MOSFET as a variable 0-5 ohm course cal resistor - Page 1
Forum section: electronics / projects

author: UPI
author_rank: Regular Contributor
timestamp: February 22, 2013, 10:48:24 pm
content: I would like to replace a calibration resistor with a MOSFET to simplify a relatively course calibration but do not have any experience with analog design.Currently, a external PCB with a set of series resistors can be tied across JP1 to put a 0-5 ohm resistance in series with the load. The resistors are 3 watts and lower and there are a total of 6.I would like to replace this batch of series resistors with a MOSFET so that I can dial in the series resistance by manipulating the gate. I have +5V, +12V and -12V to utilize in the control of this "series resistance". If possible, I would like to use IRFZ44 N-Channel MOSFETS since they are already in use in other parts of the circuit. The gate should be adjustable using a pot.Any suggestions/clues are welcome.

author: c4757p
author_rank: Super Contributor
timestamp: February 22, 2013, 10:57:40 pm
content: I'm not completely sure how you are trying to connect this, but a variable constant resistance in general is pretty easy. Just put the pot in parallel with the MOSFET, and give both of them current sense resistors. If you drive the MOSFET to take 10000 times the current as the pot, then you will have 0-5 ohms across the FET for 0-50k ohms across the pot.Obviously you have to compensate for the current sense resistors (and parallel pot resistance, if it's close enough) when you calibrate the dial.

author: Marco
author_rank: Super Contributor
timestamp: February 22, 2013, 11:21:59 pm
content: Does it have to be at the high side of the load or can it be between the load and ground? Over what frequency range should it behave as a resistor?

author: UPI
author_rank: Regular Contributor
timestamp: February 22, 2013, 11:30:20 pm
content: Quote from: Marco on February 22, 2013, 11:21:59 pmIt is possible to do it on the low side but I would prefer to do it on the high side to prevent confusion for technicians working on the equipment.It is a very slow changing 0-5V analog drive signal for a vision system's camera lighting.Based on the response from c4757p, I located the following circuit. I am just trying to fully digest how it works and then to figure out how to modify its range to 0-5 ohms.http://www.eetimes.com/design/power-management-design/4009609/Electronic-rheostat-provides-decades-of-load-resistanceThanks

author: c4757p
author_rank: Super Contributor
timestamp: February 22, 2013, 11:35:10 pm
content: Yep, that's precisely the sort of thing I was talking about.

author: amspire
author_rank: Super Contributor
timestamp: February 22, 2013, 11:44:52 pm
content: If you want something that looks like a real resistor (ie V=R*I), what you need is a dummy load circuit, but instead of using a voltage source as the input reference, you use the mosfet drain voltage as the reference input. There have been lots of circuits for dummy loads here - Dave did a great video blog on a design, and I posted a design:https://www.eevblog.com/forum/projects/switching-pre-regulator-gp-power-supply/msg102518/#msg102518If you take my design, you would remove C1 and U2, and connect the free end of R1 to the mosfet drain. The pot would then control the resistance of the dummy load. You will need a battery to power the circuit. You would probably want to reduce the current sense resistor to perhaps 0.1 ohms and if precision is required, you would need a lower offset opamp as long as the inputs can go to zero volts. The LM324 is not too bad.The IRFZ44 would be an excellent device for the mosfet as long as long as you stick to the Safe Operating Area in the datasheet. If you want a maximum current of 4A, that would mean it can operate up to 30V (with a great heatsink or for a short time).

author: Marco
author_rank: Super Contributor
timestamp: February 23, 2013, 12:02:06 am
content: Quote from: UPI on February 22, 2013, 11:30:20 pmAFAICS that one is designed for low side use. It's a tricky circuit, I think for the highest resistance values it simply uses all the sense resistors as a series shunt ... for the low resistance values it puts the MOSFET and part of the sense resistor chain in parallel with the rest of the chain.If I am reading it right then to use that one at the high side of the load the ground for the feedback loop and the sense resistors has to be connected to the high side of the load (also for the battery and opamp depending on the voltage of the power supply, if the power supply voltage is low enough it can function directly as the opamp power supply you can leave the battery out and just use the supply voltage and normal ground for the opamp ... otherwise you need to float the entire circuit, which is easiest with a battery).

author: UPI
author_rank: Regular Contributor
timestamp: February 23, 2013, 12:15:15 am
content: Quote from: Marco on February 23, 2013, 12:02:06 amI believe you are correct.I will try to reformat and digest Richard's design later tonight and see how it looks for the job. I don't see any issues at first glance but I will need to figure out how to scale it down to do just 0-5 ohms or so.

author: UPI
author_rank: Regular Contributor
timestamp: February 23, 2013, 03:08:00 am
content: Quote from: amspire on February 22, 2013, 11:44:52 pmPlease see my attached interpretation of Amspire's design and suggestions.If I understand it correctly, I either have to use it on the low side (between ground and the load) or on the high side with an isolated supply/battery.Does this look and sound right?Thanks

author: amspire
author_rank: Super Contributor
timestamp: February 23, 2013, 04:00:12 am
content: Replace VR1 with a linear 1K pot (value doesn't matter much) from ground to the drain. Connect R1 to the wiper. I just used the arrangement in the diagram as I was using junk parts including a bracket with a mounted pot with one leg chopped off.

author: UPI
author_rank: Regular Contributor
timestamp: February 23, 2013, 04:18:22 am
content: More like this?

author: amspire
author_rank: Super Contributor
timestamp: February 23, 2013, 05:07:31 am
content: Perfect. I think the output resistance would be 0.1 / ((% pot position)/100).In other words, if the pot is at 10%, it will have a 1 ohm resistance. At 3.3% it will have a 3 ohm resistance. At 100%, a 0.1 ohm resistance.If you want to limit the range to, say 1 ohm or more, add a 9K resistance from the pot to the drain, and then 100% will be 1 ohm and 33% will be 3 ohms.

author: UPI
author_rank: Regular Contributor
timestamp: February 23, 2013, 05:26:49 am
content: I will try to build and characterize this tomorrow.Thank you very much for your help.

author: Marco
author_rank: Super Contributor
timestamp: February 23, 2013, 05:27:34 am
content: Quote from: UPI on February 23, 2013, 03:08:00 amAFAICS the load is supplied by 5 volt, but you have 12 volt for the driver right? In that case the opamp can just run off 12 volt and ground and still have plenty of headroom to drive a MOSFET at the high side of the load.

author: BravoV
author_rank: Super Contributor
timestamp: February 23, 2013, 02:48:36 pm
content: Is there any minimum load (DUT) voltage for this circuit to operate properly ? Also how about max voltage ? For example the dut's is at 24 volt ? I havent read the mosfet spec yet, lets assume this 24 v is still way below it's max Vds.

author: UPI
author_rank: Regular Contributor
timestamp: February 23, 2013, 03:37:21 pm
content: Here is the entire circuit with the lamp driver and series resistance.

author: amspire
author_rank: Super Contributor
timestamp: February 24, 2013, 12:57:50 am
content: Looks good. Are you are going to breadboard the resistance circuit before a proper build? I would.Stability is the biggest issue.

author: Marco
author_rank: Super Contributor
timestamp: February 24, 2013, 02:46:37 am
content: What was wrong with c4757p's initial suggestion?

author: c4757p
author_rank: Super Contributor
timestamp: February 24, 2013, 02:56:31 am
content: Quote from: Marco on February 24, 2013, 02:46:37 amMy suggestion is identical to Amspire's - you're still forming a variable voltage divider across the load. I just kept the lower resistance fixed, while he is using one side of the potentiometer instead.An advantage to his approach is that if you do it the way I suggested, you have to take care that the minimum resistance does not draw too much current and damage anything, whereas his method keeps the total resistance, and the current and power dissipation, fixed.

author: UPI
author_rank: Regular Contributor
timestamp: February 24, 2013, 04:44:23 am
content: Quote from: amspire on February 24, 2013, 12:57:50 amI gathered all of the parts together today for a prototype but more pressing work got in the way of getting it assembled. We hope to be able to get it assembled and tested Monday.

author: UPI
author_rank: Regular Contributor
timestamp: February 25, 2013, 09:41:17 pm
content: Works like a champ! See the attached schematic for the actual values/parts used.It was very stable and would adjust from .216 ohms (including the .1386 ohm current sense resistor) through ~4 ohms (we added the 49.9 in series with the pot to limit the max resistance to 4 ohms).I did not have time to characterize how quickly it could respond to voltage changes in the 0-5V lamp driving circuit but our signal changes approx. 2.5V/sec so I don't think it will be an issue. If we decide to pursue this as an actual product for sale, we will datalog the voltage drops to make sure it can keep up.One question: If we shorted across the drain/source to eliminate this entirely from the lamp circuit, would it cause any issues? It did not appear to be a problem in the prototype.Thanks for the help!EDIT: Our max current through the circuit was ~1.3A at a max voltage of 5V from the lamp drive circuit.

author: amspire
author_rank: Super Contributor
timestamp: February 25, 2013, 11:02:14 pm
content: Quote from: UPI on February 25, 2013, 09:41:17 pmShorting the drain/source is fine. Really glad to hear you had success. Well done.