Can someone help me out with this 4X D size battery Voltage going to the PCB - Page 1
Forum section: electronics / beginners

author: axis937
author_rank: Newbie
timestamp: August 25, 2024, 09:19:51 am
content: Hiya, newbie learning how to fix thing up. A wall clock needed some TLC. The battery holder which was supposed to hold 4 D size battery was corroded due to battery leakage. Ripped it off. But before i did that, i drew how the wires were wired from the battery holder to the PCB. This clock comes with chimes and a dancing clown hence the PCB.Attached are 2 images.1 ) Firstly, there was 1 white wire dangling loose. I assume it was previously connected to Batt 1 Positive lead.2 ) There are 3 wires that heads down to the batt connector. RED BLACK & BROWN.I am confused on the voltage that is being supplied to the PCB. 2 of the batteries seems to be in Series and the other in Parallel. Can someone guide me on this?For a start, i think brown is Negative. No idea what Voltage is on the Red and Black.A little bit more of research is pointing me to 3V on the Brown 1.5V on the Red and 4.5V on the Black. Am not sure whether this is possible.

author: RoGeorge
author_rank: Super Contributor
timestamp: August 25, 2024, 09:52:55 am
content: Never seen a wall clock with that many and such big batteries. Is that the clock of a train-station, or alike?About the batteries, I will assume the 2 cells that are in parallel are for keeping the time. There are 2 in parallel, so you can change them one by one, without resetting the time. The other 2 batteries should be for display only. This is just a guess.The dangling wire might be just an antenna, I would leave it as it was, dangling (some clocks are adjusting themselves using a time beacon, in EU that is usually based on DCF77https://www.ptb.de/cms/en/ptb/fachabteilungen/abt4/fb-44/ag-442/dissemination-of-legal-time/dcf77.html).

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 25, 2024, 09:58:02 am
content: Typical D batteries have a nominal voltage of 1.5V, so two in series would yield 3V (in practice, 1.8V to 3.2V).In parallel, the voltage is between 1.6V and 0.9V, but current capability is twice that.If white is connected to black, then black is the most positive voltage (0V), red is -1.5V, and brown is -3V. That is, red(-) to black(+) would be 1.5V, and brown(-) to black(+) 3V, nominally. No part of the circuit would use brown to red, because that would discharge batteries 1 and 2 in series to batteries 3 and 4 in parallel.If it were connected anywhere else, then the load on the batteries would be uneven, draining some of the batteries before the others. Unlikely, but possible.White cannot be an antenna, because it is connected to batteries 3 and 4 positives. It also must be connected somewhere, as otherwise batteries 3 and 4 would be unused and useless.I'd say it is time to trace those three on the PCB, and see where they go. Identifying any chip, and which traces it uses for power, might tell you whether the above is correct.

author: axis937
author_rank: Newbie
timestamp: August 25, 2024, 10:01:08 am
content: Hiya, attached is the image of the clock before after i ripped off the D battery holders . 2 D size battery was on the left and 2 on the right.No antenna for this clock.

author: axis937
author_rank: Newbie
timestamp: August 25, 2024, 10:10:20 am
content: Wow...thanks. really good explanation. albeit will take me some time to digest . So may i clarify then, What is the black voltage then? 4.5v in this circuit? Will take a look at the IC.

author: RoGeorge
author_rank: Super Contributor
timestamp: August 25, 2024, 10:11:06 am
content: Well, the last picture tells the clock is very different than what I've imagined it to be.Looks like it's a standard clock with an AA battery, plus some electro-mechanically actuators for the extra figurines on the display. I would try to remove/dismantle the front cover, and follow each wire in order to figure out if the wires goes to motors or electromagnets.

author: axis937
author_rank: Newbie
timestamp: August 25, 2024, 10:52:44 am
content: Based on the explanation i have received, if i were to replace all the initial 4 batteries to a single 4.5V battery, connect to negative leads to the 1.5v and 3v and 4.5V Positive. Will it work?

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 25, 2024, 11:52:13 am
content: Quote from: axis937 on August 25, 2024, 10:52:44 amI don't think so.

author: axis937
author_rank: Newbie
timestamp: August 26, 2024, 05:57:09 am
content: So do you guys belief that the loose white wire is not supposed to be at the positive end of battery 01?

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 26, 2024, 07:38:20 am
content: Quote from: axis937 on August 26, 2024, 05:57:09 amI believe the white wire should be connected to the black wire, possibly at the positive end of battery 1. See the second and third paragraph in#2as to why.It can seem odd to have a circuit with a common positive and two separate negative rails, but because this is battery-powered, it does not matter at all. It only means whoever designed the circuit, found it useful to have a common most positive rail, and two other rails, 1.5V and 3V below that.If you showed the whole circuit board, both sides, I bet we'd see some N-channel MOSFETs or NPN transistors that use the low-voltage low-current signal from the actual clock (powered by its own 1.5V AA battery) to drive the higher-current motors. MOSFETs in particular tend to need more than the 0.9V–1.6V to fully conduct, so the 3V (1.8V–3.2V) makes sense for such circuits. I think there are small toy motors that work just fine at 0.9V–1.6V, but most tend to work better at 3V. NPN transistors would work at 0.9V–1.6V just fine, but there would be an additional diode drop (about 0.6V) on the load. Thus, I see at least two completely different ways the animatronics could be implemented: using MOSFET half H-bridge with 1.5V motor voltage and 3V logic; or using NPN transistor logic at 1.5V logic voltage and 3V motor voltage. Because of how the batteries are wired, giving double the current capability at 1.5V, I'd bet on MOSFET-based circuitry. (There are additional ways, including Darlington pairs and such, for example ULN2003 driver, though.)In essence, I believe the idea of having most positive the common rail, and two rails beneath that, -1.5V at high current and -3V at low current, is by design.