Thanks for the answer.
You'd normally use a current sense resistor and amplify the voltage across it quite a lot so
that the current sense resistor can be low wattage.
So for a 10A load you might use 0.01 ohm sense resistor (dissipation 1W, voltage out 100mV). For higher currents you get down to milli-ohm or micro-ohm shunt resistor
values to keep the power consumption reasonable, and then you need to carefully
amplify the sense voltage up (avoiding pollution from noise).
The sense resistor needs to be connected as a 4-terminal device (Kelvin connection), and
the voltage difference amplified.
Modern power FETs are often not designed for analog use and a power darlington may be
a better power device to handle the load. Either way you need to look at the SOA curve in
the datasheet (especially for higher voltage operation).
Thanks for the answer,
I have only one question left in my mind.
No resistors are used in this circuit so no power dissipation in the resistor. Why people use an op amp and a resistor instead of this circuit?
Because it will respond rapidly to change, limited only by the bandwidth of the opamp and power transistor(s). Using a software feedback loop will be slower - this may or may not matter, it depends what you're testing with the load.
There are certain limitations of the Hall sensor.
Thanks for the answer.
Thanks for the answer. Paul_B
If you are not using PWM then you don't need a driver.
But you have not shown where your load is located, above or below the fet.
Also over what range of current do you want to control.
The FET is not good at controlling very small currents as you are operating Fet's 'high' resistance region near the knee and small Vgs changes give large changes in resistance.
Here is a minimal control circuit that could be scaled up to higher currents.
It is said resistors from the same batch have very similar resistance so your compound resistor is likely to be still 1% tolerance.
Even if the resistor values were randomly (and independently) distributed over the -1% to +1% interval you may be simply unlucky and get resistors from one end of the interval - so you cannot guarantee better than 1% tolerance with any number of 1% resistors.
If the resistor values were independent and you were interested in the spread around the ideal value it is function of a square root of the resistor amount IIRC. (To reduce the spread to 1/10 you need 100 resistors.)
Correct analysis! Thank you..
Thanks for the answer,
"If you are not using PWM then you donât need a driver."
I am using PWM and then filter it to an analog voltage with the low pass filter.
"But you have not shown where your load is located, above or below the fet."
Is ıt really matter my load will be a power supply or battery?
"The FET is not good at controlling very small currents as you are operating Fetâs âhighâ resistance region near the knee and small Vgs changes give large changes in resistance."
What do you suggest for controling small currents?
No you are actually using a 'constant' voltage to control the FET. A PWM control would not have the filter and in that case (i.e. without the filter) you would normally use a high powered driver to force the Fet to switch on and off quickly
What matters is, "Is the load (battery/power supply) on the FET's Drain or Source pin?"
If the 'load' is below the FET (on the Source pin) then you will have trouble turning the FET on as you will have to get the Vgs above the load voltage. That would be a case for using a driver with a supply voltage of 30V (but need to add a zener to keep the FET gate voltage in bounds)
I did not come up with a suitable solution for the LED torch current control.
Perhaps using another FET to switch in a 'large' series resistor so the 'control' Fet has more control. Really depends on how low you are trying to control the current.
Another option would be to add a shunt resistor and FET in parallel with the load to divert the 'excess' current.
Both these get a bit messy.
Since you are operating in the 'linear' region of the FET and dissipating lots of power,
a completely different approach would be to use a transistor which is a 'current' control device rather than a FET which is a voltage control device.
See for example http://bigbro.biophys.cornell.edu/documents/Temperature_Control/Duesing/tc2.html
For example a BUV22G, 40A 250W transistor.
You will need to drive it via a darlington pair because you need about 3A base current to get 30A collector current.
Thanks for the answer,
I am using constant voltage. Arduino nano hasn't got DAC built in so I am filtering the PWM signal. Where am I wrong?
I thought the MOSFET was in saturation region. As I know saturation is the region that MOSFET acts like a current source. Am I wrong?
I will connect my battery/power supply's positive to drain and negative to source.
Re PWM, you are not 'wrong', you are just not applying the PWM directly to the FET, so not what would be normally called PWM FET control.
In the saturated region the FET looks like a small 'fixed' resistor. Close to a short circuit.
I think you need to draw that for me and show where the current flows and what is being controlled. Sounds odd.
Thanks for the answer,
I think you need to draw that for me and show where the current flows and what is being controlled. Sounds odd.
This is not my schematic. The guy who makes this schematic connected drain to positive terminal; source to negative terminal. I was planning to do connection like this.
Please draw in the 'Load' as well and power supply. Still not clear to me where the supply current and the control current is flowing.
OK with that circuit you will discharging the battery through the FET.
Is that what you are trying to do?
You are making a current controlled 'LOAD' and the there is a 5V supply for the control circuit.
That is clearer now.
The OA in the circuit is a good idea. Feed it with a heavily filtered voltage (from your PWM) and see how you go at low currents. The OA should automatically adjust the FET Gate to hold the current steady. Well at least as steady as the filtered PWM signal.
Thanks for the answer.
Hi,
You will need a circuit breaker in the battery circuit too.
Tom... 
