Hello, for my GCSE Electronic Products coursework, I am making a solar power phone charger that utilizes a 555IC. I found a circuit online but I noticed that the voltage on the 'USB Output' goes up to about 8V which isn't good for a phone. I attached a Voltage Regulator to the output to keep the voltage at 5V. I'm wondering if this would work. Sometimes when I test it on circuit wizard, it says there is a 50V Collector-Emitter voltage though I can't see why this would be.
Here is my circuit. The 100uF capacitor represents a battery/usb socket (probably quite poorly). The 12V DC power supply represents the solar cell.

Here is a link to the original circuit.

http://electronicseveryday.blogspot.co.uk/2009/07/mobile-cellphone-charger.html



Thank you.

 

hi,
On load, either a mobile or fixed 'test' resistor, the output voltage will be ~5v.
As a test, connect a 120R across the USB output and recheck the voltage.
E
EDIT:
Remove 100uF cap before using as mobile charge OR keep a 120R thru 240R permanently connected across the 100uF.

 

Voltage is at 4.72V with a current of 40mA.

 

I attached a Voltage Regulator to the output to keep the voltage at 5V.

Since you added the 7805 voltage regulator, can you explain the purpose for having the 555? Would you get the same result by feeding the 12V directly into the 7805?

 

Since you added the 7805 voltage regulator, can you explain the purpose for having the 555? Would you get the same result by feeding the 12V directly into the 7805?

The 555 acts as a bistable. When the voltage of the battery goes low, I.e it is dead, pin 2 of the 555 goes low, thus pin 3 goes high, turning on the transistor for charging process. Vice Versa is also true.

 

hi Sid,
The 120R or 240R load resistor is required in order to keep the voltage across the 100uF close to 5v.
Without a load the 7805 could charge the cap to 8v or higher, so when you plug your mobile into a powered up charger, the 100uF charged to 8v, could momentarily exceed the input rating of the mobiles internal circuitry.

The datasheet for the 78o5 states that a minimal load current should be drawn from the 7805 in order for the regulator to work correctly.

E

 

The 555 acts as a bistable. When the voltage of the battery goes low, I.e it is dead, pin 2 of the 555 goes low, thus pin 3 goes high, turning on the transistor for charging process. Vice Versa is also true.

But with the '7805 in the circuit the '555 is no longer measuring the battery voltage. It can only measure the '7805 input voltage.

 

But with the '7805 in the circuit the '555 is no longer measuring the battery voltage. It can only measure the '7805 input voltage.

Hmm. I put it in place as when no load was connected, the voltage would rise to 8V, which is obviously dangerous for mobile phone charging.