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I have a few questions about op amps, but I'm fairly new to the whole EE thing, so keep that in mind while answering. Thanks!

I am trying to use an op amp between a 6V solar panel and a 4.8v NiMh battery as a comparator. I want the solar panel to stop charging the battery when the battery voltage hits 5.6v. I'm using the comparator in a inverting circuit and I was wondering two things:

  1. Is there a voltage drop from +Vcc to Vout?
  2. If the panel outputs 200mA max, will the battery charge at the same current?

Also, I'm currently a ninth grader and I don't have any opportunities to learn about electronic engineering. I tried using the Internet as a tool for learning, but it hasn't gotten me very far. Is there any way that you would recommend going about learning the subject?

Diagram below

Thanks, Jackenter image description here

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    \$\begingroup\$ Most op amps and comparators can't output enough current to charge a battery. 200mA is not something you can expect from a normal comparator. You will need to use the comparator to drive a transistor that can switch the charge current to the battery when the voltage is high enough. \$\endgroup\$
    – JRE
    Commented Nov 29, 2016 at 6:16
  • \$\begingroup\$ There's some IC that do exactly charge control \$\endgroup\$
    – Antonio
    Commented Nov 29, 2016 at 14:33
  • \$\begingroup\$ If there is an IC, can you name it or give me a link? It would be very helpful. Thanks! \$\endgroup\$
    – MerpTheAwesome
    Commented Nov 29, 2016 at 16:33

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Your circuit has some major flaws.
You have correctly calculated the R1, R2 voltage divider, but it gives 0.396 volts above ground. If you swap R1 for R2, it will give the desired 5.6 volt.

The solar cell supply voltage of 6v is actually a variable supply - with no light source, it will sag - your circuit should work properly for any input voltage. Since your 5.6v reference voltage is derived from this supply, it will sag too. When it falls below your battery voltage, the op-amp will discharge the battery - certainly not what you want.
Most op-amps include internal current-limiting circuits. Many common op-amps limit current to about 0.03 amps. So the answer to your Question 2 is no, charging current will be less than 0.2 amps. Furthermore, even 0.03 can heat the op-amp considerably.
Addressing your question 1: Many op-amps require an internal voltage drop before charging current is allowed to flow. Some op-amps brag about "rail-to-rail" output voltage. These have much smaller voltage drop. The term "rail" refers to the op-amp supply pins (V+ and gnd).
Be not discouraged. What you are attempting might seem straightforward. Observe how complex the internal circuitry is inside a battery-charging integrated circuit:MAX712 charging IC

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  • \$\begingroup\$ Thank you so much! I know that battery charging is highly complex, but I'm trying to be able to produce this circuit for a low cost, so I watered it down to an overcharge prevention. Since there is a voltage drop, how large is it? With the normal op amps and rail-to-rail op amps? Also, how will the op amp discharge to battery if it drops the reference voltage? \$\endgroup\$
    – MerpTheAwesome
    Commented Nov 29, 2016 at 16:30
  • \$\begingroup\$ @MerpTheAwesome "normal op-amps" might drop 2v, and that's with less than 10 mA output current. Rail-to-rail improves the drop to perhaps 0.2v - learn to read data sheets. When your reference voltage drops below battery voltage, your op-amp (comparator) output pulls down toward ground, discharging the battery. \$\endgroup\$
    – glen_geek
    Commented Nov 29, 2016 at 17:36
  • \$\begingroup\$ If I'm putting 200mA through will the voltage drop be the same? Also, if I put a diode on the output line, will it limit battery discharge back into the circuit? \$\endgroup\$
    – MerpTheAwesome
    Commented Nov 29, 2016 at 18:20
  • \$\begingroup\$ @MerpTheAwesome small op-amps (like LM741) won't pass more than 25 mA, so that limits charging current - you can't force more current. Power op-amps are rare & $$. Yes, a diode prevents discharge, but adds about 0.7 volt to the op-amp's voltage drop. \$\endgroup\$
    – glen_geek
    Commented Nov 29, 2016 at 18:31
  • \$\begingroup\$ Sorry for the load of questions, but is there a rail-to-rail power op amp with a voltage drop of 50mV or less? I'm thinking of going with a Schottky diode for a 0.3V drop, so I'm wondering if I can keep the voltage above the battery voltage, 5.6V. \$\endgroup\$
    – MerpTheAwesome
    Commented Nov 29, 2016 at 19:11

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