analogcurrent-sourcejfetmosfet
Is it possible to build MOSFET current source similar to JFET current source like this:
simulate this circuit – Schematic created using CircuitLab
With MOSFET it should be something like this:
However, MOSFET current source doesn't work.
As far as I know, it should be possible to do this, since JFETS are controlled in the same manner as MOSFETs.
Or am I wrong?:)
I would like to know if it is possible to produce an n-channel jfet connected in fixed bias configuration with a high gain of 200.
For your circuit, the product of the small-signal transconductance and drain impedance at the operating point set an upper bound on the voltage gain.
Using the numbers @Andy aka provides:
\$A_v < 20k\Omega \times 6.5 \frac{mA}{V} = 130\$
Since your drain resistor and load are in parallel with the small-signal drain impedance, your circuit gain must be less than this upper bound.
So, as a first step, you must select a transistor with maximum possible voltage gain quite larger than 200.
In your first circuit V1 acts like the output of a synchronous buck power modulator, or a half bridge. No matter the state of V1, current in L1 has a good path.
In your second circuit, when M1 is turned off, current in L1 has no place to go, and an impossible situation occurs. That's what Andy is talking about, using a flyback diode to give L1 current a path to flow when M1 is off.
If you want to continue to use an N-Channel FET and switch it referenced to ground, you could change the circuit to something like this:
This is a buck with Vout referenced to V1 (12V) so, Vout (voltage across C1 referenced to ground) will be V1 - D V1, where D is the duty cycle.
Best Answer
To let current pass with a regular (enhancement) N-FET, the gate needs to be at a higher voltage than the source. This is impossible in your schematic, because of the voltage drop at R1. The mosfet is always off.
However, this circuit works with a depletion FET. Depletion FETs let current pass when gate is at 0V, and turns off when the gate drops negatively below the source voltage (under a given threshold). This is like an enhancement FET, except that the gate threshold is negative instead of being positive (this makes it some kind of "normally connected" FET).
So, with your circuit and a depletion FET: if the current increases, it increases the voltage drop across R1 and therefore lowers the gate voltage, which will turn the mosfet off a little. If the current decreases, the mosfet turns on a little. This will always find equilibrium and you have your current source.
More info about the usage of depletion FETS here: Typical use of depletion MOSFET