Sense resistor for fan current - Page 1
Forum section: electronics / projects

author: iXod
author_rank: Frequent Contributor
timestamp: August 26, 2020, 05:38:58 pm
content: I'm using a Microchip TC649B:http://ww1.microchip.com/downloads/en/DeviceDoc/21755C.pdfto PWM a 48vDC, 1.6A BLDC 2-wire fan.The examples in the datasheet use a 12v fan but since the supply for the fan and IC can be separate I'm using a 48v fan for better CFM (and I have it in my junque box).The sense input to the IC calls for a resistor in series with the fan to give a pulse sense signal in the range 50mV-90mV.I'm having a bit of a difficult time finding a resistor or shunt. 50mV would require 0.031; 90mV would require 0.056. This is a through-hole board, but a prototype, so even SMD I could adapt.I thought about using whatever resistor/shunt I can scare up and add a dropping resistor between the sense resistor/shunt and the sense input pin of the IC.Any reason this dropping resistor configuration wouldn't work?Thanks.

author: Renate
author_rank: Super Contributor
timestamp: August 26, 2020, 06:35:00 pm
content: Are you building a leaf blower?You've powered this thing up already and determined that the noise level is acceptable?Just asking.Yeah, with 48V a bit of voltage drop won't kill you.OTOH, 1 ohm is 2.5 W0.1 ohm will get you down to 1/4 W

author: iXod
author_rank: Frequent Contributor
timestamp: August 26, 2020, 07:05:38 pm
content: QuoteHence PWM...QuoteI don’t see an answer to my question there.Will there be any issue placing a dropping resistor between the sense resistor and the IC sense input?See datasheet, p.26, for example schematic. There’s no input impedance listed.

author: Renate
author_rank: Super Contributor
timestamp: August 26, 2020, 07:18:27 pm
content: Quote from: iXod on August 26, 2020, 07:05:38 pmSome fans don't take too kindly to being operated under 20% power.Quote from: iXod on August 26, 2020, 07:05:38 pmWell, your source impedance is 1 ohm or less.We don't know what the input impedance is.In worst case you will only need a series resistor instead of a divider.

author: ledtester
author_rank: Super Contributor
timestamp: August 26, 2020, 07:26:17 pm
content: See page 19 of the datasheet for information on selecting Rsense and Csense.Update: If I read Renate's remarks correctly, just try Rsense of 1 ohm with a 3 W power rating, or a 0.1 ohm resistor with a 1/4W power rating. The 90mV figure is a threshold -- i.e. a minimum -- it's okay if the voltage across Rsense is greater than that.

author: Renate
author_rank: Super Contributor
timestamp: August 26, 2020, 09:37:03 pm
content: I meant that 1.6 A through 0.1 ohm gives you 160 mV to play with.If you want a level of 50 mV and...If the input impedance is (virtually) infinite then an 11 k top resistor and a 5 k bottom resistor will make a nice divider without affectin the 0.1 ohmIf the input impedance is 1 k then a single series resistor of 2.2 k will do the job.In point of fact, I'll bet that is a comparator input and the impedance is high anyway.

author: ledtester
author_rank: Super Contributor
timestamp: August 27, 2020, 12:06:50 am
content: The functional block diagram (page 2) suggests the input impedance of the SENSE line is 10K:

author: Renate
author_rank: Super Contributor
timestamp: August 27, 2020, 02:15:24 am
content: Quote from: ledtester on August 27, 2020, 12:06:50 amAh, nice catch.So, for our hypothetical 0.1 ohm shunt and 50 mV goal, that would mean we could use a single 22 k resistor.1.6 Ampere * 0.1ohm * 10 kohm/(10 kohm+22 kohm) = 50 mV

author: Vovk_Z
author_rank: Super Contributor
timestamp: August 28, 2020, 04:29:33 pm
content: I'm not sure we can definitely rely on these 10k. This value can vary, so it is better to make outer divider with lower impedance.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 28, 2020, 09:05:31 pm
content: I have not looked at the datasheet, but I assume the sense voltage should be relatively fixed for a certain current.One easy solution is to use a potentiometer over your shunt resistor.If you use a 10 Ohm potentiometer, then it's 100x bigger then your 100mOhm or so shunt, and also 3 orders of magnitude smaller then the input of your IC, so it's input impedance effects are negligible.A disadvantage of this is that the voltage over the shunt must be bigger then the sense voltage you need, but it's very simple.Of course 2 fixed resistors also work.Alternatively, you can use a series and/or parallel combination of standard E12 resistors for your shunt to get it to just the right value.The most often used way is to use a low value shunt, and then amplify the voltage over it, but this would need an extra opamp, with very low offset voltage and surrounding components, possible bandwith problems, etc.Another option is to use a smallish (10 to 100 Ohm or so) resistor between the active side of the Shunt and the IC, and then feed a fixed current through it to add a fixed offset.If you have a relatively stable supply voltage, then a simple resistor as "current source" is probably good enough.

author: Renate
author_rank: Super Contributor
timestamp: August 29, 2020, 02:29:46 am
content: Quote from: Vovk_Z on August 28, 2020, 04:29:33 pmMaybe, that's why we should check.I'm betting that the input impedance is basically infinte and they throw on a 10,000 ohm resistor.If it checks out as 10k with the (unpowered) bare IC...