I am measuring sun irradiance captured on several small solar panels.

  1. The power generated from the panels is not consumed;
  2. The measuring device is independently powered;
  3. My focus is on high-precision measurement of current
  4. I have no concern about consuming power during measurement (unless impacting #3).

I am using the following setup:

  1. Independently powered Raspberry Pi
  2. Several ADS1115 (analog to digital) to measure voltage;
  3. Each panel is wired directly to the ADS1115 with a common ground.
  4. Each panel is 5.5 V with max poower of 1.35 W (depending on sun)

I understand Ohm's Law:

Voltage = Current*Resistance 

Current = Voltage / Resistance 

Therefore, I should be able to calculate the current by simple division by the known resistance after measuring the voltage.

I also calculate my max current as 0.25 A per cell.

I have been reading about shunt resistors on Stack Exchange and in texts, but I am unclear on where best to position the shunt. Some suggest parallel while others are series.

My understanding of resistors in parallel, would suggest a different drop in voltage; which I assume would cause a different effect of current. I am unclear how a parallel resistor will impact Ohm's law vs the formula outlined above for a series resistor.


  1. Is there a reason I need to place this resistor in parallel?
  2. Can I place the resistor on the common ground?
  3. Do I need to worry about high amps on my ground-wire if I had many panels being measured?

enter image description here

  • \$\begingroup\$ [ I found a recent article which may be of interest to you: A Simple Solar Irradiation Measurement Technique ] \$\endgroup\$ Commented Jan 17, 2023 at 18:22
  • \$\begingroup\$ I would go witn #3. \$\endgroup\$
    – Gil
    Commented Jan 17, 2023 at 18:36
  • 1
    \$\begingroup\$ If the power generated from the panels is not consumed, how do you have any current to measure? You need some sort of a load for measuring current, otherwise all you can measure is voltage. \$\endgroup\$
    – GodJihyo
    Commented Jan 17, 2023 at 19:09
  • \$\begingroup\$ Welcome! "with max current of 1.35 W" Current isn't measured in W. Do you mean power or A? \$\endgroup\$
    – winny
    Commented Jan 17, 2023 at 21:55
  • \$\begingroup\$ @winny Correct - this was an error. 1.35 W is max Power... \$\endgroup\$
    – GlennB
    Commented Jan 18, 2023 at 0:37

1 Answer 1


Version #1, Version #3 measure solar cell's open-circuit voltage. Only a tiny current flows. The resistors that have been added do almost nothing. The solar cells provide next to NO output power.

1.Is there a reason I need to place this resistor in Parallel?

I'm assuming the purpose of the added resistor is to provide a means of pulling power from the solar cell. A series resistor doesn't do that. A parallel resistor does provide a load on the cell where significant current can flow (depending on the resistor value you choose).

Version #2 does pull current from each solar cell, through the added resistor(s). It is possible to calculate power delivered to each resistor if you know its resistance. Since it likely heats up, its resistance should be independent of temperature...a power resistor of high accuracy likely meets this requirement. The ADS1115 measures voltage, but power and/or current can be found by measuring voltage across the resistor: \$ Power={voltage^2 \over resistance}\$ , \$current = {voltage\over resistance}\$

For OP's solar cells having open-circuit voltage of 5.5V, ADS1115 current flowing into the ADS1115 can be assumed so small as to be insignificant. However, there's a problem choosing the DC voltage applied to ADS1115 Vdd pin:

  • voltage across measuring resistor MUST BE LESS THAN ADS1115 Vdd!
  • serial interface to Raspberry Pi runs at 3.3V (possibly 5V-tolerant).

It is likely safer to supply +3.3V to ADS1115 Vdd for its power supply. But that means your solar cell voltage (possibly as high as +5.5V) is above Vdd. So you'll have to add a resistor voltage divider to the shunt resistor to reduce it below +3.3V.
Let's say you choose to load each cell with a 20 ohm shunt resistor. Doing so will reduce cell voltage from 5.5V to something less, but let's ensure that 5.5V will not overload the ADS1115 input range of 3.3V. Each cell is loaded with 20 ohms, yet the maximum voltage seen be ADS1115 is 3.3V across the 12 ohm resistor, when the cell's voltage is 5.5V:


simulate this circuit – Schematic created using CircuitLab

Both 8 ohm and 12 ohm resistors should be precision types that can dissipate about one watt at least. If single-ended voltage measurement is set up for the ADS1115, four loaded cells can be measured with one chip.

2.Can I place the resistor on the Common Ground?

As mentioned before, Circuit #3 only measures each cell's open-circuit voltage - next to no current flows.

3.Do I need to worry about high Amps on my ground-wire if I had many panels being measured?

Yes, you do, if your intent is to measure individual cells output power. Especially if you consider that ADS1115 resolution is quite good...any ground resistance could cause one cell's condition to influence measurements made on other cells. You want to establish a SINGLE-POINT GND connection, where all those 12-ohm resistors meet (called a STAR gnd).
Also, be aware that your R_PI must also connect to this GND point (required for the I2C serial port connection).
Since each cell's current only flows in a loop through its 20-ohm load (8 + 12 ohms) very very little GND current flows into ADS1115.
If you do this out-of-doors (where the sun shines) then be aware that any ground loops can cause a lightning strike to destroy ADS1115, and R_PI. Wiring lengths should be short, because wire resistance adds to that 20 ohms. All those 12-ohm resistors should meet at the single-point GND with ALMOST ZERO wiring length. If you cannot arrange this, then ADS1115 single-ended measurements shouldn't be attempted. Differential measurements should be done, which reduces the number of cells that each ADS1115 can measure from four (as shown in the schematic) to two.

  • \$\begingroup\$ This is helpful. in your scenario, are we saying that current is simply what I measure as voltage divided by 20? current = v/20? I was confused as i assumed that in the full nature of the circuit, does this contribute 20ohm of resistance? \$\endgroup\$
    – GlennB
    Commented Jan 18, 2023 at 0:41
  • \$\begingroup\$ There are two voltages involved: (1):voltage at solar cell, (2):voltage measured across 12 ohm resistor. Current is voltage measured across 12-ohm divided by 12 ohms. If ADS1115 measured "2.400V" across the 12-ohm resistor, then solar cell current would be 0.200 amps. At the solar cell, voltage would be 20*2.400/12 volts=4.000V. As a check, note that 4V/20ohms = 0.2A \$\endgroup\$
    – glen_geek
    Commented Jan 18, 2023 at 1:30
  • \$\begingroup\$ Thanks @glen_geek This has been very helpful. If I was to change the max current of the cell (i.e. Smaller Cell 2.5v and 200mA max current; how should i work out what ohm resistors to use? \$\endgroup\$
    – GlennB
    Commented Jan 30, 2023 at 1:54
  • \$\begingroup\$ Perhaps you'd like a resistance to load the cell near the point where it supplies maximum power. Somewhere around 70% maximum, in your new case \$2.5\times 0.2 \times 0.7\$ = 0.35 watts. A load resistor with 2.5V across it dissipates 0.35W when its value is near 18 ohms...(\$2.5^2 \over {0.35}\$). Since a loaded cell puts out lower voltage than 2.5V, choose next lower standard resistor value: 15 ohms. Since cell voltage is less than ADS1115 Vdd voltage of 3.3V, you don't need to split this 15 ohm resistor - just measure the voltage across the whole 15 ohms. \$\endgroup\$
    – glen_geek
    Commented Jan 30, 2023 at 13:50

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