# Hand crank battery charging circuit build

I'm using a 12V dc motor as our generator for a hand crank power generation system. We've tested our motor without a load we thankfully can generate a maximum of 9V. Our project requires us to generate enough power to charge a phone for a 2min call.

We're almost done, but I'm somewhat struggling with the circuit. I'm using two LED's one to indicate we are charging the batteries when cranking the generator, and the other to indcate the batteries are discharging.

Our input mechanical power that we've calculated would be at least 3.5W(assuming low efficiency), we're using four NiMH rechargeable batteries that is 0.8Ah at 1.2V, they will be connected in series so that the voltage would be 4.8V acceptable for the phone charger. Here is our initial circuit schematic:

I'm assuming the input voltage from the generator when connected to the batteries as the load, has to be in the range or maximum of 4.8V. The resistors will dissipate some power for the LEDs(20mA at 3V) Using a blocking diode to block any reverse current.

My concern of this circuit is that when I crank the generator both LEDs will function, and the batteries are being charged, how can I change this so that only LED1 is functional and the batteries are being charged? And when we strop cranking the batteries will light up LED2 and power the load, and the motor I think, should we use switches? While charging/discharging?

Our load will be a phone battery, and we will use this charger to connect to it:

This is an introduction class project, it's quite simple. But while reviewing the circuit realized some mistakes...

EDIT : Final circuit, after help.

simulate this circuit – Schematic created using CircuitLab

• Your battery is connected all wrong, and leave out the "charger" because its not a charger, is a DC-DC step-down converter which just feeds your 4.8V straight through. Commented Dec 9, 2015 at 14:41

First, you really, really need to check how well your generator works when under load.

As brhans pointed out, you've got your battery connected wrong. Fixing this and adding a MOSFET will produce

simulate this circuit – Schematic created using CircuitLab

M1 is a p-type MOSFET, and when the generator output is zero the FET will be turned on. When the generator is charging the battery, the gate of the FET is high, and the FET is turned off. D4 compensates for the voltage drop in D2, and keeps the FET gate voltage from going the wrong way.

• I can't get the MOSFE in time, I need to finish this circuit tomorrow, would this switch work: radioshack.com/products/pk2-slide-spdt-sw?variant=5717513477 Commented Dec 9, 2015 at 16:17
• Thank you so much, I think I can get a MOSFET from a store 11 mins away from my home, radioshack.com/products/mosfet-transistor?variant=5717613189 would this work? Commented Dec 9, 2015 at 16:47
• If you want to use a switch instead of a FET, that will work. The Radio Shack MOSFET may work and it may not. Your gate voltage will be marginal. And on second thought, add a 10k resistor from the FET gate to ground. Commented Dec 9, 2015 at 17:16
• Although a switch is primitive in comparison to the MOSFET :p, I'll use a switch. BTW, is the battery connected to the wire between M1 and the wire having D4 and D1, or is it over it? I'll replace with M1 with a switch thanks again! Commented Dec 9, 2015 at 17:20
• @M.A - If you're using a switch ignore that wire, and get rid of D4. If you want to try the FET (and I recommend this as a learning experience), the wire crosses over. Also, if you're going to be at RS, get a DMM while you're there - the cheapest they've got. Commented Dec 9, 2015 at 17:22

Your lighter socket charger requires a 12 volt input as labeled, so your circuit output will be too low to drive it. The charger runs on 12 VDC and has an internal 5 volt regulator which is applied to the USB socket.

Consider wiring your own USB socket to your battery output and not using the dc-dc converter plug.bypass the

• Although that charger is looking for 12V nominal, it will very likely work over a much wider voltage range, since it's designed for automotive use. It may well be a boost/buck converter internally. Since it sounds like it's a supplied component for class use, the instructor probably tested it already. Commented Nov 3, 2018 at 0:02