You can use a digital potentiometer to drive a transistor which then acts as a kind of variable resistor. The same is true if you use PWM with an RC filter to drive the transistor. You don't need a MOSFET as the current and voltage is within the range of a standard transistor.
You can use a digital potentiometer to drive a transistor which then acts as a kind of variable resistor. The same is true if you use PWM with an RC filter to drive the transistor. You don't need a MOSFET as the current and voltage is within the range of a standard transistor.
Since the original circuit uses a potentiometer, you do not need a transistor. All you need to do is remove the original potentiometer and replace the three leads from a digitally controlled potentiometer. The original circuit will behave as it used to, with the digital pot replacing the function of the old mechanical one.
V_king:
I also considered a digital potentiometer, but I can only find ones, which work with 5V limit
The 5V limit usually refers to supply voltage, which would be what you use for the arduino and the logic control signals (SPI, I2C). The A/B connections on the actual pot can be higher. The only real limitation is current handling. With the information you provided (15V with a minimum of 1K of resistance) the maximum current through the digital potentiometer when it is at 0 ohms will be about 15ma. Shouldn't be tough to find a digital pot to meet your needs.
The datasheet of the digital potentiometers I have specify the supply voltage as the maximum voltage to apply to the A/B pins too. I don't know how other types are in this aspect.
pylon:
The datasheet of the digital potentiometers I have specify the supply voltage as the maximum voltage to apply to the A/B pins too. I don't know how other types are in this aspect.
Just checked the data sheet on mine and you are correct; however, I have used that pot with a 12volts across the pot with no problems...
Connect the wiper to the input of a non-inverting op-amp circuit with gain = 2, powered from 15V
Connect the output of the op-amp to the analog controller's input
Unless there's a negative power supply available that you haven't told us about, the op-amp has to be a "rail-to-rail" device, or one that's otherwise capable of driving its output very close to its negative supply. A general purpose op-amp can only get its output down to a couple of volts above its negative supply, and I think that you want to get closer to 0V for this application.
With 15V available, there's no problem getting 10V out of the op-amp. Close to 0V output, though, you might find unexpected behavior. A rail-to-rail op-amp will get a lot closer to the rails than a general purpose device, but it won't actually go all the way to the rails. You could expect to see an output as low as about 15mV if the analog controller doesn't ask you to sink a lot of current. Be sure to check the data sheet of the op-amp you select for its output low voltage limits and sink current capability. On a 10V scale, 15 mV is about one part in 667 - better than the resolution of the single-byte PWM outputs, and better than most digital pots. If that's not suitable, you'll need some other solution, or you'll need to do something to help drop the op-amp's output that I haven't tried. If there's a negative supply available, powering the op-amp with +15V and - is a possible solution - and, you could use a general-purpose device rather than a rail-to-rail unit.
If you want to try this cheaply and quickly - without waiting for a rail-to-rail op-amp to come in the mail - you can test it with the potentiometer you already have and a general purpose op-amp. Connect the existing analog pot where the digital pot would go, set up the op-amp as non-inverting, gain = 2, and try it. Just be aware that you won't get an output below about 2V with a general-purpose op-amp.