# Can I use MOSFET to output big current with small voltage?

I search online for the usage of (power) MOSFET and it leads to some chip like STE250NS10. I read the manual and it show the schematic diagram like

In the manual, there are several tables to illustrate the output characteristics. One of them tells the change of drain current against voltage of drain and source if we fix the voltage between gate and source. I am going to output about 20A current with small voltage. I didn't use MOSFET before and I don't know if I should fix the voltage of gate and source then change the voltage b/w drain and source to output the desire current. But if I did that, I need the drain-source voltage about 0.14. However, the curve is so steep so even I increase the drain-source voltage by 0.2, the current change to about 100. So I wonder what's the good way to use the MOSFET for current control. I don't have any experience in using MOSFET, so which pin is the gate, source and drain in about diagram and if 1 and 4 are connected, why we need to separate them into two pin? Thanks.

• Usually BJTs are used to control current. That's why they're considered current amplifiers, and why hfe characterizes this. Commented Jan 25, 2014 at 22:52
• Link to the datasheet where you found that? Commented Jan 25, 2014 at 22:55
• updated with link post Commented Jan 25, 2014 at 22:59
• The image in the datasheet makes more sense than the circuit diagram, I changed it. Commented Jan 25, 2014 at 23:15
• 2 = Gate; The down arrow more or less indicates the bulk (the semiconductor chip itself) and with a single transistor like this one it is usually connected to Source (1,2). That leaves 3 for Drain. Interesting read: mcmanis.com/chuck/robotics/projects/esc2/FET-power.html Commented Jan 25, 2014 at 23:20

If you want to switch a big current, at a small voltage, it's easy, just whack plenty of volts on the gate to make sure it's fully on (in this case 10V is where it is specified, so 10 or 12V would be a good voltage to pick).

If you want to control a big current, at a small voltage (for example, to get 100A +/- 5% into a small load resistance), you'll want to close a control loop around the MOSFET with an error amplifier. Either a low-value shunt resistor or a hall-sensor could be used to detect the current. That way the amplifier will be responsible for controlling the gate voltage (which will lie somewhere between the threshold voltage and 5 or 10 volts) and will alter the gate voltage to maintain the current as the MOSFET heats. The amplifier should have a supply of perhaps 12V to allow it to drive the gate fully on.

The concept is illustrated here:

Typically $R_g$ would be about 100 ohms, R would be perhaps 1K, and C might be 10nF. Rs will determine your transfer function $I_{out}$=$V_{in}\over R_s$, and must be rated to not overheat with the maximum possible load current.

A suitable resistance might be something like $1m\Omega$ or $500\mu\Omega$. Care needs to be taken in the layout at such currents (Rs will need a Kelvin connection).

An approximate calculation for C is 0.2 * Cin, assuming Ro is 100 ohms, Rg is 100 ohms and R is 1K. So the values I show will be stable for loads of up to 50nF at the MOSFET gate. That should be okay for your purposes $C_{ISS}$ = 31nF typical (no maximum given), since you state very low voltage, but for a high voltage Miller capacitance will add to the $C_{ISS}$ and you might want to increase R a bit if the step response overshoots at all. When the loop is operating properly, the MOSFET gain will reduce the effect of $C_{ISS}$, but it's better to have it stable under all conditions.

The loop compensation reduces the frequency response into the audio range-- if you need to drive a huge 200A MOSFET at many kHz, an ordinary op-amp is going to need some help driving the gate.

• hi Spehro, I am reading your analysis carefully though I don't understand that carefully. But from the result you show there, if I want the output current from MOSFET's drain to be around 20A, and I want the input Vin to be 4V, does it mean I have to make Rs=0.2$\Omega$? Also, I have a question what's the main reason to use the amplifier to connect to the gate. If I want Vin=4V, can I directly connect the DC voltage to the gate instead of using amplifier? Thanks. Commented Jan 26, 2014 at 1:41
• Divide the input down to something reasonable like 100mV! Commented Jan 26, 2014 at 2:22
• The main reason to use the amplifier is to employ negative feedback to control the current accurately. You have rightly noted that the gate voltage will not be easily predicted (nor will it be stable). The amplifier reduces those errors by a factor proportional to its gain, which is enormous. Commented Jan 26, 2014 at 2:24
• The point of C is only to keep the amplifier stable. Typically, a few 10s of pF should be all that is required. I think you'll have a hard time finding a opamp that isn't stable with 100 pF for C. 10 nF sounds way overkill. The circuit will work with that, but the response will be considerably more sluggish than necessary for stability. Commented Jan 26, 2014 at 14:30
• No, I hadn't looked at the particular FET. I agree that lots of capacitance on the right side of Rg would necessitate a higher C. I already gave you a +1 earlier anyway. Commented Jan 26, 2014 at 15:01