Hi everyone. I am using the attached FET (n-channel enhancement mode) circuit as a switch to turn on an LED. The input pulses from my other circuit are about 3V, however when I probe the output, the output pulses drop to about 2V peak (VDD is 3V). Is there some sort of impedance mis-matching going on here? Why does the load drop? Thanks for your help.

 

Where are you measuring output? And what are the stats on your FET and your LED?

 

Gee, FETs generally have extremely high impedance.

I don't know why it might be loading your circuit. It would help a great deal if you could sketch it out and post it.

Until then, you probably won't get much useful input.

 

Your LED will be off when the MOSFET is on. When the MOSFET is off, the voltage across the LED will be something less than 3V, the actual value depending on the specs of the LED. It is generally more efficient to use series instead of shunt switching, but you will have to invert the logic level of your input pulse.

 

Most Mosfets need 4V or more on their gate to turn on a little. They need 10V to completely turn on. You have only 3V.

Your supply is only 3V. Your LED might need 2.2V. Then its current when the mosfet is turned off is only (3.0V - 2.2V)/1k ohms= 0.8mA which is very dim.

 

Some logic level MOSFETs are pretty much full on with Vgs=3V.
Baseball07, what is the part number of your MOSFET?

 

The symbol is for a small signal MOSFET. It's not the right device for that application. something like a VN10LP would be more like it, although the 3 volts is pretty marginal.

 

Hi everyone. I am using the attached FET (n-channel enhancement mode) circuit as a switch to turn on an LED. The input pulses from my other circuit are about 3V, however when I probe the output, the output pulses drop to about 2V peak (VDD is 3V). Is there some sort of impedance mis-matching going on here? Why does the load drop? Thanks for your help.

Are you sure its an enhancement mode mos? form the schematic, i can see its a depletion mode.

Its normal for the output pulses to drop at 2 volts for a normal 20mA led (if you measure the voltage across the led).
Another think to mention is that if your mos is a depletion mode one, when you have zero Vgs then the output wont be near 3 volts but at a lower value depending on the Idss of the mos.

 

You guys are giving our OP (or whoever drew the schematic) way too much credit for knowing proper symbols. He didn't even get the LED right. I doubt you can infer the type of MOSFET (other than NMOS) from the schematic he drew.

 

It is a low Vth FET from Supertex.

I am a newbie, sorry about the incorrect schematic symbols.

Someone told me to add another resistor from the gate to ground. What will this do?

 

It is a low Vth FET from Supertex.

I am a newbie, sorry about the incorrect schematic symbols.

Someone told me to add another resistor from the gate to ground. What will this do?

MOS have an internal capacitance between the gate and the source. If you turn it on and then remove the gate signal, a charge will be still stored in this capacitance so you need a resistor from gate to source to remove this charge and turn it off.

 

Your 270k gate resistor is way too large. You need to make it smaller (100 ohms), and in any case, you probably won't need a resistor to GND unless your source is a SPDT switch to +3v. If you use 270k, the transistor will turn on and off very slowly.

You need to address some of the other questions asked (or implied):
What is the part number of your transistor?
Where are you measuring this 2V pulse?
What kind of LED are you using?
Why are you using shunt switching (transistor in parallel with LED) to turn it on and off?