non Inverting OP-AMP configuration do not work as expected - Page 1
Forum section: electronics / projects

author: Nikos A.
author_rank: Regular Contributor
timestamp: May 14, 2021, 10:18:40 am
content: Hi everyone,I have a circuit with a non inverting Op-Amp configuration as shown below. I apply a 0.4V ref. voltage on the non inverting input.I am using this Op-Amphttp://ww1.microchip.com/downloads/en/DeviceDoc/20005367B.pdfI supply the Op-Amp with 1.8VIn my case Rs is a variable resistor.This is the formula I am using to calculate the output voltage:Vout = (Vref x Rf) / Rs +VrefThe circuit works fine for high resistor values. For example whenIf Rf = 90Ω and Rs 680Ω I measure around 0.456V which is correctIf Rf = 90Ω and Rs 180Ω I measure around 0.616V which is correctMy problem is that when Rs < Rf I do not have any gain and the output voltage decrease instead of increases... For exampleIf Rf = 90Ω and Rs 56Ω I measure around 0.736V which is incorrect (it should be around 1.05V)If Rf = 90Ω and Rs 33Ω I measure around 0.624V which is incorrect (it should be around 1.50V)Rs (Ω)Vout (V)6800.4562700.5361800.6161000.768820.808560.736330.624100.512As you can see for resistor values less than 80Ω (that is close to Rf) the Vout stop increasing and instead decreasesI do not have this problem for high Rf and Rs values ( in the order of kΩ).Any advice?Nick

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 14, 2021, 10:32:31 am
content: Quote from: Nikos A. on May 14, 2021, 10:18:40 amPossibly, it isn't intended to work with such a low Rload. Datasheet data contains typical load "RL = 20 kΩ" - quite a large value, which may confirm my guess.Typical opamp minimal Rload usually has to be more than 600 R (if it isn't clearly stated).

author: tooki
author_rank: Super Contributor
timestamp: May 14, 2021, 10:47:04 am
content: Quote from: Nikos A. on May 14, 2021, 10:18:40 amCalculate the current that flows from the output through the resistors and compare it to the datasheet.

author: Zero999
author_rank: Super Contributor
timestamp: May 14, 2021, 11:20:54 am
content: Quickly looking at the graphs of the output voltage headroom vs current, gives an output resistance of roughly 100Ω, when the output stage is saturated hard on and the power supply voltage is 1.8V. This output resistance forms a potential divider with the resistors in the feedback network, limiting the maximum output voltage.

author: Alti
author_rank: Frequent Contributor
timestamp: May 14, 2021, 12:54:20 pm
content: You cannot expect the output to drive just any load.It has some limitations, especially when output is close to the rails (sinking near Vss and sourcing near Vdd).

author: Nikos A.
author_rank: Regular Contributor
timestamp: May 14, 2021, 01:07:50 pm
content: Ok when I am working with higher resistors e.g Rf = 100k and Rs = 20k - 40k the Op Amp works as expected..So the problem is with the low resistance values but I cannot figure out the reason..Quote from: Zero999 on May 14, 2021, 11:20:54 amXmmm I am not sure that this is the problem.. When Rs is open circuit and Rf = 100Ω then I can mesure 0.4V at the output (that is the reference voltage). So I think that something else is going on here.Maybe reduced resistance values mean an increased power consumption and additional loading effect on the output? Maybe this limits the output voltage?Quote from: Alti on May 14, 2021, 12:54:20 pmCould you just elaborate more on that? Where can I see the limitation on the datasheet?Thanks

author: jmw
author_rank: Frequent Contributor
timestamp: May 14, 2021, 04:22:19 pm
content: Quote from: Nikos A. on May 14, 2021, 01:07:50 pmIf Rs is replaced with an open circuit then there is no load on the op amp output in the circuit you've drawn above. It's operating as a simple buffer, so it easily matches the IN+ voltage level.I think what you're missing is a real op amp has a non-zero source resistance. There is effectively a series resistor immediately at the output and it's maybe around 100 Ω (it's not a fixed number really).Also, generally you want the feedback loop to draw negligible current. Is there a reason to use resistors in the 10 - 1000 Ω range, when 10 - 1000 kΩ can give you the same voltage divider ratio?

author: Alti
author_rank: Frequent Contributor
timestamp: May 14, 2021, 04:56:08 pm
content: Quote from: Nikos A. on May 14, 2021, 01:07:50 pmYou can start from Figure 2-22 which shows output short-circuit current. You cannot get beyond those ranges, ever.

author: TimFox
author_rank: Super Contributor
timestamp: May 14, 2021, 05:17:35 pm
content: Any circuit will have a finite output current capability. One must check that before designing in any components that draw current from the output. As pointed out by Alti, op amps usually have well-defined short-circuit current limiting. Other circuits may be limited by their power supplies, fuses, or safe operation without smoke.

author: jmelson
author_rank: Super Contributor
timestamp: May 14, 2021, 05:28:51 pm
content: Quote from: Nikos A. on May 14, 2021, 10:18:40 amThe op-amp power supply pin is at 1.8 V? Check the data sheet for how close to the supply rail it can go. Most op-amps cannot go closer than 1-1.5 V to the supply rails. A few so-called rail-rail op-amps can get much closer with limited output load.Jon

author: Zero999
author_rank: Super Contributor
timestamp: May 14, 2021, 09:23:27 pm
content: It seems reasonable to me. I tried simulating it, with an op-amp, with an output resistance of 230R and got similar results. The op-amp model used in the simulation has an output resistance of 10R and doesn't seem to be adjustable, so I put 220R in series with the output to model it.

author: Nikos A.
author_rank: Regular Contributor
timestamp: May 14, 2021, 10:37:09 pm
content: Thank you all for your responces!!Quote from: jmw on May 14, 2021, 04:22:19 pmYes, actually the Rs is a pressure sensorhttps://lcsc.com/product-detail/Pressure-Sensors_UNEO-GD25_C466673.htmlI want to measure kq from 0-22kg and the for 22kg the resistance accross the leads of the sensor is 30-40Ω (varies from sensor to sensor). Then I calculated the feedback resistor Rf so that the output voltage of the op amp to be 1.5V when the applied pressure on the sensor is 22kg (30-40Ω). This is what I want to achieve but it doesn't workQuote from: jmelson on May 14, 2021, 05:28:51 pmThe op-amp power supply pin is at 1.8 V? Check the data sheet for how close to the supply rail it can go. Most op-amps cannot go closer than 1-1.5 V to the supply rails. A few so-called rail-rail op-amps can get much closer with limited output load.This is a rail to rail op amp and I have confirmed that it's out can get almost to 1.8VQuote from: Zero999 on May 14, 2021, 09:23:27 pmThank you for you simulation Zero999, so what is the problem with my circuit? The reason I used this circuit is because I had tested it in the past with other pressure sensors. The diffrence is that those sensos have had a much greater resistance value for the same amount of pressure. If I remember correctly at 22kg the output resistance was around 20k and I was using a 100k feedback resistor. This circuit was warking like a charm.So what is the problem with this one? I do not care to design a new circuit, but I care to learn from my mistakes.Thanks again

author: jmelson
author_rank: Super Contributor
timestamp: May 14, 2021, 11:31:23 pm
content: Looking at the datasheet, I see LOTS of specsmanship!They rate it as worst-case 121 mV from supply rails, but that is with a 2K Ohm load! But, then, they say that's with 0.5 V "overdrive". So, it only does that well when fully saturated, not in the linear range!Also, they rate short circut current at 1.8 V supply as 7 mA. Your readings with the 90 and 56 or 33 resistors come out to 5 mA just in the Rf and Rs, not counting any load on the output.Jon

author: ledtester
author_rank: Super Contributor
timestamp: May 15, 2021, 01:01:03 am
content: QuoteI would try raising the supply voltage of your op-amp to like 5V.

author: Zero999
author_rank: Super Contributor
timestamp: May 15, 2021, 11:29:05 am
content: Quote from: Nikos A. on May 14, 2021, 10:37:09 pmThe problem with your circuit is the op-amp has too lower load resistance. The op-amp IC has an open-loop output impedance of around a couple of hundred Ohms, modelled by Rout in my simulation. It drops a voltage across it, thus limiting the maximum output voltage.The solution is to reduce the op-amp's output current, so its output can swing nearer the positive rail. Suppose the op-amp has an output resistance of 250R and we want the output voltage to reach 1.55V, with a 1.8V supply. 1.8-1.55 = 0.25V, so the output current, can't be any more than 0.25/250 = 1mA. If the minimum value of Rs is 10R, then Vref can't be more than 10mV. Suppose we set Vref to 9mV, just to be on the safe side. We want the output voltage to be 1.55V, so the voltage across Rf is 1.55 -0.009 = 1.541. Select Rf to give that voltage, when the current is 0.9mA. R = 1.541/0.9*10-31k71, the nearest E96 value is 1k69.The problem now is 9mV is very low, but fortunately the op-amp you're using has a low offset voltage, so it's not so much of an issue.

author: tooki
author_rank: Super Contributor
timestamp: May 15, 2021, 12:46:24 pm
content: Quote from: jmelson on May 14, 2021, 05:28:51 pmYup! This is a rail-to-rail op amp, but graph 2-20 shows how the headroom quickly rises with load. The OP’s circuit is literally off the chart here!