-N-channel Power MOSFET 30v/60A logic level( with a threshold voltage of 2.5v)
-10k POT
-Adafruit Super bright purple LEDs( threshold voltage of 3.2v)
-220-ohm resistor
-86-ohm resistor
-5v power supply
(I will have attached pics below)
Hello, I am using a POT to control the gate of the MOSFET to turn on an LED with a 220-ohm resistor.
I am using an 86 ohm resistor in series between the gate of the MOSFET to the wiper pin of the POT. When I use my Multimeter to check the voltage on the POT, my LED will only turn on (barely on) after I reach 4.75v. I realize that voltages in series add up. In my situation am I adding the threshold voltage of the MOSFET which is (2.5) and the threshold voltage of the LED which is (3.2)?? That would add up to 5.7 which is more than my supply voltage. Can someone please help me make sense of this.
The part number or datasheet of the MOSFET would be more useful since threshold voltage tells you nothing useful. Also a circuit diagram showing connections, pins etc. because they're difficult to see.
But I guess you're trying low-side switching so the only thing you have to worry about is the MOSFET gate-source voltage. But "threshold" is (roughly) where it just starts to think about letting a little bit of current through. It take a lot more voltage to switch on fully.
You need to stop and think for a minute. The gate on a mosfet is isolated from the source/drain, and while the two can influence each other, they are separate.
The mosfet also acts as a variable resistor that is voltage controlled (gate). The datasheet tells you the Vgsth voltage, which is when the source/drain starts to conduct-in your case 2.5V in respect to the source voltage (normally gnd potential)
On the gate side, your pot wiper should be connected straight to the gate, and a 10K resistor should be added from the gate to source. This resistor is what removes the charge from the gate and prevents it from floating. On the drain/source side you need to set the circuit for the max voltage and determine the resistor from there. With 5V, the Vf of the led being 3.2V, you still need to know the If (forward current) needed. To be safe, I will just use 20mA. That requires a 90ohm resistor in series with the led. Your 86ohm will work just fine.
The circuit for the led side should then be 5V-> led anode and led cathode -> 86ohm resistor then resistor to drain and finally source to ground.
Keep in mind, without the actual datasheet for the mosfet, it is unknown what VVgs on is. That is important, and what you need to know because the mosfet is acting as a resistor in this transitional area. That means it is shedding that current as heat. The more current, the more heat.
tinman13kup:
You need to stop and think for a minute. The gate on a mosfet is isolated from the source/drain, and while the two can influence each other, they are separate.
The mosfet also acts as a variable resistor that is voltage controlled (gate). The datasheet tells you the Vgsth voltage, which is when the source/drain starts to conduct-in your case 2.5V in respect to the source voltage (normally gnd potential)
On the gate side, your pot wiper should be connected straight to the gate, and a 10K resistor should be added from the gate to source. This resistor is what removes the charge from the gate and prevents it from floating. On the drain/source side you need to set the circuit for the max voltage and determine the resistor from there. With 5V, the Vf of the led being 3.2V, you still need to know the If (forward current) needed. To be safe, I will just use 20mA. That requires a 90ohm resistor in series with the led. Your 86ohm will work just fine.
The circuit for the led side should then be 5V-> led anode and led cathode -> 86ohm resistor then resistor to drain and finally source to ground.
Keep in mind, without the actual datasheet for the mosfet, it is unknown what VVgs on is. That is important, and what you need to know because the mosfet is acting as a resistor in this transitional area. That means it is shedding that current as heat. The more current, the more heat.
Thanks for the response. The 10k resistor between the gate and source worked and made it a smoother dimming fade. I noticed that i origianlly had the load(LED) between the source and ground but when i tried it your way( LED in between drain and 5v) it worked alot better and brighter. can you explain that? and does it matter where i decide to put the load (Drain or source)??
tjones9163:
Thanks for the response. The 10k resistor between the gate and source worked and made it a smoother dimming fade. I noticed that i origianlly had the load(LED) between the source and ground but when i tried it your way( LED in between drain and 5v) it worked alot better and brighter. can you explain that? and does it matter where i decide to put the load (Drain or source)??
Because of the simple fact that the gate voltage is referenced to the source, not the drain. Yes, it matters a lot. In electronics you can assume that everything matters unless you have a well researched reason why it doesn't.
aarg:
Because of the simple fact that the gate voltage is referenced to the source, not the drain. Yes, it matters a lot. In electronics you can assume that everything matters unless you have a well-researched reason why it doesn't.
Thanks for the response, but what does that mean the gate voltage is a reference to the source? could you explain? Thanks again.
hello, below I will have 2 different circuits,( I now realize that for my MOSFET circuit that I have the load hooked up between source and ground instead of between drain and 5v)
My question is why for my first circuit with just the POT and LED why do I only need 2 wires on the Pot pins, but all 3 hooked up when I am controlling a MOSFET?
tjones9163:
Thanks for the response, but what does that mean the gate voltage is a reference to the source? could you explain? Thanks again.
It's a literal statement. Vgs means voltage between gate and source. That 10k resistor ties the gate to source, and using the mosfet as a low side transistor, means having gnd (0V). With a Vgsth of 2.5V, the gate starts to open the channel when it is 2.5V above the source voltage. If you have a resistor between source and ground that drops 2.5V, then you would need 5V on the gate
tjones9163:
Thanks for the response. The 10k resistor between the gate and source worked and made it a smoother dimming fade. I noticed that i origianlly had the load(LED) between the source and ground but when i tried it your way( LED in between drain and 5v) it worked alot better and brighter. can you explain that? and does it matter where i decide to put the load (Drain or source)??
I'm trying to picture this, but my head is a bit fuzzy at the moment. Trying to break out of it, so if I don't get it right let it be known.
I'm seeing it as you had the led in parallel with the mosfet drain/source. Initially, with the mosfet gate below Vgsth, the led should have been lit up brightly as long as there was no channel formed from the floating gate. As soon as you crossed Vgsth, the channel formed would have started allowing more current to bypass the led, as you effectively had a voltage divider. As the gate would have reached 5V, the channel might have been fully formed, allowing maximum current to pass from drain to source. This was basically a short circuit to ground. The led would have been off long before this. Your additional resistor on the pot wiper might have saved you.
tinman13kup:
I'm trying to picture this, but my head is a bit fuzzy at the moment. Trying to break out of it, so if I don't get it right let it be known.
I'm seeing it as you had the led in parallel with the mosfet drain/source. Initially, with the mosfet gate below Vgsth, the led should have been lit up brightly as long as there was no channel formed from the floating gate. As soon as you crossed Vgsth, the channel formed would have started allowing more current to bypass the led, as you effectively had a voltage divider. As the gate would have reached 5V, the channel might have been fully formed, allowing maximum current to pass from drain to source. This was basically a short circuit to ground. The led would have been off long before this. Your additional resistor on the pot wiper might have saved you.
I have the picture above. Isnt my LED in series with the Drain/Source, ccould you clarify?
tinman13kup:
You need to stop and think for a minute. The gate on a mosfet is isolated from the source/drain, and while the two can influence each other, they are separate.
The mosfet also acts as a variable resistor that is voltage controlled (gate). The datasheet tells you the Vgsth voltage, which is when the source/drain starts to conduct-in your case 2.5V in respect to the source voltage (normally gnd potential)
On the gate side, your pot wiper should be connected straight to the gate, and a 10K resistor should be added from the gate to source. This resistor is what removes the charge from the gate and prevents it from floating. On the drain/source side you need to set the circuit for the max voltage and determine the resistor from there. With 5V, the Vf of the led being 3.2V, you still need to know the If (forward current) needed. To be safe, I will just use 20mA. That requires a 90ohm resistor in series with the led. Your 86ohm will work just fine.
The circuit for the led side should then be 5V-> led anode and led cathode -> 86ohm resistor then resistor to drain and finally source to ground.
Keep in mind, without the actual datasheet for the mosfet, it is unknown what VVgs on is. That is important, and what you need to know because the mosfet is acting as a resistor in this transitional area. That means it is shedding that current as heat. The more current, the more heat.
Can i put the 10k resistor between gate and ground since thats where the source goes too?
tjones9163:
I have the picture above. Isnt my LED in series with the Drain/Source, ccould you clarify?
Yes. It is just as Leo posted. You tried using an n-ch on the high side, and your results show why it is a bad idea as a general rule. (there are exceptions).
tjones9163:
I have the picture above. Isnt my LED in series with the Drain/Source, ccould you clarify?
Drain/source?
In your picture with the mosfet, swap the green wire with the LED/resistor.
So the source is directly connected to ground, and the LED/resistor is between 5volt and drain.
Forget about a 10k resistor. Just connect the wiper of the pot to the gate.
Leo..
In your picture with the mosfet, swap the green wire with the LED/resistor.
So the source is directly connected to ground, and the LED/resistor is between 5volt and drain.
Forget about a 10k resistor. Just connect the wiper of the pot to the gate.
Leo..
Thanks for response. Im still trying to wrap my head around this. Can you explain the difference between using it in high side or low side. And i dont need the 10k resistor anymore?
In your picture with the mosfet, swap the green wire with the LED/resistor.
So the source is directly connected to ground, and the LED/resistor is between 5volt and drain.
Forget about a 10k resistor. Just connect the wiper of the pot to the gate.
Leo..
Thanks for the response, I'm still trying to wrap my head around this. What is the difference between using it in high side or low side?? and i don't need the 10k resistor anymore?
The idea for that resistor is to pull the gate to a particular state, which for you application is the same as the source (gnd). You have your pot wired correctly, with 5V on one side, gnd on the other, with the wiper being variable between the two.
I picked 10K because it passes little current to gnd with voltage being applied to the gate, but would quickly pull the gate voltage to 0V when the voltage from the pot was removed.
In your case, the gate is connected to gnd through the pot. With the full 5V being applied to the gate, the resistance from gate to gnd is the value of the pot (10K). With the wiper all the way to the other end, 0V is sent to the gate and the gate is connected to gnd. Regardless of where the wiper is positioned, the current for the gate remains fixed at .5mA.
Sorry, I told you I was a bit fuzzy right now. My wife passed yesterday.
To clarify- no, you don't need the additional 10K fixed resistor as long as the pot is connected to ground.
High/low side refers to where the switch (component) is. High side is between the power supply and the load( your led) and low side is between the load and ground. High side controls the positive voltage supply, whereas low side controls the ground side.
Generally, n-ch mosfets are used in low side applications and p-ch mosfets are used on the high side.
An n-channel mosfet starts to conduct when the gate is about 2volt higher than the source.
If you have a (2volt) LED connected between source and ground, then you need to overcome two threshold voltages on the gate. Threshold voltage of the fet and threshold voltage of the LED.
Leo..
