Design of an AC adpater for hot-shoe flash - where am I wrong?

  • Thread starter Thread starter ZapoTeX
  • Start date Start date
  • Tags Tags
    Ac Design Flash
AI Thread Summary
The discussion focuses on designing an AC adapter for hot-shoe flashes, which typically rely on batteries. A user shares their unsuccessful attempt to power a flash with a low-current universal power supply, highlighting that the required discharge current during reload is around 15 A, necessitating a large and expensive capacitor. Suggestions include using an external battery to provide the necessary peak current or connecting directly to the high-voltage section of the flash. The conversation also touches on the cost disparity between flashes that accept external power and those that do not. Overall, the feasibility of powering flashes with external supplies remains a key concern, with potential solutions being explored.
ZapoTeX
Messages
2
Reaction score
0
Hi everyone!

First post on this really awesome forum!

I'm thinking of an AC-adapter for hot-shoe flashes, which are usually powered with batteries.

Here a guy describes his unsuccessful attempt: he built a wooden battery pack with contacts at the right place and connected to a universal power supplier giving 300 mA @ 6V.

The flash does not even switch on. The reason is simple: the current is not enough. My first thought was: high current is only needed to charge the capacitor of the flash. How about I add an external capacitor that stabilizes the 6V at higher currents? But how big does it need to be?

Easy to calculate. A flash takes about 5 seconds to reload after a full discharge. A flash can flash about 120 times with 4 AA batteries of 2500 capacity connected in series. Then, the discharge current during reload must be about 2.5 Ah / (120 x 5s / (3600 s/h) ) which makes 15 A (massive!).

So, even if I allow a drop of 2 V, I still need a 7.5 F capacitor, which is pretty expensive (although easily available on car hifi stores).

Am I doing something wrong? Is what I'm trying to do impossible? Is it necessary to SKIP the low-voltage stadium and connect the output of the AC/DC converter to the high-voltage section of the flash, so that the required currents are lower? (there is a product on the market, which I can't mention because it would be spam, which does exactly that, but it only works with flashes that have a plug for an optional external battery pack)

What sucks, though, is that flashes that can use power from an external supply are twice as expensive as the ones that don't! Is there a way to "hack" into the flash and connect an external power supplier even if the flash does not natively accept it?

Thanks a lot!
 
Engineering news on Phys.org
Years ago, when a hot-shoe was nothing more than a switch, I installed an external power jack on my strobe. I simply wired it to the original battery terminals. I made a belt hook for a 6 volt lantern battery and plugged it into the jack. Worked flawlessly.

ETA: I realize that I haven't answered any of your questions -- I'm very tired. I'll try to come back to this later.
 
Last edited:
Hi Pantaz, thanks a lot for your reply!

It does answer some of my questions! You proved that it is possible to use a self-made device and external battery to power a flash, which is not obvious given the currents involved.

It would be even more convenient to connect the flash straight to the AC, but maybe I can use an external battery as if it was a capacitor, just in parallel to the flash.

Thanks again & bye bye!
 
ZapoTeX said:
Hi everyone!

First post on this really awesome forum!

I'm thinking of an AC-adapter for hot-shoe flashes, which are usually powered with batteries.

Here a guy describes his unsuccessful attempt: he built a wooden battery pack with contacts at the right place and connected to a universal power supplier giving 300 mA @ 6V.

The flash does not even switch on. The reason is simple: the current is not enough. My first thought was: high current is only needed to charge the capacitor of the flash. How about I add an external capacitor that stabilizes the 6V at higher currents? But how big does it need to be?

Easy to calculate. A flash takes about 5 seconds to reload after a full discharge. A flash can flash about 120 times with 4 AA batteries of 2500 capacity connected in series. Then, the discharge current during reload must be about 2.5 Ah / (120 x 5s / (3600 s/h) ) which makes 15 A (massive!).

So, even if I allow a drop of 2 V, I still need a 7.5 F capacitor, which is pretty expensive (although easily available on car hifi stores).

Am I doing something wrong? Is what I'm trying to do impossible? Is it necessary to SKIP the low-voltage stadium and connect the output of the AC/DC converter to the high-voltage section of the flash, so that the required currents are lower? (there is a product on the market, which I can't mention because it would be spam, which does exactly that, but it only works with flashes that have a plug for an optional external battery pack)

What sucks, though, is that flashes that can use power from an external supply are twice as expensive as the ones that don't! Is there a way to "hack" into the flash and connect an external power supplier even if the flash does not natively accept it?

Thanks a lot!

If your problem is that the PSU won't supply a high enough peak current, then why not float a rechargeable battery across the system (or even a large capacitor) to help you with peak current? The battery solution would be cheaper than a large capacitor.
 
In the thrift stores there is usually a basketfull of these "Wall Warts".

A transformer type capable of an amp will be pretty large and heavy.

A "switcher" will be much lighter.

Look carefully through them and find one marked 1000ma or higher that seems too light to be true.

either of these would probably work okay


http://www.mpja.com/5VDC-1A-Regulated-Plug-Supply-USB-Mini-B/productinfo/18611+PS/
or

http://www.mpja.com/6VDC-800mA-Unregulated-Plug-Supply-Otres/productinfo/16228+PD/

though at light load it'll put out slightly more than 6 volts.
 
While I was rolling out a shielded cable, a though came to my mind - what happens to the current flow in the cable if there came a short between the wire and the shield in both ends of the cable? For simplicity, lets assume a 1-wire copper wire wrapped in an aluminum shield. The wire and the shield has the same cross section area. There are insulating material between them, and in both ends there is a short between them. My first thought, the total resistance of the cable would be reduced...
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
Back
Top