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Topic: Confused about DataSheet, need help find a cure to a problem. (Read 11716 times) previous topic - next topic

crullier

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.

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?

Grumpy_Mike

Quote
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.

michinyon

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.


crullier

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).


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.


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.

westfw

Quote
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.


crullier

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.

Grumpy_Mike

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.

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

Quote
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.


Paulcet

Quote
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  :smiley-grin:

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.

crullier

I am glad I am making progress.
Thank you both for the answers, so based on that lets see if I got PaulsC problem right this time.

I am still unsure about the voltage across the LED and ground.


But here we go.  Essentially there is 12V across the Source and ground, which means that if I am powering the mosfet with 12V there is a difference of 0V which is below the threshold Vds(th). This Mofet will never turn on. I would need at least 14V to make it start to turn on which is not ideal. please please correct me if I am wrong or please tell me if I am right.




I am still unsure about the voltage across the LED and ground.

If I am correct, I would like to post a schematic in bit to make sure I am on the right track if its okay with you guys.

CrossRoads

Seems to me you are using the wrong type of MOSFET there.
N-channel MOSFETs "like" to sink current from a load to Gnd.
P-channel MOSFETs "like" to source current to a load (to Gnd).

N-channel MOSFET can be turned fully on by Arduino if they have Logic Level gates (Vgs <= ~4.5V)
P-channel MOSET needs gate pulled to Vs (12V in this example) to turn off, and gate is pulled low (Vgs shown as a  negative #, as Vg is less than Vs), so a simple NPN is used to perform that while buffering the Arduino pin from 12V.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

CrossRoads

IRF540 has a "Standard" level gate, needing Vgs = 10V to fully turn on.
http://www.irf.com/product-info/datasheets/data/irf540npbf.pdf

Look where Vgs is called - it's always shown at 10V.

Where Vgs is shown lower:

Page 2, Electrical Characteristics @ TJ = 25°C (unless otherwise specified):
VGS(th) Gate Threshold Voltage, 2V to 4.0V Max, VDS = VGS, ID = 250μA  << hardly any current flow

Figure 1,2,3 do show it turning on with Vgs = 4.5V, but not nearly as much as with Vgs at higher voltages, and only brief pulses.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

Paulcet

There you go.  I think you got it.  The LED Vf=2V is just a typical value used in this case to test out a theory.  The actual value depends on the LED and conditions in use (found in the datasheet).

One thing for me to point out since others may be confused:  The schematic I posted, the values represented, and the voltages calculated by Crullier are for study only!  It won't actually work!

Paulcet

Just picking nits here, but the P-channel mosfet looks like this in a schematic:





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