Hello, this is my first post on this forum.

I'm experimenting with 2N7000 MOSFET, and I'm seeing some very strange behaviour; well - I think it's strange, but it might just be me!

I'm feeding in 7.5 volts at the drain
I'm placing 5 volts on the gate
The voltage I see at the source is 3.69 volts.

I expect to see the full 7.5 volts at the source, especially since I have no load connected. The gate threshold voltage is specified to be >3V, so I would expect the MOSFET to be fully switched on. Is my expectation correct? Are these simply faulty/fake 2N7000s? I've tried a bunch of them, and they all exhibit the same behaviour.

Basically, I'm getting less voltage "out" of the FET (at the source) than I'm putting in. It's dropping ~3.81 volts "through" the FET. No load on the fet. Bizarrely, when I measure voltage across the drain and source, the reading is 0 volts!

I'm at a loss. Any insight would be greatly appreciated.

Regards

Mark

 

It would be better if you post a schematic of your circuit and the way you are taking measurements.

The voltage I see at the source is 3.69 volts.

How are you measurig that voltage? What is your reference point or ground? On a typical configuration, the source is connected to ground.

I expect to see the full 7.5 volts at the source, especially since I have no load connected.

You would see the full voltage between drain and source if the MOSFET is off, in a typical low side switching configuration. I'm assuming it, because you haven't posted schematic.

when I measure voltage across the drain and source, the reading is 0 volts!

If the FET is fully turned on, this measurement makes sense.

 

All is exactly as it should be.

Assuming your circuit is what I think you are describing, it is a "source follower" and the source voltage will be equal to the gate voltage minus the gate-source voltage required to allow the channel current that is flowing. If the current is very small, as it would be with a meter as the load, the difference will be close to the gate threshold voltage specified in the datasheet. The ON Semi sheet spec's 0.8 to 3.0 V, characterized at 1 mA. With 5 V on the gate, the source voltage would therefore be expected to be in the range of 2 to 4.8 V at 1 mA, and somewhat more for a current of a few hundred nanoamperes as would be the case with a 10 M meter as a load.

If the source is completely open circuit, the drain to source voltage will be zero, since there is no place for current to flow.

 

Why do you have the source open circuit?
This has to be connected to power common in order to conduct.
Your load side is drain to power + through the load!
Max.

 

I think it's strange, but it might just be me

It's just you.
Everything you are seeing is perfectly normal.
Look at the data sheet.
The gate threshold voltage is the voltage from gate to source to start turning it on.
Thus for a 3V Vgs, the source will always be at least 3V below the gate when it is on.

If you want to use the MOSFET as a switch, connect the MOSFET source to ground and put a load between the 7.5V and the MOSFET drain.
Then when you apply 5V to the gate, the transistor will turn on and the load will have 7.5V (or so) across it.

Note that to turn off the MOSFET you must apply 0V to the gate (from the signal or with a resistor to ground).
You cannot leave the gate floating as it will go to some arbitrary voltage.

 

Thank you all for your replies. I'm trying to use this device like a switch. Or, perhaps more specifically, a relay, such that when a 5V input is received at the gate, the 9V battery is connected "through" the MOSFET. As in the following:


So, when the MCU drives the gate, there should be 9V available on the header connector, for powering a load.

Am I a total failure?

Regards
Mark

 

Thank you all for your replies. I'm trying to use this device like a switch. Or, perhaps more specifically, a relay, such that when a 5V input is received at the gate, the 9V battery is connected "through" the MOSFET. As in the following:
View attachment 158562

So, when the MCU drives the gate, there should be 9V available on the header connector, for powering a load.

Am I a total failure?

Regards
Mark

For your purpose, this typical set up should work.

 

That is a classical "source follower" or "common drain" configuration. The voltage across the load will always be less than the gate voltage by some magnitude greater than the gate to source threshold voltage, so again, something in the range of 0.8 to 3 V for the 2N7000 for a current of 1 mA. If you refer to the ON Semiconductor datasheet Figure 2. Transfer Characteristics will give you a good idea of typical behavior. For example, for 100 mA of drain current, the gate-source voltage would be between 2 and 3 volts, depending on temperature.

When an N-channel MOSFET is to be used as a switch it is usually used as a "low side" switch where the load is in the drain circuit. This is a "common source" configuration. Now the output voltage from the processor will be the actual gate to source voltage and the FET can be fully turned on for some reasonable amount of current. Returning to Fig. 2 of the datasheet, 5 volts gate to source would be sufficient for about a third of an ampere, however, that does not mean that the drain-to-source voltage will be very low, and it is higher than the continuous current rating of the 2N7000.

[EDIT]: In order to make your original circuit work to put nearly 9 V across the load, the gate-to-source voltage would still need to be at least 0.8 to 3 V for 1 mA, which would mean that the gate to common ("ground") voltage would need to be the voltage at the source plus that 0.8-3, so approximately 9.8 to 12 volts. For a load of 100 mA, the gate would probably need to be at least 13 volts. "Approximately" and "probably" are used here not to avoid doing some arithmetic, but because there is fairly large variation in the characteristics of the FET from one unit to another - as that 0.8 to 3 V span shows.

I believe there are some applications notes at ON Semi, and certainly at other FET manufacturers' sites that explain the basics of power MOSFETs.

 

I'm trying to use this device like a switch. Or, perhaps more specifically, a relay, such that when a 5V input is received at the gate, the 9V battery is connected "through" the MOSFET.

You should use an N channel MOSFET to do low side switching as shown in post #7.

A word of caution. 2N7000 are susceptible to damage by ESD. I've damaged a few in the past few years; and I was trying to handle them properly. As I recall, the devices still worked (sort of), but exhibited high leakage current.

 

Note that we're all assuming you're handling few mA as the OP haven't specified at any moment the load current.

 

Also note fig 3 & 4 on the last page.
This was the original bulletin by Siliconix.
Max.

 

Thank you all for your replies. I have decided to change tactics and somewhat and am going to try the ASSR-1281 solid state relay instead.

Thanks

Mark

 

Thank you all for your replies. I have decided to change tactics and somewhat and am going to try the ASSR-1281 solid state relay instead.

Well, you have your reasons. I can't tell much since I don't know what your load is.

 

Load is tiny, but basically unknown at this time. It's driving the digital input on a PLC.

Having spoken to the PLC manufacturer, I think there was no hope of making my previous car-crash design work. This is what happens when you let amateurs loose with a soldering iron.

The digital inputs on the PLC side are actually powered. There's 24 volts on them (very weakly tied high). For the PLC to register an input, you switch it's input to the PLC 0V line, which drags it low, and thus the PLC registers an input. So the PLC digital inputs are "sourcing" if I have the correct nomenclature.

Thusly I have come up with the following:

Which I believe will satisfy the requirements. The 1K resistors should allow 5mA through the optos (5V supply from CPU) which is the recommended value according to the data sheet (see note 3 on page 10 of the datasheet linked elsewhere above).

BAT_9V is the power to the CPU. It runs to a HT7350 voltage regulator, the output of which is VCC for the CPU. All the CPU side of this stuff is already working. It's just interfacing to the PLC that was/is causing a headache.

What a marvellous web site this is. I am very grateful to all the respondents above. As a novice I'm finding some of the datasheets very difficult to digest so the sagely advice from all respondents is very welcome.