Testing Quantum Turbos without batteries

[fdonadio]

Hello everybody!


I have a handful of Quantum Turbos that were given to me by a friend and I want to test them. Every single one had bad batteries (weren't holding any charge anymore), so I had to take them to recycling.

Now I want to test them and the damn 8V batteries are nowhere to be found in Brazil. I have several 12V batteries, but that's 50% more current and that would certainly cause problems. I can solder and have fixed stuff by replacing failed components, but I lack the theoretical knowledge to regulate the output of a 12V battery to 8V. I've read about using resistors to do that, but I am still not sure enough to try.

It would be only a way to test them, not a definitive solution to use them.

Any ideas?


Cheers,
Flavio

 

[wiltw]

So you cannot obtain batteries like these for the original Quantum Turbo?

https://www.amazon.com/PowerStar-PS-832-17-Quantum-Battery-Replaces/dp/B00ZY4KNIG

It seems this company can provide them 'in the Americas'

https://www.trustedparts.com/en/part/power-sonic/PS-832F1

All you need for testing is a DC power supply that can provide
8V 160mA

 

[fdonadio]

So you cannot obtain batteries like these for the original Quantum Turbo?

Yes, these PS-832 batteries are impossible to find here. Shipping from overseas is more expensive than the battery itself, so it only makes sense to buy more than one battery to save on shipping (I mean, per battery). So, I must know how many batteries I should get before ordering, and that’s why I want to test all the Turbos.

I don’t believe a 160mA power supply can feed a flash connected to the Turbo. The battery charger that comes with the Turbo is spec’d like that, though.

 

[wiltw]

Yes, these PS-832 batteries are impossible to find here. Shipping from overseas is more expensive than the battery itself, so it only makes sense to buy more than one battery to save on shipping (I mean, per battery). So, I must know how many batteries I should get before ordering, and that’s why I want to test all the Turbos.

I don’t believe a 160mA power supply can feed a flash connected to the Turbo. The battery charger that comes with the Turbo is spec’d like that, though.

The 160mA is the specification for the current delivered by the PS832, the rate which can be delivered; OTOH the Ah is the total capacity of how much current for how many hours

 

[UnderwoodPhoto]

Look at some of the smaller emergency lights or stores that sell them. They will have either a 4 volt or 6 volt SLA or AGM battery inside. Two of the 4 volt batteries fit perfectly.

 

[fdonadio]

The 160mA is the specification for the current delivered by the PS832, the rate which can be delivered; OTOH the Ah is the total capacity of how much current for how many hours

I have access to “beefier” power supplies. I guess this will work on my bench. Thanks for the tips!

 

[fdonadio]

Look at some of the smaller emergency lights or stores that sell them. They will have either a 4 volt or 6 volt SLA or AGM battery inside. Two of the 4 volt batteries fit perfectly.

I’m gonna take a look at those. Thanks!

 

[AgX]

I have several 12V batteries, but that's 50% more current and that would certainly cause problems. I can solder and have fixed stuff by replacing failed components, but I lack the theoretical knowledge to regulate the output of a 12V battery to 8V. I've read about using resistors to do that, but I am still not sure enough to try.

In any good workshop there needs to be lab power-supply.

But as you hinted at your 12V battery at hand: if you put 1 or more resistors in series at a voltage source, you can take over each resistor a different voltage than the primary one delivered by the battery directly:

https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRdSqrdDqWWekNLdwE11-c4YgP2-odd5--cCg&usqp=CAU

Maybe someone else now can come up with the respective math behind it...

 

[fdonadio]

In any good workshop there needs to be lab power-supply.

Yes, I can borrow one of those from a friend. I should invest in one, but...

https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRdSqrdDqWWekNLdwE11-c4YgP2-odd5--cCg&usqp=CAU

Maybe someone else now can come up with the respective math behind it...

The page where you got this image from, most probably, has the math. I have found several explanations through my Google searches, but my problem is knowing how much power the Turbo will draw from the battery when cycling the flash unit (seems to be a lot more than 160mA). I don't know what happens to the circuit if the draw exceeds the resistors' rating.

I am not very fond of the idea of having things exploding in my face.

Thanks for the help!

Cheers,
Flavio

 

[fdonadio]

I think I have just understood a little more about the issue: resistors are only useful when the current draw is constant. In my case, the Turbo will consume pretty much nothing when "idle", but a lot when cycling the flash. When idle, the resistors would be dissipating a lot of power. I need a better solution.

 

[wiltw]

If you think about charge times for flash units, some batteries (alkaline) are slower in chage time than others (NiMH). IOW, the amount of current that CAN be delivered by a battery chemistry only changes the amount of time necessary to charge the power capacitor...so there is not an absolute rate at which the PS832 has to provide current (although it does permit full power recharge of the power capacitor in 2 sec). For your testing you only need to validate function, not recharge time (which the battery is the determinant).

 

[UnderwoodPhoto]

The Turbo at 8 volts provided 2 volts more than the flash it was powering. It depended on resistive losses in the cable to limit current which would exceed 2 amps easily. The circuit in flash units is not voltage critical within reason so a few extra volts wont hurt.

 

[BradS]

Fresh ordinary D cells deliver about 1.55 volts. Five of those in series would supply about the correct voltage. I might even try six in series. The internal resistance may be high enough to make the flash cycle slowly but it should be good enough to test them.

 

[fdonadio]

@wiltw Uhhmmm... I didn't about that. I thought the flash would "pull" too much energy and, because of that, could damage the power supply. I got this idea from RC battery discharge rates and the (maybe false) common-sense that high loads could make the battery overheat.

@UnderwoodPhoto If the Turbo already supplies more than the flash unit needs, I guess it wouldn't be very wise to make that even higher. But I will consider that.

@BradS 6 alkaline cells would total around 9V, which is close enough. I also have some NiMH cells (18650), with which I could make a pack. Seven of them will total 8.4V.

Thanks all of you for the ideas. It's always good to acquire good knowledge.