5V charger's output is 9V!

[Electroenthusiast]

I have a mobile charger of a wellknown & a renowned mobile company(all over the world), was using it till now, but now its a spare charger. It is rated as ' 5.0V 350mA ' at the back. But when i measure its output voltage using a multimeter, it shows an output of 9V. How can such a huge difference happen?I made a extension cable with pins that would help me charging iPod. I've been charging gadgets using it, would it be problem?
How reliable are the chargers from that 'Reliable Company'!
I'm afraid my iPod would burn out, i sensed heat while charging...

Another Simple Question, Just dont want to open a New thread for this one.

Communication Ques: Confusion with Phase Locked Loop
When are two signals said to be 'in Phase'?
I'm asking this because i think as
' the phases cannot be same, without frequencies of each other being same'
Ref: Phase (waves) - Wikipedia, the free encyclopedia

 

[Reloadron]

Did you measure the output loaded or unloaded?

Ron

 

[ben7]

The power supply output is not well regulated. That is why it measures 9volts when it is unloaded.

-Ben

 

[Electroenthusiast]

Yes, I measured it unloaded... Even i thought the same.
But i dont know why there's that difference in value while 'Loaded and Unloaded'.
Is it a problem as Multimeter Impedance being High?.
Can anyone explain? Thanks in advance.

Edit:

Another Simple Question, Just dont want to open a New thread for this one.

Communication Ques: Confusion with Phase Locked Loop
When are two signals said to be 'in Phase'?
I'm asking this because i think as
' the phases cannot be same, without frequencies of each other being same'
Ref: Phase (waves) - Wikipedia, the free encyclopedia

 

[blueroomelectronics]

Who is the "wellknown & a renowned mobile company"? Is it genuine or counterfeit? What's the model #?

 

[Reloadron]

Yes, I measured it unloaded... Even i thought the same.
But i dont know why there's that difference in value while 'Loaded and Unloaded'.
Is it a problem as Multimeter Impedance being High?.
Can anyone explain? Thanks in advance.

Most of these devices are not, as mentioned, regulated output. What you generally have is a transformer, rectification and a capacitor to act as a filter in there. That cap will charge to the peak value of the RMS voltage. So we take 5 volts times 1.414 and that gets up to about 7 volts. Now if the transformer primary is a little high, that adds to the transformer secondary RMS value out to the rectifiers. What you are seeing is not at all unusual. Under a load of 350 mA is what you want to see.

Ron

 

[audioguru]

The "charger" is just a cheap and simple AC/DC power supply. The charging circuit is inside the phone.
The simple power supply is made to produce 9V without a load then its internal resistance causes the output voltage to drop to 5V when it has its rated load current.
A better power supply has a voltage regulator.

 

[Electroenthusiast]

Comparitevely a OLD thread, but Topic remains the same.

Hi,
I now bought a USB 5V adaptor(Rated 5V,1A) for my iPod.

Did you measure the output loaded or unloaded?

Ron

I measured the unloaded output, and multimeter shows it as 5.6V, and i require 5V loaded output.
I feel the output of 5.6V(unloaded) will not burn my iPod, moreover the loaded output will be less than that value. I want someone to say whether it's OK to use it.

What (optimal)resistance should i use to measure the loaded output voltage?

 

[audioguru]

To measure the output of your AC/DC adapter at its full rated output of 5V/1A then simply use Ohm's Law to calculate a load resistor:
5V/1A= 5 ohms. The power in the resistor will be 5V x 1A= 5W.

5 ohms is not a standard value so use two 10 ohms/5W resistors in parallel.

 

[Electroenthusiast]

To measure the output of your AC/DC adapter at its full rated output of 5V/1A then simply use Ohm's Law to calculate a load resistor:
5V/1A= 5 ohms. The power in the resistor will be 5V x 1A= 5W.

5 ohms is not a standard value so use two 10 ohms/5W resistors in parallel.

Hi AG, i couldn find a 5W resistor in my collection, but i got a 1/4W(or 1/2W) resistors. So experimented with it, by momentarily(for few secs) connecting the multimeter (and Resistor) to the adapter.
For 10ohms load, i got the voltage o/p as 5.07V i.e., (500mA).
For 100ohms load, i got it as 5.32V (50mA).
Since i used 1/4W resistors, i felt an huge heat when i touched the resistor after that.

This means that: as drawn current decreases, the o/p voltage increases.
I dont know the max current drawn by the iPod, but it would probably be around 500mA(not sure though).
But, is there anyway i can find the current drawn by the iPod(i mean by calc the iPod's res(i.e., Load))?
I sense that there will be no prob in using this USB Adapter. TY

 

[audioguru]

I don't have an iPod.
You must find out the current drawn by one when it is operating and when its battery is charging at the same time.

 

[Electroenthusiast]

I don't have an iPod.
You must find out the current drawn by one when it is operating and when its battery is charging at the same time.

Yes, but there's no chance of measuring the current drawn(iPod as it is). It completely fits into the adaptor, there's no wires/ pins that help me do that. Maybe i need to use some extra wires, and do some work to connect it. It would have been easy if we could have measured the impedance/resistance of Ipod.
Specifications in the Website shows that:
The battery is a lithium-ion battery, Charge time:Fast charge in about 2 hours (80% capacity), full charge in about 4 hours.
Specs: ( 3.7V/600mAh - from another website).
So calculating, i get the value of max current drawn to be around 300mA.

That means the output voltage will be between 5.1V to 5.3V.

 

[Electroenthusiast]

The Charger has Model No: A1265, and it also printed there that it complies with UL Standards. So, i'm Okay with it [] .