Discussion in 'Homework Help' started by Runs212, Apr 26, 2015.

1. ### Runs212 Thread Starter Member

Mar 12, 2015
56
1
I am trying to build transmitter/receiver circuit to charge a phone. I tried using a mosfet to build the DC/AC but didn' quite work. So I settled for the below(see attached file) and was able to generate a sine wave in my the Rx circuit.

Experimentally, using the same parameters, my voltage reading was 22V, when connected to a load (iphone4), the voltage drops to around 1.4V. Current was 2.2mA. It does seem to charge the phone. However, I'm sure it's not getting enough current. I need at least .8A. What could I be doing wrong in the circuit?

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2. ### WBahn Moderator

Mar 31, 2012
20,232
5,755
It's unclear what you are trying to do. Are you trying to use the power in the RF signal to charge the cell phone's battery?

If so, how much power is the transmitter putting out?

Why do you think you'll be able to receive that that kind of power (0.8A at a few volts) from the RF signal?

Runs212 likes this.
3. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,071
Do you think that you can get a coupling coefficient of one using separated air-core coils?

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4. ### Runs212 Thread Starter Member

Mar 12, 2015
56
1
Probably not Mike. I'd take that into consideration.

5. ### Runs212 Thread Starter Member

Mar 12, 2015
56
1
I
I am trying use inductive coupling to charge a phone. At input pulse of 5vpp, 50% duty cycle, I am getting a voltage reading of about 9v on my experiment. At input pulse of 10vpp, it's 21v. When I place a 5v dc regulator at the output, I get a reading of 4.1v which shouldn't be.

I do suspect I'm not getting enough current. 5W is what is expected at the output. Per the schematic, the resistor is supposedly the phone.

6. ### MrAl Distinguished Member

Jun 17, 2014
3,747
791
Hello there,

Actually, this is just another typical circuit in these modern times, even with air core coils. It's meant to be an inductively coupled circuit where the magnetic field transfers the energy from one coil to the other.

One of the key points is the distance between transmit coil and receive coil. With 1mm distance you might get as much as 500ma with 40cm diameter coils, but with 10mm distance only 50ma. That's a big difference so you can see how the distance makes all the difference. Even a 2mm distance might bring it down to 300ma already.

5 watts at 5 volts means 1 ampere of current is flowing. Good luck with that

0.8 amps would take a good coil size and short distance. I would think 500ma would be a more reasonable first goal.

1N4148 diodes will never be good enough for 800ma, use 1n5817 or better.

The coils must be large enough too, so what are the sizes of your two coils?

Last edited: Apr 26, 2015
7. ### Runs212 Thread Starter Member

Mar 12, 2015
56
1
2.10" L x 2.10" W x 0.24" H (53.3mm x 53.3mm x 6.0mm) for primary coil (6.3u)
1.96" L x 1.96" W x 0.13 " H (50.0mm x 50.0mm x 3.5mm) for the secondary coil. (24u)