# Peak Power Tracker

Discussion in 'The Projects Forum' started by milon, Mar 28, 2009.

1. ### milon Thread Starter New Member

Mar 28, 2009
5
0
Hi,

I'm wanting to build a peak power tracker for a solar powered car that I am building. Using my basic knowledge of electronics and help from other sites I have come up with a circuit design shown in the attached picture.

The panel is 24V nominal and with peak output of 175W (though obviously only in the Sahara!) at 35V. The microcontroller will be a rabbitcore that takes in values of voltage and current both upstream and downstream of the MOSFET. The DC-DC conversion is handled by a buck converter.

Here is where my knowledge falls short. The voltage sensors and the OPAMP current sensor needs to measure the voltage drop across a very small resistor in series with the solar panel and MOSFET. In theory there will be 5A @ 35V running through this resistor and I'm looking for it to only be in the region of a milliohm (if that). Say its 1mΩ, then it will dissipate 25mW of power. Does this mean that I only need a resistor that can dissipate this power? What about the other 5ish amps running through, won't this burn out the resistor?

Any other ideas as to the circuit design would be greatly appreciated.

Cheers,

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2. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
Without component values and labels, the schematic is very uninformative.

As far as your sensing resistor goes, power dissipated by it will depend on the square of the current times the value of the resistor (P = I^2 * R). So 25 (I^2) times .001 = .025, or 1/4 watt. Use a 1/2 watt resistor instead. You never want a component running right at the edge of any parameter (in this case, power dissipation).

3. ### milon Thread Starter New Member

Mar 28, 2009
5
0
Ok, but my question was will the other 5ish amps running through the resistor not burn it out?

4. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,349
731
Measure the voltage drop across the MOSFET. Once the RDSon is known for the gate drive, your current sensing resistor is already in place. The downside is if using PWM drive, as the peak voltage across MOSFET will be supply voltage.

Second option is to measure the voltage drop across the supply wire to the motor. The current is high enough that a measurable amount of voltage drop will appear if the wires are over 6" long.

An analog "peak hold" circuit might work easier if only the peak power is wanted.

5. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
Whichever that "other resistor" turns out to be, the power it will dissipate is calculated in exactly the same way as the example I gave you.

6. ### milon Thread Starter New Member

Mar 28, 2009
5
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But I'm not asking for how much power it will dissipate.

Say I go with the first setup, then the resistor is dissipating the 25mW, then there will still be (175W - 0.025W) 174.975W passing through the resistor without being dissipated.

Obviously the 25mW will be given off as heat, but presumably there is an upper limit on the power that can pass through (ie NOT dissipated) without giving the resistor problems?

7. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
I believe we have accounted for the totality of the current passing through the resistor. That is the 5 amp figure. I believe that we have also arrived at the power dissipated ( in other words, lost as heat) in the resistor as the product of the square of the amperage times the value of the resistor.

The resistor "sees" only the power it has to dissipate - the I^2 * R figure. The load may receive power in the kilowatt region, but that resistor is only heated by the current passing through it. That is one very good reason to keep current sensing resistors at a low value - they have negligible effect on the circuit that way.

By the way, I missed the decimal point in my original figuring - a 1/4 watt resistor (250 mw) can easily handle that level of dissipation (24 mw).

8. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,349
731
ΩI believe beenthere is measuring a value of a resistor in parallel with a smaller shunt resistor. The shunt will carry the majority of the current, and the parallel sense resistor will carry only enough to measure the relative amount in the shunt.

Even a 0.1Ω resistor would be dissipating 2.5W at 5A.

--ETA: I now see he is speaking of a 1/1000th Ω resistor, which would dissipate 25mW

9. ### milon Thread Starter New Member

Mar 28, 2009
5
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Great - thank you very much.

10. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
8
These terms can indeed be confusing when one is learning them. I'll try to help...

Watts don't flow through anything. Watts are a rate, like miles per hour or gallons per second.

Current is also a rate - it is the rate of the electrons flowing through your resistor. Five amps means roughly 31.2 billion billion electrons per second.

Ohm's Law and a little algebra gives us the voltage drop across your resistor: E=I*R, so E = 5*.001Ω = five millivolts across the resistor.

There are no "other 5 Amps." Kirchhoff's current law assures us the current through anything in series will be the same.

The other 174.975W get dissipated by the load.

Obviously, the previous assumes 1000W/m$^{2}$ at 25°C and 1.5 air mass spectra. Previous also assumes load resistance equal to V$_{mp}$ over I$_{mp}$.