hi
m new in this forum and need ur help. m developing a current n voltage sensor for the solar PV panels. using two similar pv panels i have received maximum 40 volts in series configuration and 8 A current in parallel configuration. so i have limits of 40v 8A. adc 0804 has input bias current of 3 mA so i have to bypass the remaining current using shunt resistor. the circuit i designed is as follows

problem is that when i practically implement this circuit; connecting 5 ohm resister always considers it short. Why

where as its not so in simulation. Please guide me for the complete circuit of the sensors for voltage and current.
tabular data for the simulation is belowView attachment manual data.pdf

 

I'm not sure what your schematic is about but the way I get voltage is just put the panel (or battery) voltage through a voltage divider to get it under 5V so I can put it into an ADC pin on a microcontroller. Using a hall effect sensor is the efficient way to get current these days. Hall effect sensor, check out: http://www.sparkfun.com/products/8882

Voltage divider into a PIC, take a look at: http://2.bp.blogspot.com/-fcShHZt2i2Q/TYTvzgT5tsI/AAAAAAAAAGQ/CIQEfRTRF5g/s1600/2-12-11%2BSolar%2BBoard%2BSchematic.png

I use a 10K and 1.8K resistor for the divider then put a 5V zener diode just in case someone plugs in something with too high a voltage.

 

problem is that when i practically implement this circuit; connecting 5 ohm resister always considers it short.

5 ohms IS almost a short. As you've drawn it, the resistor will have to dissipate 320W of heat. You haven't said how you're making your connections; panels in series? in parallel? No other load in series with the 5Ω resistor?

 

Are you are using a load resistance equivalent to the series open-circuit output voltage of two panels divided by the current obtained with the panels in parallel? If so, you would hardly expect to get much output voltage if you connect this resistance to the two panels in series.

To obtain a good power yield from a photovoltaic battery, there is an optimum load resistance, which depends on the panel design and the strength of illumination. Because of the shape of the battery voltage / current characteristic, reducing the load resistance too far results in a quite dramatic fall in output voltage.

You will not see this effect in a simulation where the photovoltaic panels have been replaced by an ideal voltage source.

 

A standard Voltage and Current graph of generic silicon PV panel

 

A standard Voltage and Current graph of generic silicon PV panel

That's interesting. I didn't know that the peak power of a PV panel is obtained so close to the peak voltage. Looks like ~80% of Vmax gives you Pmax.

My thinking is skewed by working with alternator coils. With a decent winding - to give low DC resistance - peak power is output at quite a low voltage but with a whopping current. That is, power peaks when the load impedance matches the coil's impedance.