Need help with going from 6V to 5V

[Rockworthy]

Hello there helpful forum! I have this costume I'm making for halloween. I'm using LEDs and lasers and things, and I'm a real new guy to electronics in general. I have a problem because I can't use the AC adapters that they come with for power. I need to have a simple way to make this battery powered so that it's portable. The AC adapters provide 1500mA of power at 5V. What is the simplest way to make (or buy) a portable battery system? It needs to be only a couple of pounds in weight, at most. I have these big 6V lantern batteries. What's the simplest way to drop the voltage to 5V? Also I don't have time to order any thing by mail. Can anyone help? Thanks so much!

 

[davelectronic]

Hi there Rockworthy.
If you only need to drop a volt, from 6 volts to 5 volts you could use a single 3 amp silicone rectifier diode foward biased, this will drop the voltage to 5.3 volts approxamatley, i cant see the extra 0.3 volts as a problem, no need for regulators, if your not happy with 0.3 volts over, add a second diode in series with the first and under power at 4.6 volts, but i would opt for 5.3 volts.
Dave.
PS, if the batteries are in series you will need a regulator circuit like a LM7806, 6+ volts regulator.

 

[davenn]

Appolagies post appears twice on a PS, Edit

you can edit it yourself dave
just delete the first post

 

[davelectronic]

Thanks Dave.
Still showing my newbie status, Thanks.
Dave.

 

[KJ6EAD]

The single diode trick is a good one. It would be useful to know how much current the circuits actually draw in use; obviously less than 1.5A but how much less?

 

[Rockworthy]

Hmm, our diode didn't work...

Well I went to the store and got a 3A rectifier diode, epoxy-type, and it doesn't seem to be doing anything. It only drops the voltage like 200 mA or something. Do I need to specifically use the silicon ones? Radio Shack only has the epoxy type...

 

[Resqueline]

So 200mA (or is it actually 200mV) is nothing? And how did you measure this? With no load perhaps? Another question is how accurate do you need the 5V to be?
Dave did mention that you could use two diodes if you wanted to, to increase the drop. Did you get only one diode? Silicon is inside, epoxy is the casing material btw..
And to repeat KJ6EAD's question; how much current do you plan on drawing?

 

[Rockworthy]

So 200mA (or is it actually 200mV) is nothing? And how did you measure this? With no load perhaps? Another question is how accurate do you need the 5V to be?
Dave did mention that you could use two diodes if you wanted to, to increase the drop. Did you get only one diode? Silicon is inside, epoxy is the casing material btw..
And to repeat KJ6EAD's question; how much current do you plan on drawing?

The current draw is 1500mA. I'm not sure how close to 5V the laser needs to work right. What is weird is how it's only like $2 on eBay to buy a 12V cigarette lighter-to-USB-5V converter. I only need to go from 6V to 5V! I'm sure there is a really simple solution, I'm just uneducated when it comes to electronics and stuff

 

[davelectronic]

6 volts to 5 volts

the single silicon diode will give you 5.3 volts if forward biased, posative lead out, or from the battery through the diode, look up silicone diode forward biased. If your batterys are in series, look up voltage regulator circuit, it will work, if volts is critical, you need a close tolerance regulator circuit, all resitors should be 1 % in this circuit.
Dave.

 

[Resqueline]

So you're running the AC adapters to the max as it is, or perhaps the devices actually draws a little less?
Why don't you connect two lantern batteries in series to get 12V and use those cig-to-USB converters.
It's "trivial" to make a step-down converter (buck) when you have plenty of overhead voltage. The market for those is big too.
It's a little more complex to design a buck/boost converter. Remember that you get 6.6V when the battery is new, dropping to 4.4V when it's spent.

 

[KJ6EAD]

The current draw is 1500mA.

I don't believe it's that high. You need to measure it, not read the supply capacity labels. If it really is that high then carry a small backpack 12V battery and use two or three of those 1A USB buck converters as suggested unless you can find one rated for 2A.

I only need to go from 6V to 5V! I'm sure there is a really simple solution.

No, there isn't. Well, there was, but you weren't satisfied with the simple 0.7V silicon diode drop solution that was offered and linear voltage regulators require at least 1.2V of input to output differential. If you had a giant 9V battery then there would be a "really simple" solution. Rigging 12 volts and a buck converter is not difficult and it has good efficiency. I still suspect that you could get by with an 8-pack of AA batteries but I'll never know unless you measure the current drawn by the circuits.

 

[Rockworthy]

So you're running the AC adapters to the max as it is, or perhaps the devices actually draws a little less?
Why don't you connect two lantern batteries in series to get 12V and use those cig-to-USB converters.
It's "trivial" to make a step-down converter (buck) when you have plenty of overhead voltage. The market for those is big too.
It's a little more complex to design a buck/boost converter. Remember that you get 6.6V when the battery is new, dropping to 4.4V when it's spent.

YAAAAY THANK YOU Rosquiline! This was the tasty little nugget of info I was looking for. I think we can make it work now