3v problem

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wejos

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i was thinking to supply a working electronic flash but instead of 2 AA batteries 9v battery connected to a 5v regulator then through a resistor divider using 20k and 30k (that i thought would give me 3v output).

checked it with my multimeter the reading says near 3v. this is just experiment though but at a loss why the flash isn't charging.

i already soldered the on/off switch of the flash inside so as soon there's power the flash should start to sound like it is charging.

here's another thing, when i used two AA batteries to charge, the electronic flash charges

what am i missing here?

thank you in advance.
 

hi,
Try measuring the 3V WHILE its charging, I believe you find its not 3V, the 20K/30K divider is too high a resistance.
 
I’ve had problems like this before with other circuits. The 9v transistor battery can’t supply as much current as the little 1.5 volt cells. I fixed my problem by connecting two 9v batteries in parallel. If you go this way, use a double poll on/off switch or diodes to isolate the two batteries so they don’t self discharge.
 
so the issue here is current thanks gary plus the divider has too high resistance as eric pointed out. i'll experiment more reduce the resistance of divider still using the 9v batt, and maybe use ac-dc adapter that had 500 miliamps.

thanks for your guidance guys, i appreciate all your help (from the start)
 

hi,
In place of the 7805 use a LM317, set it for 3V output.
 

A resistive voltage divider cannot supply enough current to run the photoflash. You need to use an LM317 adjustable regulator, where the adjustment resistors are set up so that it makes a regulated voltage of 3.3V from the 9V source. Also put a 220uF 5V electrolytic capacitor across the output of the regulator. The photoflash power supply draws a large current in short pulses, so the capacitor will store energy, and recover the photoflash faster than if the capacitor were omitted.
 
yes you're right eric it's 7805 thanks.

i don't have that regulator yet. adjustable regulator very interesting.

first thing tomorrow.

hope electronic stores here sell them.

and mike thanks for the additional details.

many thanks for all the help guys
 
thanks viz quite a huge file there but i'm sure tons of useful info. i just opened it using online pdf player and had a quick look at the contents. i might need to install adobe also.

went to your blog cool site : )
 
A 9V battery uses six AAAA cells which are much too small to deliver enough current for a camera's flash circuit more than a few times.
You will be throwing away 6V by making heat in the voltage regulator. Two AA cells produce much more current than a 9V battery and will last much longer. They also cost less.
 
Wejos,

To make this a learning experience, I suggest you do some reading on Thevenin Equivalent Circuit. Starting with your voltage divider of 20K/30K and a 5V voltage source, figure what the short-circuit current is. When you understand Thevinin's Theorem, it will be obvious why the voltage divider did not work for your application.
 
AG thanks for the heads up. point taken. energy is wasted.

mike i had read it need more time to grasp the idea so i'll keep on rereading until i have clear view.

i think i ran into this from one book before but did not try in actual breadboard guess all that VAB and IAB gave me cold feet lol.

ill prep some components for visual aid.

many thanks guys
 
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