If you've seen my previous posts, you know I've been having trouble with making a circuit for a mobile phone charging circuit that uses a 555. I decided instead of using a 'complicated' one of the internet, I will try to make my own. Here is my circuit.

The 5V DC power supply is supposed to represent a 5v USB Socket which a phone charger cable can plug into. I want my circuit to flash LEDs D1 and D3 at approxiamtely 1Hz when SW1 is on and a load is connected to V1 (i.e. a mobile phone charger). As you can see, transistor Q1 is not connected anywhere. I was wondering if anyone knows how I can connect the transistor to the LEDs so that they will only begin to flash if there is a mobile phone battery connected.


Thank you.

 

Is this really homework?

 

Is this really homework?

It's part of my GCSE Electronic Products coursework.

 

It's part of my GCSE Electronic Products coursework.

If it's homework, you need to show your best effort. We'll provide guidance based on what you're attempting.

 

Have you considered how you will compensate for the voltage drop across D2?
Have you checked the datasheet of the 7805 to see how it is typically used?
Have you read up about the USB current limits?

 

Okay, in that case. Here were my original ideas.


Here was my attempt number 1:

The problem was that the base was still connected to 5v, regardless if a load was connected or not, thus the LED (in this example only 1) would flash regardless of whether there was a load connected or not.


Alternatively, I tried this:

This one faced the same problem, the LED could still get power even if there wasn't a load connected.



I'm specifically trying to find a place where I can connect the base of the transistor so that the LED will only flash if there is a load (hence turning on the base).

 

Think about how you would sense that the load is drawing current.

 

Have you considered how you will compensate for the voltage drop across D2?
Have you checked the datasheet of the 7805 to see how it is typically used?
Have you read up about the USB current limits?

If the device connected to the USB were to pull 250mA, and the voltage drop across the diode of about 0.7 volts, then 0.25*0.7=175mW. It is my understanding that any 7805 except for '78L05' will output 1A max. Looking through the internet, it seems I should connect some smoothing capacitors in parallel to the battery.

USB 3.0 can handle outputs of 900mA whereas 2.0 can handle 500mA.

 

Think about how you would sense that the load is drawing current.

I replaced the 5V DC Power Supply with a 20ohm resistor which pulls approximately 200mA.

 

Hello,

And what will be the voltage after the diode?

Bertus

 

What role is D2 supposed to be playing?

Have you learned about current mirrors?

 

I replaced the 5V DC Power Supply with a 20ohm resistor

So how will that guarantee 5V for charging the phone, bearing in mind that the charging current, and hence the voltage drop across that resistor, will depend on the state of charge of the battery?
Have you read up about USB current 'negotiation'?

 

I'm not sure, but I don't think the negotiation is a factor. As far as I know and can tell, many USB chargers and power supplies for stand-alone USB-powered devices don't even have the data pins brought out and simply 5V. I don't know that all of them even provide specific current limiting, though I'm pretty sure at least a lot of them do.

 

I'm not sure, but I don't think the negotiation is a factor. As far as I know and can tell, many USB chargers and power supplies for stand-alone USB-powered devices don't even have the data pins brought out and simply 5V. I don't know that all of them even provide specific current limiting, though I'm pretty sure at least a lot of them do.

Yes, I'm basing the idea of using a 7805 on the fact that many circuits use this. For example, Afrotechmods designed a mobile phone charging circuit using only a few capacitors and a 7805.

 

That was Afrotechmods' circuit and the regulator shown on it is the 7805.

On my circuit I've noticed that when I turn the switch off and back on again, the IC555 and R2 say that they are receiving reverse voltage, thus exploding. I have no idea why this is.

 

What role is D2 supposed to be playing?

Have you learned about current mirrors?

D2 is to make sure that the current goes into the phone battery occurs and not vice versa. I suppose phones already have a protection circuit that prevent this anyway...

 

That was Afrotechmods' circuit and the regulator shown on it is the 7805.

On my circuit I've noticed that when I turn the switch off and back on again, the IC555 and R2 say that they are receiving reverse voltage, thus exploding. I have no idea why this is.

How does a 555 or a resistor "say" that they are receiving reverse voltage? Why would a reverse voltage on a resistor cause it to explode?

 

How does a 555 or a resistor "say" that they are receiving reverse voltage? Why would a reverse voltage on a resistor cause it to explode?

Sorry, I meant the resistor is getting too hot from the reverse voltage. The reverse voltage to both the IC and resistor are 22V. That is what my software, circuit wizard, is saying.

 

With a 12V input and no inductive components , how can you get a reverse voltage of 22V? Is Circuit Wizard lying?

 

With a 12V input and no inductive components , how can you get a reverse voltage of 22V? Is Circuit Wizard lying?

It may well be.