Confused about DataSheet, need help find a cure to a problem.

Grumpy_Mike: PS: Two transistors in a darlington arrangement are considered to be in series correct?

no they are in "cascade"

LOGIC LEVEL MOSFET

This is not a logic level FET, it requires a gate signal of 10V to turn it on fully. You need a logic level FET, one that only requires 5V to turn it on fully.

Ah, (urgh) how did I miss this. I wish the info on the package was a bit more clear. Or I should say I wish they had noted that on the package and not Vgs 2-4V.

BTW,

They are real components, but packaged and priced to a ridiculous extent.

By this I am taking that you are saying they are way overpriced?

Yes way over priced, many times more than a top end distribuitor. But then you have to pay for a shop and the people in it.

Hi, if that's an irf510 in the bag, take it back because the specs on the back are not for irf510, even radio shacks web site says that Vgs is +-20V, not 2 to 4V.

Tom...... :)

I think part of the challenge is also my inexperience..

I want to add that after some research I realized that Rds(on) is what you guys are looking at when you say the the FET I bought need 10V to be on. However my statement about Vds(th) still applies, to start conducting there needs to be a 2-4V differential to "start" to turn on the mosfet.

But for instance I do not see that listed anywhere on the data sheet other then as a Test condition for "Drain-Source On-State Resistance".

I am quoting myself: You guys are also looking at Rds(on).

Now with a logic level mosfet all the above goes out the window since it turn on with either 3.3V or 5v.

Hey guys, I found a good resource that explains things very well for newbs like me,

http://blog.oscarliang.net/how-to-use-mosfet-beginner-tutorial/

the only parts I dont understand which I am hoping you guys can explain is:

Where to put the load to a MOSFET? Source or Drain?

Because load has resistance, which is basically a resitor. For N-channel MOSFET the reason we usually put the load at the Drain side is because of the Source is usually connected to GND.

[u]If load is connected at the source side, the Vgs will needs to be higher in order to switch the MOSFET, or there will be insufficient current flow between source and drain than expected.[/u]

If I draw a schematic of either scenario, they both seem the same to me. Vcc stops at the gate...

Is this guys solution correct?

http://letsmakerobots.com/content/mosfet-jokes-why-load-goes-always-drain

Can't read that link because it is not formatted correctly for my iPad now the forum software has changed. But the reason is that with the load in the source the voltage on the source can never be more than the voltage on the gate. It is called a source follower.

Maybe you can work it out for yourself:

See image, and suppose we could somehow turn on the mosfet with 12V (this is over the required level, right?) and it would have Rds of .01Ω (Disregard the part #, I’m just pulling these numbers out of the ether!)

What is the voltage from A to Gnd?
What is the voltage from B to Gnd?
What is the voltage from C to B? (This is Vgs)

Paulcet,

Correct, we only need 10v to drive the mosfet, but let’s say we will use 12v as you suggested.

A to ground = -2v
B to ground = -2.2v
Gate to Source is 12v - 2.2v = 9.8v

I hope I am doing this right.
But yes I see how is this case Vgs less than Vsd.

Not quite right. Ohm's law --- V=I*R Voltage across the resistor is .02 * 500. Add that to the LED forward voltage.

How did you come up with 2.2V B to Gnd?

we only need 10v to drive the mosfet,

No you need 10V between the gate and the source. If the source is being lifted up then the gate needs to be even higher.

You normally have the high-current circuit so that the current flows from the power source through the load ( your electromagnet ) and then through the switching device ( your transistor ) and then to ground ( back into the power supply ).

So you control the current coming out of the load, not going into it.

In this aspect, the diagram you drew in your very first post, is the wrong way around.

The reason you do this, is because the arduino ground is the same ground, and you need to have the base/gate voltage of your transistor device at a level close to ground, that the arduino can manage.

If you have a 12 V ( or higher ) power supply, and you want to put your transistor device on the high side of the load, then you have to be manipulating the base/gate terminal of the transistor around 9 or 11 or 12 volts, which is possible, but more complicated, using the arduino. You probably also need to use a P type transistor.

Grumpy_Mike: No you need 10V between the gate and the source. If the source is being lifted up then the gate needs to be even higher.

Mike, I have two questions I would love for your to answer so we don't go in to many circles:

  1. Where is the datasheet are you seeing that it takes 10v to run on the mosfet ?
  2. When you say "you need 10V better the gate and source.." What does "between" mean? are you taking about putting the voltmeter probe on the gate and source and seeing 10V? or are you saying there needs to a 10V difference between the gate and source? meaning if the source were at -10V the gate would need to be at 0V (please entertain my example).

Paulcet: Not quite right. Ohm's law --- V=I*R Voltage across the resistor is .02 * 500. Add that to the LED forward voltage.

How did you come up with 2.2V B to Gnd?

I was in the middle of a meeting trying to respond on my phone. But regardless my math was wrong.

So lets do yours. V=i*r 0.02 X 500 = 10V From across the Resistor.

So the answers to your prompt:

A to ground = 2V There is only the VF of the LED (right?) B to ground = 10V + 2V = 12V Gate to Source = Not sure, I thought we said 12V.

Please tell me what I am missing. I don't want to waste your time or annoy / upset anyone.

I think the schematic is wrong. The resistor should be connected to 12 V, the led to the resistor, and the mosfet should sink the current. That would allow you to apply 12 V to the gate of the mosfet and it would be 12V above the source.

If I draw a schematic of either scenario, they both seem the same to me. Vcc stops at the gate...

That's in the "off" state; you also need to consider the "on" state. When the "high side" switch (MOSFET or BJT), the output of the ON transistor will be approximately at Vcc, because the transistor voltage drop is small compared to the load voltage drop (hopefully!) The gate or base voltage is also at about Vcc, making the difference (Vbe or Vgs) much smaller than you want it to be.

raschemmel:
I think the schematic is wrong. The resistor should be connected to 12 V, the led to the resistor, and the mosfet should sink the current. That would allow you to apply 12 V to the gate of the mosfet and it would
be 12V above the source.

The whole point is the the Source needs to be higher than the source by at least the threshold value. (lets forget about the specified 10v Rds)

The schematic that PaulC drew was to show me how by putting the load on the Source the voltage at the gate “becomes less” thus I would need more than 12v to turn it on. Unfortunately I don’t understand why even after doing the math he asked me to do. Maybe he will take a few minutes to give me why it does work.

As current flows through the load in the source it develops a voltage across it. This raises the voltage on the source and thus reduces the voltage between the gate and source, thus starting to turn the FET off.

This then stops the drain / source drawing any more current.

So an equilibrium point is reached where you can not get any more current down the drain / source because if you did it would turn the FET more off. Any less current on the drain / source and the voltage between the gate and source rises and turns the FET on more. What you end up with is the balance point.

When you say "you need 10V better the gate and source.." What does "between" mean? are you taking about putting the voltmeter probe on the gate and source and seeing 10V?

Yes.

or are you saying there needs to a 10V difference between the gate and source?

again yes.

meaning if the source were at -10V the gate would need to be at 0V

No that would not be a difference of 10V it would be a difference of -10V. Fo an n-channel FET the gate has to be more positive than the source.

Grumpy_Mike:

meaning if the source were at -10V the gate would need to be at 0V

No that would not be a difference of 10V it would be a difference of -10V. Fo an n-channel FET the gate has to be more positive than the source.

I think he's got it right, Mike. 0V is 10V more than -10V :grin:

crullier: The whole point is the the Source (you meant Gate?) needs to be higher than the source by at least the threshold value. (lets forget about the specified 10v Rds)

The schematic that PaulC drew was to show me how by putting the load on the Source the voltage at the gate "becomes less" thus I would need more than 12v to turn it on. Unfortunately I don't understand why even after doing the math he asked me to do. Maybe he will take a few minutes to give me why it does work.

Good, I think you got the point I wanted to make. I would have given you the "why" now, but Mike beat me to it. His explanation is a good one for "source follower" or "emitter follower". The gate volatage (or base current) "follows" the source current (or emitter voltage).

That balance point is not what we want for a switch. So an N-channel mosfet must have the source at 0v to use it as a switch.