# Capacitor Charger

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#### Hyperion

##### New Member
Hello all,
(I'm new here so I apologize in advance for any mistakes)

I recently obtained a flash charger from a camera, and I'm looking to decrease its charging time. I have a general understanding of how the oscillator and transformer physically function, however I haven't ever worked with them in this type of circuit. Thus I have a couple questions regarding the charge time.

1. How would an increased supply voltage affect the charge time? I figure the oscillator would need to be modified to run off a higher voltage, but how would the transformer react?

2. What effect would changing the transformer have? I actually wouldn't opt to do anything with the transformer at this time but I'm curious.

##### Banned
An increase in supply voltage will decrease charge time right off the bat, however the extra current is going to hurt the circuit, I wouldn't do more than .5-1 volts. If you want more than a trivial increase in charge time you'd have to replace pretty much everything in the circuit. If you try to charge it fast this way too many times you'll burn out the transformer. I've done this on two similar circuits.

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#### colin55

##### Well-Known Member
If the circuit uses one cell, you can use 2 cells and the charge-time will decrease appreciably.
If the circuit uses 2 cells, you can add another cell.

#### Hyperion

##### New Member
Ah alright I see. Actually what exactly is it in the circuit that determines the charge time, is it the transformer itself or the output of the oscillator or the oscillator frequency or what?

##### Banned
The transformer and one or two transistors is the oscillator. Using an extra cell for a better charge time will cause the circuit to fail sooner. It's not any one specific component that ultimatly determines the charge time, it's the ENTIRE circuit.

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#### microtexan

##### New Member
capacitor charge time

Maybe I'm missing something here, but can someone please explain how increasing supply voltage increases a capacitors charge time?

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##### Banned
It's called a typo micro =)
Pardon.
Edited.

#### colin55

##### Well-Known Member
If the circuit uses 1 cell, there is a certain percentage of the supply voltage that cannot be used.
For example, the base needs 0.7v before the transistor turns on. When the supply is increased from 1.5v, to 3v, you can see the available (effective) voltage increases from 0.7v to 1.3v. This allows the transistor to turn on much harder and the collector-emitter voltage is lower.
This means we have more voltage available for the transformer. In addition, the transistor turns on harder and a higher current flows though the primary of the transformer.
The effect of all these improvements is enormous and the capacitor will charge in less than half the previous time.
Nothing will be damaged. As the capacitor charges, the load on the circuit decreases and the current drops, so nothing overheats or explodes. The 120u capacitor is rated at 330v. It will charge to about 450v with a 3v supply. It will produce an emourmous flash from the tube.

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#### Hero999

##### Banned
nothing overheats or explodes. The 120u capacitor is rated at 330v. It will charge to about 450v with a 3v supply. It will produce an emourmous flash from the tube.
That 330V capacitor charging to 450V might cause an explosion.

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