Electronics help needed.

[Wayland]

I have managed to stretch my woeful electronics knowledge to put this battery pack and lead combination together for my Fuji X10

That was straight forward as the DC coupler required 5V and there was a 5V output on the pack.

Now looking at the pack I can see that there is a 9V and a 12V output socket as well.

I cannot help but wonder if there is a way to use the 9V socket to run my Canon SLR but the DC coupler for the Canon needs between 7.4V to 8V.

How do I go about reducing a DC voltage from 9V to the right amount? I presume it's not just as simple as just putting a resistor into the circuit?

 

[bikething]

How do I go about reducing a DC voltage from 9V to the right amount? I presume it's not just as simple as just putting a resistor into the circuit?

You'll need some kind of voltage regulator to drop the voltage. In order to spec one we would need to know how regulated the voltage out of the battery pack is, what current it's capable of providing and how much current does your camera need. (which will depend on things like what lens you have attached, are you using auto focus or Image stabilisation, etc.)

The data on the battery pack should give you the first 2, the third one may be a bit trickier to define - is there anything in the camera manual relating to it?

 

[Mikey P]

You could just drop a resistor in but you'll waste a bit of power doing this. Having said that, a regulator also needs some power to operate. Suspect that regulator is the way forward though, as suggested above, as it will also offer a little bit of protection. Suggest diode protection (Zener) in output circuit too if you can spare the voltage drop.

 

[bikething]

You could just drop a resistor in but you'll waste a bit of power doing this.

not that easy - the voltage dropped across the resistor, and hence the voltage available to the camera, will vary depending on the current draw. The current will vary - being highest while the lens auto focuses (motors) and dropping once the lens locks on and stops moving.

Having said that, a regulator also needs some power to operate. Suspect that regulator is the way forward though, as suggested above, as it will also offer a little bit of protection. Suggest diode protection (Zener) in output circuit too if you can spare the voltage drop.

Linear regulators are easy to use but waste power as heat. Switchmode regulators are more efficient but trickier to design, though I've seen some switch mode regulator chips lately that only require an external inductor and maybe a capacitor to work.

 

[HWMBLT]

Easy way to drop 1.4 volts is just to put two silicon diodes in series. Each diode drops about 0.7 volts. Don't need to calculate current or anything. If you use something like a 1N4001 they can carry about 1A. If you need more current just use a bigger diode. The diodes go in series with the lead.

 

[bikething]

Easy way to drop 1.4 volts is just to put two silicon diodes in series. Each diode drops about 0.7 volts. Don't need to calculate current or anything. If you use something like a 1N4001 they can carry about 1A. If you need more current just use a bigger diode. The diodes go in series with the lead.

That would work - provided the battery can deliver enough current.

The volt drop across the diodes is largely constant, so if the battery pack starts delivering a lower voltage due to current limiting / poor regulation / flat batteries, the voltage going to the camera would drop also - which may or may not damage the camera.

apart from that it's an elegant solution

 

[Wayland]

Easy way to drop 1.4 volts is just to put two silicon diodes in series. Each diode drops about 0.7 volts. Don't need to calculate current or anything. If you use something like a 1N4001 they can carry about 1A. If you need more current just use a bigger diode. The diodes go in series with the lead.

That sounds like an easy option. Anyone see any problems with this?

 

[HWMBLT]

I don't think low voltage will damage the camera after all batteries run down!

 

[bikething]

I don't think low voltage will damage the camera after all batteries run down!

And the camera switches off automatically before it gets too low - not sure if it does the same for an external power supply though.. might have to try it on my 10D if i can find a suitable connector to power it off a bench supply.

 

[bikething]

.. might have to try it on my 10D if i can find a suitable connector to power it off a bench supply.

maybe not. The 10D doesn't have an external power connection, nor does the 50D

(and why does it take 4 or 5 goes to post a reply in a thread without the site locking up :aargh4

 

[Wayland]

I don't think low voltage will damage the camera after all batteries run down!

The power pack has much higher capacity than the standard batteries so the voltage shoud lessen quite slowly anyway.

And the camera switches off automatically before it gets too low - not sure if it does the same for an external power supply though.. might have to try it on my 10D if i can find a suitable connector to power it off a bench supply.

maybe not. The 10D doesn't have an external power connection, nor does the 50D

(and why does it take 4 or 5 goes to post a reply in a thread without the site locking up :aargh4

I'm using a coupler from a cheap Chinese Ebay AC adapter.

The darn thing arrived faulty so I couldn't check the voltage output from the transformer unfortunately but the coupler works fine and I can hack the lead from the transformer to use the plug so no great loss.

 

[Teepee]

I have a little switching buck/boost unit that I used to use on my battery packs from a robotics company. Weighs 3.5g and will convert any voltage from 3v-12 input to 3v-14.4. -regulated , adjustable and 80-90% efficient.

If you have a voltmeter, the voltage can be dialled in to suit.

 

[Wayland]

Easy way to drop 1.4 volts is just to put two silicon diodes in series. Each diode drops about 0.7 volts. Don't need to calculate current or anything. If you use something like a 1N4001 they can carry about 1A. If you need more current just use a bigger diode. The diodes go in series with the lead.

I'm looking at two types in the Maplin Catalogue. 1A Glass Passivated Rectifiers or 1A Silicon Rectifiers. Both similar Spec.

Specifications:
Max. average forward rectified current:	1A
Peak forward surge current:	30A
Max. DC reverse current:	5uA
Junction capacitance typ.:	15pF
Temperature range:	-65 to +150°C
Part	Max. recurrent	Max. RMS	Max. DC
Number	peak reverse	voltage	blocking
	voltage		voltage
1N4001G	50V	35V	50V

Any thoughts on type?

 

[Wayland]

I have a little switching buck/boost unit that I used to use on my battery packs from a robotics company. Weighs 3.5g and will convert any voltage from 3v-12 input to 3v-14.4. -regulated , adjustable and 80-90% efficient.

If you have a voltmeter, the voltage can be dialled in to suit.

Any info or links?

 

[HWMBLT]

Silicon rectifier would be fine. I've got plenty. It would be no bother to post you some.

 

[Wayland]

That would be very kind of you.

I'll drop you a PM.

 

[Teepee]

You can have a play on the 16th if you want Its unused ATM.

Maplins do an adjustable regulator with a dial, Como drills I think makes them.

Mine were made by Dimension engineering-linky for any volt micro. This gives 05A. Check the voltage on this, its been a while and I honestly can't remember the exact specs.

Also, they make a step down with a 1A output. I use it to step 12v (8x AA cells) down to 5v. http://www.dimensionengineering.com/products/de-swadj

 

[Teepee]

Thinking about it, you can borrow the battery pack if you want. It takes any number of AA's / AAA's/ 9vs up to 8 and has the DE-Swadj wired in. The switch position is; off, full output/ regulated output.

 

[Wayland]

I wouldn't mind a look at it Pete.

It sounds like a useful bit of kit.

 

[Teepee]

I'll bring it with your PE sheet