I am trying to accurately measure the power a solar panel produces over a day as cheaply as possible. To accomplish this I am trying to build a "dummy load" that will hold the panel at its Vmp and immediately dissipate the energy via some load (avoiding the need for a battery). The challenge here is that the voltage will remain constant but the current will change as the panel's illumination changes. Any suggestions as to how to do do this? Should I use a different approach?

The panel I am working with has a Vmp of 17V and a Pmax of 75W.

  • \$\begingroup\$ If you keep the voltage constant it won't be at its Vmp very often. \$\endgroup\$
    – user16324
    Commented Oct 18, 2016 at 15:57
  • \$\begingroup\$ @BrianDrummond I agree but I'm happy with it being close. Some Linear Technology ICs just hold the panel at Vmp \$\endgroup\$
    – RKsquared
    Commented Oct 18, 2016 at 20:23
  • \$\begingroup\$ Vmp doesn't change very much with lighting level. The plan is OK, I think. You need a buck type converter, except instead of regulating Vout, you want to regulate Vin to Vmp. Vout will be connected to your 75W resistor (which must dissipate 75W at a voltage slightly lower than Vmp). You will need to measure the current and voltage delivered to the resistor over time to calculate total energy dissipated by the resistor over one day. it might be easier to use a battery than a resistor. Less heat to deal with (as long as the battery capacity is adequate). \$\endgroup\$
    – user57037
    Commented Oct 19, 2016 at 4:14
  • \$\begingroup\$ @mkeith are you suggesting to put a regulator between the panel and the load so the voltage at the load is constant? \$\endgroup\$
    – RKsquared
    Commented Oct 19, 2016 at 15:34
  • \$\begingroup\$ Not a normal regulator. A normal regulator senses vout and regulates current to maintain vout. What you need is a regulator that senses Vin and regulates Iout to maintain Vin at the assumed maximum power point. Just like the Linear documents describe. \$\endgroup\$
    – user57037
    Commented Oct 19, 2016 at 22:16

5 Answers 5


A solar panel is a current source but has a Voc limit where Vmp =85%Voc at full current and drops to 70~75%Voc at <10% rated short circuit current with overcast or 10% solar power input.

This applies at 25'C to many, but not all PV's.

I like to model PV's like diodes so that the voltage will rise with current to match the Vmp for increasing solar power as close as possible without tracking.

  • if PV is rated Vmp=17V @75W

    • then Imp = 4.41A and ESR = 17/4.41=3.85 Ohms
    • Let's estimate the 70%Voc at 10% solar input of 0.441A and thus the equivalent Zener active load will have an ESR and a Vth threshold voltage.

    • if Voc max in = 17V/85%=20V

    • then Voc 10%in= 70%*20V=14V
    • thus ESR = (17-14)/4.41-0.44)=0.75 Ohm

So your Active load should be 14V regulated sink+ 0.75 Ohm

How you dissipate 85W and regulate this load is up to you and there are many many ways to implement this after you have a datasheet with these parameters to verify ( if possible please)

Series drop resistor or two 50W 14V headlamps drops to 5A LDO (quasi PNP emitter out) to a 100mV current shunt at 5V to ground. 20 milliohm calibrated wire.



simulate this circuit – Schematic created using CircuitLab

Adjust the Regulator to ~12.7V such that PV is 17V at max solar power. Then LDO input is ~13.8V and R12 such that I shunt reads 4.4Amp*20mOhm = 88mV


Maximum Power Transfer Function Theorem wiki

the resistance of the load must equal the resistance of the source as viewed from its output terminals. Moritz von Jacobi published the maximum power (transfer) theorem around 1840; it is also referred to as "Jacobi's law".

  • Applies to source impedances that cannot be changed but Load is variable
  • Load is the conjugate of the source impedance

    • for passive resistive loads we say source equals load ( taking the absolute value) even though a power source has a negative differential slope or ESR

    • Below is showing load lines for a Voltage source with a 5 Ohms ESR and a load of 5 Ohms where the maximum power transfer (MPT) point is the intersection of two identical slopes but conjugate polarity.

enter image description here

Below is the OP's BP 275U I-V Curves 75W 19V Solar Panel conjugated by rotating and flipping the graph.
enter image description here

- The same is true for reactive source and load impedances in RF antenna power optimization with conjugate  matching 
-  perhaps same is true for battery charging with pulses where a capacitive low ESR battery should be charged with a conjugate matched source ( inductive).
  • \$\begingroup\$ The panel I am using is a BP Solar 275U. I initially thought of doing something like this but was having trouble understanding how the regulator would operate with a current source (solar panel). I am still trying to wrap my head around this. \$\endgroup\$
    – RKsquared
    Commented Dec 4, 2016 at 19:47
  • \$\begingroup\$ @RKsquared When you match the non-linear load impedance profile to the non-linear source profile the output is always the maximum power transfer. Think of it as corollary to the "linear" matched impedance Law of Maximum Power Transfer and it is called Jacobi's law \$\endgroup\$ Commented Dec 4, 2016 at 20:27
  • 1
    \$\begingroup\$ This only applies the external resistance can be varied, and the internal resistance is cannot be changed. \$\endgroup\$ Commented Dec 4, 2016 at 20:33

The circuit that sinks current to maintain constant voltage is called "shunt regulator". Google it, there are many, many designs around.

  • \$\begingroup\$ sorry @theamk, but CC is not the optimal MP load line \$\endgroup\$ Commented Nov 17, 2016 at 20:54
  • 1
    \$\begingroup\$ @TonyStewart.EEsince'75: yes? but what does it have to do with shunt regulator? The shunt regulators do not do CC, they do CV \$\endgroup\$
    – theamk
    Commented Nov 18, 2016 at 16:29
  • \$\begingroup\$ If you follow my design , it will emulate a perfect MP load that rises in voltage from 14V at sunset to 17V at 4.4A in full sun, if you adjust as I indicated. Then tweak Vout for your PV to get perfect match specs. 25W R is close and may need 10% tuning more or less since 45W tolerances are unknown. alternative to lamp is a 50W R \$\endgroup\$ Commented Nov 18, 2016 at 17:11
  • \$\begingroup\$ Upvoted because even though this is a basic answer, it mentions an important term no other answers did mention. \$\endgroup\$ Commented Dec 4, 2016 at 22:30
  • \$\begingroup\$ Well I can disagree but a better match is neither shunt nor series but a constant slope regulator to match the ESR of the Source for max. reduced sensitivity to solar fluctuations when adjusted. \$\endgroup\$ Commented Dec 4, 2016 at 22:37

You wont be able to measure the peak power from the panel with a fixed resistor OR fixing the voltage and your panel won't be operating at maximum efficiency. To operate the panel at maximum efficiency you have to have a maximum peak power tracker. Power peaks when your not drawing too much current from the cell as shown below.


If you don't want to run a MPPT tracker with your panel, then just use a fixed load and you will only need to measure (or datalog over the course of the day) the voltage, since you know the load you can calculate the current and power from the cell.

A good way to find out what your cell can do is get a lamp and calibrate it. Then find the I-V curve by varying the load (usually people get a digital load) at different light levels to the cell, but that's probably more time than you want to spend.

  • \$\begingroup\$ Some MPPT devices operate by regulating panel voltage to a pre-entered Vmp value. Maybe this is what the OP is proposing to do. Not use a fixed resistor. Hard to tell. \$\endgroup\$
    – user57037
    Commented Oct 19, 2016 at 4:17
  • \$\begingroup\$ @mkeith I'm open to the idea of using a fixed resistor. Measuring the voltage and current I can get an idea as to how illuminated the panel is which I can then use to roughly calculate the Pmax under typical operating conditions for a given level of illumination. I would prefer to be able to measure Pmax more directly but this may be suitable. \$\endgroup\$
    – RKsquared
    Commented Oct 19, 2016 at 15:39

I've been contemplating the exact same question. I concluded the best system would be to use a microcontroller to track the maximum power point.

I would use an Arduino, with a minimum of external parts:

  • one ADC pin to measure current, voltage drop over a (say) 0.25 Ohm sense resistor
  • on the next ACD pin, use a resistor divider to measure panel voltage
  • (you need to subtract the resistor voltage drop, from the apparent panel voltage
  • have the Arduino drive a large MOSFET, or several, on a large heatsink
  • smooth the PWM "analogue" output with an RC network to drive the MOSFET gate
  • write a simple algorithm to track the mppt


simulate this circuit – Schematic created using CircuitLab

(sorry about the gaps in the wires, I couldn't get it to jump)

The arduino can also report the power back to my computer, or log it, or integrate it up over the day, or something

If you're just interested in tracking the mmpt, there's no need to correctly scale the voltage or current. As long as the sense resistor and voltage divider resistors don't drift in value.

  • \$\begingroup\$ The only problem is the lack of choice in MOSFET. The 50W into a heatsinkmax only \$\endgroup\$ Commented Nov 17, 2016 at 22:30
  • \$\begingroup\$ Yes this is a problem. Also I'm not certain that 5 V on the gate will get RDS to the values required.. I suppose you could use many in parallel to increase the power handling. You could also build a high power D/A convertor with several large resistors in a range of values, and use MOSFETs to switch them in. \$\endgroup\$
    – tomnexus
    Commented Nov 18, 2016 at 5:42
  • \$\begingroup\$ thats why I used LDO and dummy load using old car Headlamps to a make a precision adjustable Active Load \$\endgroup\$ Commented Nov 18, 2016 at 6:20
  • \$\begingroup\$ You will also need capacitors in parallel to the panel to filter the ripple. As said you will also have issue dumping the power to a single mosfet. Another issue is driving a mosfet on it's analog region is challenging, you have a few mV to play and the temperature dependency makes this region float in a scale of Volt. You definitely rather buy something like this ebay.com/itm/… and add measurement to it. \$\endgroup\$
    – Damien
    Commented Apr 25, 2017 at 6:00

As the other post stated, you will need to implement some sort of MPPT tracking (Maximum power point) to have any significant data.

You can use a dummy load that you can size the be close to the Vmp @ STC but it won't be very accurate.

One cheap way to do this though, is to use a micro-inverter that you can find for 20-30 USD that will take care of this for you, and you just need to measure the voltage and current at the panel lead.


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