Hello,
Happy Christmas.

I have a sony HX100V sony camera. The power adapter/charger (ACL200D) is not functioning properly.
The input to the adapter is 230v and the output is 8.4V, 1.7A.
the adapter few times fell down and is not charging my camera now. I tested the output voltage and is 4.3v.
I opened the charger and there is no physical damage observed. I want to make a substitute for that.

I made a small experiment. I have connected a 7809 (without heat sink) to a 12v rail of an ATX. The output is connected to my camera and is charging. I tested this for just 3 seconds. 7809 is heating Of-course it is not damaged. As my camera battery is completely discharged, it is taking more than 1 Amp which makes 7809 to get heated.

I have an adjustable flyback converter in my workshop. I can adjust its voltage to 8.4V. I don't remember the current rating (its definitely more than 1.5A). Can I use this to charge my camera? Or do i need any current controlled buck converter?

 

I would say yes, constant voltage of 8.4V with current upto 1.5A will do,

 

The final voltage that the 7809 will charge the battery to is 0.6V higher than your original charger. I would put a 3A Silicon diode in series between the 7809 output and the + input to the camera. That will drop the voltage to ~8.4V

With 12V in, and 9V out, the 7809 will be dissipating (12-9)*1 = 3W, so will get hot enough to burn your finger unless it is mounted on a suitable heatsink. About 10 sq. cm of aluminum should do it...

 

I would say yes, constant voltage of 8.4V with current upto 1.5A will do,

Thanks for your reply.
What if my power supply is capable of driving more than 1.5A?
If the camera battery is at minimum voltage and if it draws 2A from my flyback converter, i hope my camera port may get damaged.

 

The final voltage that the 7809 will charge the battery to is 0.6V higher than your original charger. I would put a 3A Silicon diode in series between the 7809 output and the + input to the camera. That will drop the voltage to ~8.4V

Thanks. I have 1N4007 diodes. I'll use 4 in parallel.

With 12V in, and 9V out, the 7809 will be dissipating (12-9)*1 = 3W,

Why did you take 1A current? It may be around 1.4A.
7809 is rated for 1A. Can I use two in parallel?

 

Thanks. I have 1N4007 diodes. I'll use 4 in parallel.

You cannot parallel diodes and expect to increase their current carrying capacity without putting low-value resistors in series with each diode. If you don't, the diode with the lowest forward drop will end up carrying most of the current..

Why did you take 1A current? It may be around 1.4A.
7809 is rated for 1A. Can I use two in parallel?

Yes, your heatsink will need to be doubled. Both TO220 tabs can be bolted to the same piece of aluminum.

Since (per the linked datasheet), the LM7809 current is internally limited to 1A, you can put a 1N4007 1A diode behind each 7809.

 

Thankyou MikeML