I believe that these are the questions that are preventing me from understanding. Sorry Sage....

Hello,

Don't worry. All understanding comes in time, and plenty of things I don't understand too.

The voltage across a resistor is *always* equal to the current that is passing through * the resistance. V = I * R is always true for a normal simple resistor.

Suppose you have a 6V supply with its (-) on the Ground, and a 1 MegaOhm resistor from the (+). The other side of the resistor connects to the drain of an N-channel MOSFET, and the source of the MOSFET is connected to Ground.

If you turn the MOSFET on by bringing the gate voltage high. Then the MOSFET will conduct, equivalent to a very low resistance like maybe 20 mOhm or so... in any case it can pass a lot of current very easily. In this case, then the 1 MegaOhm resistor will have a voltage of 6V. It will "stretch" from the (+) all the way to almost perfectly Ground. In this case, then the resistor is limiting the current. In this case, there is a maximum of 6 uA that will pass. It is limited by the resistor value, and the 6V value of the power supply.

But, if you turn the MOSFET "off" by bringing the gate to Ground, then there may still be a tiny amount of current able to leak through the MOSFET. In this case, let's say that around 1 uA can leak through. Then, in this case, the MOSFET is limiting the current, not the resistor. So with 1 uA passing through, the 1 MegaOhm resistor will measure 1 Volt across it. Then, the MOSFET will measure 5V from drain to source. The MOSFET is not a simple resistance, but it is in this case *equivalent* to about 5 MegaOhms, since it measures 5V while passing 1 uA.

Then you may say that the MOSFET's Rds(off) -- resistance from drain to source while off -- is around 5 MegaOhm.

(This is not all *exactly* correct because the MOSFET is not exactly like a resistor. It may not be perfectly linear in current versus voltage like a simple resistor. But it is close enough to speak of it like this.)

Now, here is another thing. Suppose that you tried to use a 100 Ohm resistor to measure the leakage current instead of 1 MegaOhm. If you try to measure 1 uA of leakage by the voltage across a 100 Ohm resistor, then you will be trying to measure 0.1 mV. Unless you have a very, very excellent voltmeter, this will not be a good measurement. So that is why one must choose an appropriate resistor when sensing current. You could probably use 10K Ohm or 100K Ohm just as well as 1 MegaOhm. But you see that 1 Ohm would not work, for example.

But if the leakage was more than 6 uA then 1 MegaOhm would be too big because it would be limiting the current instead of the MOSFET leakage limiting the current.

Does this help to make more sense?

 

Thanks Sage this makes more sense now. I have an attached file and was wondering how i could measure this offstate current. Would i have to split the 2 BJTs up and individual measure each leakage current?

 

Heres the file

 

im measuring 1 BJT off state current but i cannot get a steady ready of the voltage across the resistor. It lies around .3 mV but fluctuates a lot. Is this an acceptable reading?

 

and one more question is the BJT resistance in parallel with the 1 Mohm resistance? Also when i turned on the Hi-Z feature of the multimeter i get a more steady reading, which fluctuates from .03 mV to .15 mV.