Current Source with MOSFET not working as expected

[Mickkkk]

Hello,

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Aim to build a current source using an op-amp, a MOSFET and a sensing resistor. The input reference voltage Vin is applied to the non-inverting input terminal of the op-amp. The voltage at the inverting input terminal of the op-amp does not track the Vin at the non-inverting input terminal of the op-amp.

Question-1
What are the possible reasons for the substantial difference (approximately 0.6V) in voltages at the inverting and non-inverting input terminals of the op-amp?

Question-2
If the op-amp operates in linear mode, is it normal to have a substantial difference (approximately 0.6V) in voltages between the inverting and non-inverting input terminals of the op-amp?

Question-3
If the op-amp operates in linear mode, is it normal to have such a substantial amount of current (approximately 0.5 mA) flowing into the inverting input terminal of the op-amp?

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Thank you.

[antercreeper]

A adjustable current source by changing Vin?

AFAIK, The reason is that devices aren't ideal in the real world. (instead, fulls of features, for example MOS channel length modulation, The Op amp's input's current, etc..)

From some little previous knowns(maybe not correct  :'( ),
That mos cannot provide enough impedance(some called gs, about dozens of Siemens).(both Vgs and Vds can affect how many currents flow), so you need to build a feedback, input is the error between how many current should flow and how many current really flow, output can be Vgs.
and then do some loop analysis and compensation on it.

[Mickkkk]

A adjustable current source by changing Vin?

AFAIK, The reason is that devices aren't ideal in the real world. (instead, fulls of features, for example MOS channel length modulation)

From some little previous knowns(maybe not correct  :'( ),
That mos cannot provide enough impedance(some called gs, about dozens of Siemens).(both Vgs and Vds can affect how many currents flow), so you need to build a feedback, input is the error between how many current should flow and how many current really flow, output can be Vgs.
and then do some loop analysis and compensation on it.

Yes... aims to build an adjustable current source by controlling the Vin... and hope that the output sourcing current is simply given by Io = Vin/Rs

[pqass]

Yes... aims to build an adjustable current source by controlling the Vin... and hope that the output sourcing current is simply given by Io = Vin/Rs

That can never be.  It's Io = (VDC-Vin)/Rs

What's the expected current range?
What specific P-FET are you using? 
I think VDC=4V is low for most general purpose FETs.

I've taken a crack of modeling it here but can't produce higher than 522mA before running the opamp output into ground.

[Zero999]

Yes... aims to build an adjustable current source by controlling the Vin... and hope that the output sourcing current is simply given by Io = Vin/Rs

That can never be.  It's Io = (VDC-Vin)/Rs

Not with that circuit, but it is possible. Use potential dividers to form a Howland topology.

R3 = R1
R4 = R2

A_V = R1/R2

I[R_L] = V2/R5*A_V

[Mickkkk]

Yes... aims to build an adjustable current source by controlling the Vin... and hope that the output sourcing current is simply given by Io = Vin/Rs

That can never be.  It's Io = (VDC-Vin)/Rs

What's the expected current range?
What specific P-FET are you using? 
I think VDC=4V is low for most general purpose FETs.

I've taken a crack of modeling it here but can't produce higher than 522mA before running the opamp output into ground.

Thanks pqass... That's my mistake... should be Io = (VDC-Vin)/Rs

Actually my friend threw me this question. He did not share much about the details, e.g. the expected current range or the specific P-FET that my friend is using. However, my friend did tell me that the Vgs(th) of the P-FET he's using is -1.5V (typ).

I agree with you that a VDC ~= 4V is quite low for most general purpose FETs.

[Mickkkk]

Yes... aims to build an adjustable current source by controlling the Vin... and hope that the output sourcing current is simply given by Io = Vin/Rs

That can never be.  It's Io = (VDC-Vin)/Rs

Not with that circuit, but it is possible. Use potential dividers to form a Howland topology.

R3 = R1
R4 = R2

A_V = R1/R2

I[R_L] = V2/R5*A_V
(Attachment Link)

Thank you very much... A great solution... Will look into the LTspice model soon...