1
\$\begingroup\$

I am trying to create a small low cost circuit that will allow me to log the output of a solar cell over the course of a month as a student project.

Currently I have a 5V Silicon PV cell and am looking to log the current over the course of a month. I have a circuit that I have designed to allow me to log the voltage and therefore be able to work out the current using Ohms law (see below) but I am unsure it will work.

If anyone could offer any advice or pointers that would be much appreciated. On a similar note if anyone could recommend a low cost voltage data logger that could be used to take readings once or twice every minute that would also be a great help.

\$\endgroup\$
  • \$\begingroup\$ Are you looking for an off-the-shelf data logger solution or might you consider something you can make yourself using a cheaper MCU and memory to store values. Playback is important - how do you plan to look at the data once it is stored? \$\endgroup\$ – Andy aka Apr 3 '13 at 12:17
0
\$\begingroup\$

I know this is dumb , I would use a thermos plot it's calory loss and drop in a appropriate resistor and heat the water. May not be as accurate but would be fun for a school project, a lot of simple math to get the job done cheep.

\$\endgroup\$
  • 2
    \$\begingroup\$ How does this answer the question exactly? \$\endgroup\$ – clabacchio Jan 14 '15 at 17:57
0
\$\begingroup\$

I would use a transimpedance amp, known also as a Current to Voltage converter. The advantage of these circuits is the virtual short between the current in and current out pins.
Trans-Impedance amplifier
Typically when I have read about these, the most common application people point to is Photo Voltaic cells and measurement of their current.

\$\endgroup\$
  • \$\begingroup\$ Finding one that can output 81mA might restrict the search a bit. It would also need local power to do measurements and unfortunately it can't use the cell because of it being used as a trans impedance amp with input connected to one of the battery pins. They are used a lot in photo-diode amplifiers. A common application in measurement of current in photo-voltaic cells is a new one on me but I'm here to learn. \$\endgroup\$ – Andy aka Apr 3 '13 at 12:22
1
\$\begingroup\$

Your method to measure current will work fine. The 'usual' way to measure current is to measure the voltage drop across a known resistor value, and that is precisely what you are doing.

However, what you measure isn't going to be well correlated with how much energy the cell is able to generate. This has to do with how solar cells function and how you can extract energy from them. I'm guessing that is what you hope to do, as opposed to "How much energy is the cell supplying to this very artificial load of 62 ohm".

A solar cell is able to produce a certain amount of energy, based on a number of environmental factors. Consider this a first level of cell efficiency, where simply shining X Watts of sunlight onto the cell, the cell is able to produce eX Watts of power. E is typically to the order of 10-30% depending on the quality of cell you're using. Given you probably got your cell off-the-shelf and for less price than an arm and a leg, I'd expect your efficiency to be between 10-20%.

The cell is able to generate this power at a certain voltage. When you've got no load attached (the circuit in your case is open), you can measure a certain voltage across the cell. This is called the open circuit voltage. This voltage is theoretically a function of the physics of the cell, and has nothing to do with illumination, though in real situations there is some small effect. The cell can not produce more than that voltage. It can, however, produce a lower voltage if the load is too heavy (More on that later).

Illumination and other factors determine the power the cell is capable of putting out, which is given by product of Voltage and Current (as is usual). So we've fixed the Voltage because of the physics of the cell, and the power because of the power the cell is capable of putting out in this conditions. Therefore, we've effectively fixed the current that we need to take out of the cell to fully utilize the energy.

Consider the case where the load is less than what the cell can supply under some specific conditions of lighting, temperature, etc. This means that the cell, at it's open circuit voltage, can 'power' the resistor without crossing the limits of it's power output. The amount of current the resistor needs can be calculated from ohms law. This means that the cell puts out that much energy, the excess is dissipated as heat from the cell. What this means is that once the cell is able to supply more that V(open-circuit) / R(load), you're not asking for any more from it, and hence won't know if its able to produce any more.

Now consider the reverse case, where the load is too heavy (resistor is too small). This means that to sustain open circuit voltage, you'd need a higher current than the cell is able to supply. The cell then compensates by lower the output voltage. The physical characteristics of an ideal cell don't allow this to happen, so instead what happens is that non-idealities are exploited to make the cell output voltage come down. In doing so, efficiency takes a reasonably significant hit, so again what happens is that you aren't getting a decent picture of how much energy the cell is able to produce.

As, hopefully, you've gathered so far, for each specific set of conditions, you'd need a different load resistor. The way this is handled is to use a switching mode regulator that basically presents itself as a variable load to the source (the cell). The load that is presents itself as is controlled used a scheme known as MPPT (Maximum Power Point Tracking). In your case, depending on what kind and quality of information you're after, you could try to bring your experiment slightly closer to what is usually done by perhaps trying out different load resistors at each data point, if you're doing it manually.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.