Current Source with MOSFET not working as expected - Page 1
Forum section: electronics / beginners

author: Mickkkk
author_rank: Newbie
timestamp: April 13, 2024, 01:21:01 pm
content: Hello,#####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-1What 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-2If 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-3If 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?#####Thank you.

author: antercreeper
author_rank: Contributor
timestamp: April 13, 2024, 01:43:51 pm
content: 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.

author: Mickkkk
author_rank: Newbie
timestamp: April 13, 2024, 01:46:33 pm
content: Quote from: antercreeper on April 13, 2024, 01:43:51 pmYes... 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

author: pqass
author_rank: Super Contributor
timestamp: April 13, 2024, 03:23:29 pm
content: Quote from: Mickkkk on April 13, 2024, 01:46:33 pmThat can never be. It's Io = (VDC-Vin)/RsWhat'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 ofmodeling it herebut can't produce higher than 522mA before running the opamp output into ground.

author: Zero999
author_rank: Super Contributor
timestamp: April 13, 2024, 03:52:22 pm
content: Quote from: pqass on April 13, 2024, 03:23:29 pmNot with that circuit, but it is possible. Use potential dividers to form a Howland topology.R3 = R1R4 = R2AV= R1/R2I = V2/R5*AV

author: Mickkkk
author_rank: Newbie
timestamp: April 13, 2024, 04:56:20 pm
content: Quote from: pqass on April 13, 2024, 03:23:29 pmThanks pqass... That's my mistake... should be Io = (VDC-Vin)/RsActually 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.

author: Mickkkk
author_rank: Newbie
timestamp: April 13, 2024, 05:00:52 pm
content: Quote from: Zero999 on April 13, 2024, 03:52:22 pmThank you very much... A great solution... Will look into the LTspice model soon...