alright well i was working on the power regulator for my new board and i had 4 AA's hooked up like nobodys business, and im getting a steady 4.24V. The AA's themselves are not fully charged and are 5.07V combined. I want to add two additional AA bats, using the original 28x1 board case (normally 3 slotted) with a jumper, im not sure if this makes a difference but ill mention it. When i tried adding the other two i smelled a burning smell and then burnt my finger on one of the springs, the V while this was happening dropped to .8V or something. I'd rather not risk my new capacitors or my batteries (or finger) so ill take as many pictures as i can.
using Odd's schematic for the 6v
Im not sure if you can see all the connections, also the black wire doesnt always correspond to ground. On the last picture its V2
To connect the other batteries two the circuit i had two wire stick out of the + and ground rails and just holding them on the heads of the battery
Also my board. Im just going to solder in the circuit exactly like the schematic. As of now i connected all the + terminals on the board, im not sure if thats how you combine all the elctricity to one line.
Right now the middle line is where all the V+ is connected two, is that how it works? from there im going to make it look like the breadboard and use the one V+ line as V1.
Do i have to connect all the grounds? or can i wire them to anyone? I feel im missing very basic electronics knowledge : (
Hi. All linear regulators have a minimum voltage dropout which means they need that extra voltage from the battery to operate properly.
A low drop out regulator may only need 0.5-1.5 volts over the output voltage. So if its a 5v regulator the battery must be a minimum 5v + dropout = about 6 volts. Hence the 6v battery in the schematic.
If your battery voltage is 5.07v it`s too low for the regulator to work properly.
As for the burning… it looks like you have the capacitors connected backwards in the 2nd last picture of the regulator. There is a black wire going from the +ve rail to the -ve lead of the capacitor. In fact you have all the wire colours backwards. Hook up the colours around the right way to save confusion.
alright finally got it alright finally got it sorted out, thanks. I rechecked everything, now when i add the extra batteries it jumps by like .02 and thats all. Nothing heats up
Here is how the batteries should be connected to provide power to the regulator :
Your description sounds like the positive of one pack connected to the positive of the other, making a parallel rather than series connection. With that, one pack would try to charge the smaller, and less power is available. ALso, it is best to use rechargable batteries, these look like alkalines.
Ahhh, i was wondering why Ahhh, i was wondering why the voltage would drop when i plugged in the smaller pack, it would also explain how the battery burned me. I assume this also applies if the V differential of two packs of 4AA’s were connected.
If you have 2 equal packs, then there probably won’t be much of a problem connecting them parallel, you’d just never get a higher voltage. In parallel, 4.8 + 4.8 volts = 4.8 volts, but much more Amp running capacity, twice as much.
But connecting positive of first to negative of second and using the second pack positive as output, 1st pack negative as ground, the volts add to 9.6, but the current capacity remains the same.
The problem occurs when connecting unequal packs in parallel, as one tries to maintain 4.8 volts while the other is trying to hold 3.6 volts. THe 4.8 will discharge into the 3.6, sometimes in a quick hot manner.
