmeter, no LEDs or anything else connected to it, and I see the same voltage drops.. so it's not the load dropping the voltage.
 
rackley said:
They aren't swapped, I've checked and re-done the circuits many times They do indeed turn off with 0V on the gate - I get 0.00v on the source.
Click to expand...
You aren't making sense. If the source is grounded, you will ALWAYS get 0.00V on the source. Sounds to me like you have the load connected to the source, with the drain at +5V. Is this the case? If so, you are seeing the gate threshold voltage. If Vgs<Vth0, the transistor will be off.
 
Well, if you are measuring these voltages (Vgs & Vds) right on the FET package and your voltmeter is accurate then it is most likely that the FETs are defective. I assume the current through them is less than an amp or so.
 

Errr...late night. I just corrected the post. I'm getting 0.00v on the *drain* when I have source connected to GND and gate connected to +5v.

Load is not connected to source, it is connected to drain. Source is going straight to ground. Gate is being driven at +5v, which is the potential on the other side of the load as well.

Quick diagram, hope I did this right
 

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The output SHOULD be at zero volts when the gate is at +5V. This means the MOSFET is on. You have your LED symbol upside down, and you need a current limiting resistor.
 

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kchriste said:
Well, if you are measuring these voltages (Vgs & Vds) right on the FET package and your voltmeter is accurate then it is most likely that the FETs are defective. I assume the current through them is less than an amp or so.
Click to expand...

Now you see why I'm confused I've tried 3 different FET models and multiple parts from each model and I see the same thing. My voltmeter may be cheap, but it will read 4.98 volts for my +5, and 4.5/4.19/3.79 on the drain, so I know there's a drop. I can also see it in the brightness of the LEDs.
 
Ron H said:
The output SHOULD be at zero volts when the gate is at +5V. This means the MOSFET is on. You have your LED symbol upside down, and you need a current limiting resistor.
Click to expand...

I was thinking of a different circuit than the one I just illustrated. The one I was referring to of is one I have on PCB, not the test circuit. For the one I have on PCB I have the drain connected directly to +5v and the load is connected to the source. So when I turn on the gate I see +4.19/4.20 volts on the source but when I turn off the gate I get 0.00v. On the drain I have +4.98 or so volts. But whether the load is on the drain or source side I get the same voltage drop...
 
But the consensus is this relatively large voltage drop is not normal? If not, what "feature" of these FETs could be causing it and how can I select one that doesn't do this?
 
The voltage drop is normal when you have the load connected to the source.
Are you sure you don't have it connected like this?
 

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Yep, that's how I have it connected - my original diagram was off. It was a late night last night

So it's normal to drop voltage like that when connecting a load to the source side, but not the drain side? That'd odd - why does this happen?

Is there some other style of FET or other device I can use to switch a load that exists on the source side?

Thanks,
Ray
 
Use a P-channel Mosfet to pull a load up to the positive power supply voltage. Then its gate is grounded for it to turn on and is at the positive supply voltage like its source for it to be off.
 
Both sides of what?
A load is usually connected to the positive supply, and the drain of an N-channel Mosfet grounds its other wire to turn it on. If the load is connected to ground then the drain of a P-channel Mosfet can connect its other wire to the positive supply to turn it on. The Mosfet is a switch.
 
For NMOSFETs, the gate must be 5 volts (or whatever the spec says) more positive than the source in order for the FET to be fully on. If you tie the gate to the drain (as in the schematic I last posted), the transistor will be off when the drain (and gate) are at zero volts (actually, any voltage below the threshold). In fact, this is negative feedback, and the drain voltage will be dependent on the individual transistor's threshold voltage, and on the load. Threshold voltage varies from one part number to another, and even varies considerably in those with the same part number.
 
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