Trying to measure IV curve of solar cell in nanoamp range, does my circuit work? - Page 1
Forum section: electronics / beginners

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 03:32:20 am
content: I've got a tiny solar cell operating low-light conditions and i tried to measure the current with my multimeter but it was below 1uA. Since i wanted to plot the current versus voltage i made a circuit i've included below.Now i'm not an electrical engineer so i was wondering if anyone can let me know if my circuit actually works the way i think it does.The battery, the resistors (R3 & R4) and the potentiometer (R1) form a voltage source and divider that lets me impose a desired voltage across the solar panel.I don't have an ammeter that can measure nanoamps so I'm using a 1 megohm resistor (R2) and voltmeter (M1). A nanoamp going through a 1 megohm resistor should give me a 1mV voltage that i can read off the voltmeter. Since the voltmeter (M1) has a 10 megohm internal resistance i calibrate the device first by measuring the combined resistance of R2 and M1 with an ohmmeter. I got a total shunt resistance value of 892 K ohms. I divide the shunt voltage by the shunt resistance (892k) to get the shunt current.Now to measure the voltage across the solar cell I use M2. It's connected outside the ammeter circuit (M1 & R2) because i found the current draw of the voltmeter (M2) (due to its internal resistance) was enough to throw off the results of ammeter if i connect it to the solar cell only. I know the way I set it up to gives me the voltage of the ammeter and solar cell together but I correct for that mathematically by subtracting (or in my case adding, since i connected it backwards) the ammeter voltage and the total voltage.To get my IV curve i sweep the potentiometer across the voltage range of the solar cell and read off the voltages from the voltmeters. Then i perform the necessary mathematical conversions.It seems to work, i get the curves of the right shape an everything, but I'm not an engineer. Is my circuit and methodology sound or am i missing something huge (like bias currents totally destroying my accuracy or something)?I know opamps could make my life easier, but all of my stock ones had bias currents of tens of nanoamps, so i couldn't use them to resolve single nanoamps.Thanks for any an all help.

author: Paul Price
author_rank: Super Contributor
timestamp: November 13, 2016, 04:02:12 am
content: Since you seem so willing to make many calculations anyway to get the voltage across the solar cell that you want I see no reason why you could not also just connect the M2 voltmeter directly across the solar cell array and make a calculation to cancel out the constant (10Meg input resistance)shunt resistance of the M2 voltmeter.Otherwise, if your curves fit, wear it. There are no other bias currents other than what your biasing circuit creates.

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 04:09:45 am
content: Quote from: Paul Price on November 13, 2016, 04:02:12 amI'm incredibly lazy!I don't have a means of accurately measuring the 10Meg range so i avoid the problem by putting the voltmeter outside the ammeter. At that point i just need to add the voltages together.That's actually mathematically simpler than compensating for the internal resistance of the M2 voltmeter...... or at least i think that's how it works. Is measuring the way i have it in the diagram actually NOT additive like i think it is?

author: Paul Price
author_rank: Super Contributor
timestamp: November 13, 2016, 04:21:13 am
content: Anyone who bothers to make this kind of circuit can never be called lazy, but what you are trying to and do for what purpose with the great possibility that your goal might better be achieved otherwise, I donno, all this trouble seems to me to be just futzing around for GNW reason.In the absence of light upon the solar cell array, the only current flowing through the solar cell array would be due to the biasing voltage across the solar cell by your circuit.As you increase the voltage with the adjustment pot you will reach the exact voltage of the voltage divider you've created on the right. At this point there is no current flow through the solar cell and no bias current to measure.Whether or not the M2 voltmeter is connected or not will not make any difference.Exciting the solar cell array a very small amount of light will then produce a voltage change that will induce a current to flow through M1 meterbut the voltage across M2 will not change measurably at the very low light level you are trying tomeasure at.What you have here is a galvanic bridge type of measuring circuit.

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 04:25:03 am
content: Quote from: Paul Price on November 13, 2016, 04:21:13 amIs this for the case where the voltmeter is connected outside like in my diagram? or for the case where it's connected directly to the solar cell?

author: Paul Price
author_rank: Super Contributor
timestamp: November 13, 2016, 04:29:01 am
content: This is the case for both placements of M2, with no light applied, and the bridge has been balanced by adjustment of the pot. At this point it also does not matter whether the solar cell array is left in the circuit or taken out of the circuit and put into your pocket.

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 04:34:27 am
content: Quote from: Paul Price on November 13, 2016, 04:21:13 amMy apologies i just saw this and i don't know why i missed it.To answer: I'm trying to measure the IV curve of the solar cell. I don't have equipment sensitive enough to measure nanoamps so i made my own from two voltmeters and some resistors and batteries i had floating around.Quote from: Paul Price on November 13, 2016, 04:29:01 amMakes sense. Would i still be able to use my circuit to measure the rest of the IV curve? Would i get larger errors the further i went away from the balanced case you mentioned?

author: Paul Price
author_rank: Super Contributor
timestamp: November 13, 2016, 05:04:09 am
content: Depends on the direction you move away from the balance point, if you are injecting current into the solar cell array then you are not measuring what you think you are measuring.Let me put it simply: You could use just about any op-amp instead. If the op-amp has bias currents, you could measure them by looking at the output voltage without anything but a very high resistance load for the solar cell array connected to the inputs to grouind. If you use the same value of resistance on both the + and minus inputs of the op-amp, the effect of bias currents on both input pins will almost completely cancel each other out. What small error left can be consistently and accurately calculated out.After you've measured the effect of both the offset voltage and biasing currents of the op-amp, you would then connect the solar cell.In the case of using an op-amp instead, you will not be inducing anything near the biasing current(both forward and reverse)l created with your circuit.If the load resistance for the solar cell is small in value, then the effect of biasing currents will be very small on the solar cell array. If you connect the solar cell array to the positive input of the op-amp and use the negative input and a feedback resistor from the positive output to the negative input to create a precise voltage gain, then you can get, with just a single calculation, a conveniently amplified and accurate voltage versus current curve. If you bother to find a FET input op-amp, your expected bias currents will be in the pico or femto-amp range.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 13, 2016, 05:14:45 am
content: Notice the structure you've developed: a series ammeter (for all intents and purposes), with a parallel voltmeter outside that.This is placed inside a nulling bridge, so you can measure the voltage and current for various applied voltages.You have only one alternative: the series-parallel transformation, i.e., voltmeter in parallel with the DUT, ammeter in series with that. (A series-parallel transformed bridge looks the same, but requires a current source instead of a battery. But the bridge's resistance is nearly constant, so we don't care.)Which one is appropriate for your test depends on the parameters of the meters, and what V/I range you need to test.The voltage error, due to the ammeter, results in this calculation:V(actual) = V(meas) + R(ammeter) * I(meas)I(meas) = I(actual), of course.For the other arrangement, you get:I(actual) = I(meas) - V(meas) / R(voltmeter)V(meas) = V(actual), of course.Now, R(voltmeter) is often poorly defined, and variable. My DMM has a "high Z" mode under 400mV, and is about 10M on the higher ranges; but it's some oddball value, and it's tricky to measure (set up a voltage divider?).That said, as long as the shunt voltage drop is in the "High Z" range, you can rest assured that the ammeter's resistance will be precisely the value of the shunt resistor you used. If the voltage drop is insignificant, you can go ahead and ignore it anyway.This makes this design quite handy for high impedance tests: if you're measuring nA and V, that's gigs of impedance, and your 1M shunt doesn't matter much.In the other direction, if you had mV and uA, you'd have kohms of impedance, and the parallel-series form would be better.Tim

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 05:27:03 am
content: Quote from: Paul Price on November 13, 2016, 05:04:09 amSo my circuit doesn't work like i think it does and i gotta use the opamps. bummer. oh well. I'm glad you told me before i presented this. It's clear to me i really don't know what i'm doing at all. The data i collected seemed almost too good. Guess i'll hit the books again.

author: Paul Price
author_rank: Super Contributor
timestamp: November 13, 2016, 05:28:23 am
content: You have only one alternative.. I disagree with this comment.You have many alternatives.When you use an op-amp you can forget about the impedance of any single voltmeter you will use and you are measuring what you want, the characteristics of your solar cell array output voltage versus a constant load resistance(giving you the current value) and without significant biasing voltages or currents.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 13, 2016, 05:32:59 am
content: Quote from: Paul Price on November 13, 2016, 05:28:23 amI meant topologically. But yes, there are other solutions to the overall problem.Tim

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 05:37:30 am
content: Quote from: T3sl4co1l on November 13, 2016, 05:14:45 amAs said in my opening post I think i'm eliminating the voltage error by indeed adding the ammeter measurement as you show inV(actual) = V(meas) + R(ammeter) * I(meas)As for the internal resistance of the voltmeter being poorly defined. I think i solve that by directly measuring it along with the 1M shunt. I get a value of 892k which is clearly not 1M and should be the combination of the two resistances.By putting the second voltmeter M2 in parallel with the ammeter and the DUT i can ignore its internal resistance since the ammeter is still measuring entirely the current in the solar cell..... At least that's what i think i'm doing. If i'm wrong let me know.I'm reading that i should be using op amps so i might be completely off in my design.

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 05:52:06 am
content: Here is the data after i processed it.(processed by basically finding the solar cell voltage by adding the shunt voltage and the total together, then dividing the shunt voltage by the shunt resistance to find the shunt current. power is solar cell voltage times solar cell current)What error is my circuit causing? what SHOULD the curve look like? How do i fix my math?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 13, 2016, 05:59:47 am
content: Yes, that is correct.The V-I curve should indeed have a knee, because it's a diode. The effect of light is to shift the curve to the right, so that instead of reaching 0A at 0V, there's a zero offset. And so, between 0A at <some> volts (here, about 2V), and 0V at <some> amps (900nA here), there is some amount of powerdeliveredfrom the device. And since the knee is towards the high-voltage side, the peak power point is also there. Usually something like 80% of open circuit voltage.The ideal linear condition, for a Thevenin source, would have the power peak exactly in the middle. So you can see, diodes are weird.TimP.S. This isn'ttheNurdRage, is it..?

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 06:01:59 am
content: Quote from: T3sl4co1l on November 13, 2016, 05:59:47 amSo i'm using my circuit correctly and processing my data correctly? (just confirming i haven't totally messed this up and not realized it)

author: TheAmmoniacal
author_rank: Unknown
timestamp: November 13, 2016, 11:24:33 am
content: Quote from: T3sl4co1l on November 13, 2016, 05:59:47 amIt istheNurdRage of YouTube chemistry fame.

author: 2N3055
author_rank: Super Contributor
timestamp: November 13, 2016, 02:26:46 pm
content: Quote from: NurdRage on November 13, 2016, 06:01:59 amNope, those curves look pretty much perfect...Good, quick overview:http://www.ni.com/white-paper/7230/en/P.S: My son loves your channel... He studies chemistry and says you're great !! He started a program on his university called "Traveling scientists", they go to elementary schools and promote science, inspired by your work...

author: NurdRage
author_rank: Contributor
timestamp: November 13, 2016, 06:45:35 pm
content: Quote from: 2N3055 on November 13, 2016, 02:26:46 pmThanks.

author: TheAmmoniacal
author_rank: Unknown
timestamp: November 13, 2016, 06:49:25 pm
content: I also started out as a chemist before I realized my university only had ChemE, and the2nd and 3rdsemester had no chemistry at all - just a bunch of math and physics (Calculus I, Calculus II, Linear Algebra and Differential Equations, Fluid Dynamics, Thermodynamics, Kinetics and Statics ... ). Dropped out of that shit and pursued a more research-oriented biotechnology / biochemistry degree. It's a good compromise

author: 2N3055
author_rank: Super Contributor
timestamp: November 13, 2016, 06:51:34 pm
content: Quote from: TheAmmoniacal on November 13, 2016, 06:49:25 pmWell I wasn't very precise, he also studies biotechnology but he loves chemistry....

author: TheAmmoniacal
author_rank: Unknown
timestamp: November 13, 2016, 06:59:28 pm
content: Quote from: 2N3055 on November 13, 2016, 06:51:34 pmI'm aiming for the millions of dollars I'll get as a Monsanto shill!

author: 2N3055
author_rank: Super Contributor
timestamp: November 13, 2016, 07:48:11 pm
content: Quote from: TheAmmoniacal on November 13, 2016, 06:59:28 pmLOL you and him both...He's actually more of a "I'm gonna cure cancer AND make money..." kids...Good luck to you anyways, it is a good vocation, good prospects...Take care!!