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| Solar panel regulator design question |
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| JoeP:
I'm designing a circuit to deliver power from a 25W solar panel (which will never output very much power here in the UK) to a 24V lead acid battery. Since the power is so low, I'm not particularly concerned about trying to keep up a constant current phase, I just want to keep the PV at its optimum voltage (18V). I've designed the circuit shown below. The theory is that if the voltage is >18V, the transistor turns on and more current is delivered to the battery, reducing the PV voltage, and vice versa. What I'm concerned about is unpredictable oscillations, as the change in current is having to propagate back through an SMPS, which could take a long time. I've considered a few possible solutions, such as putting a capacitor across the B-E junction of the BJT in order to keep it in its active region, but I'm not sure whether this would work very well (if the PV takes a while to respond, it could take a huge capacitor). Efficiency isn't vital, so I don't mind if some power is dissipated in the transistor. Does anyone have any ideas? |
| NiHaoMike:
Just limit the duty cycle of the boost converter so that it doesn't try to pull the input voltage too low. |
| JoeP:
Interesting idea, but I'm struggling to see how that would keep the PV voltage at 18V. It would cap the output power, but depending on the light intensity, the duty cycle limit would need to vary (correct me if I'm wrong, I'm just guessing). Also, it wouldn't be very easy to do as I'm using a boost IC and not a microcontroller. |
| mikerj:
Configure the error amplifier as an integrator and make the bandwidth much lower (order of magnitude) than the switcher. You need to think about protecting the Vbe junction from reverse breakdown if/when the output of your op-amp falls well below battery voltage. |
| David Hess:
It would be just as easy if not easier and more efficient to just use an 18 volt buck-boost converter. --- Quote from: mikerj on July 02, 2019, 07:10:41 pm ---You need to think about protecting the Vbe junction from reverse breakdown if/when the output of your op-amp falls well below battery voltage. --- End quote --- A diode in series with the base will protect the transistor however base-emitter breakdown is not a serious condition for a power transistor which will have low gain at high current anyway. |
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