I have a question about general electronics, I'm sorry if it's not about arduino, but you guys know a lot about electronics, hope it's ok.
I have an old toy WowWee robot (RS media), almost never used. It uses 6x large D batteries and I have bought new rechargeable ones, but the robot shuts down due to "empty batteries" and they said that due to low voltage (1.2V instead of 1.5V), the unit thinks the batteries are depleted and shuts down.
Can anyone think about a hack around this? Either to increase the voltage somehow or possibly modify the hardware? Is it possible there is an obvious hardware part that measures the voltage I could rewire?
A possible fix is to add one more rechargeable cell, in a single cell holder, in series with the original battery pack.
Ah yes, this would increase the voltage from 7.2 V to 8.4V which is close to 9V. I'm not sure if the reqired voltage is 3x1.5 or 6x1.5 since it has 3 batteries in each leg.
Could this damage the electronics due to lower resistance of the batteries?
I wonder if there exists rechargeable batteries with 1.5 voltage, li-ion perhaps...
No, and no.
I'm not sure if the reqired voltage is 3x1.5 or 6x1.5 since it has 3 batteries in each leg.
Post a picture that includes the wiring or carefully trace it yourself.
If the battery holder is center tapped or in parallel, then you will probably have to add a cell to each side.
Well, there's no place for new batteries, specially another D (R20) battery. But perhaps I can glue a battery holder somewhere else on the robot, since its large. And rewire the whole thing.
I do not quite get that it has 3xD batteries in each leg + 2xAA batteries. Perhaps they do different functions (D ones for movement and the AA ones for talk and leds..?) I'll trace the wiring...
I have to check which ones are causing problems, the big ones or the AA ones.
Another idea: could I just stick two 5V power banks in each leg (I think they should fit or I can make them fit) and wire them to the 3xD battery holder? Then each leg would have 5V instead of 4.5V and it would be easy to charge via two USB's. Any downsides here?
6hearts:
Can anyone think about a hack around this? Either to increase the voltage somehow or possibly modify the hardware? Is it possible there is an obvious hardware part that measures the voltage I could rewire?
So cool that we live in an age where there is such a thing as an "Old Robot" ;D
How about a DC-to-DC converter. It would boost the battery voltage to the nominal voltage, regardless of the battery voltage.
Like this one: https://www.ebay.com/itm/LED-DC-DC-Digital-Boost-Step-up-Voltage-Converter-Board-LM2577-3V-34V-to-4V-35V/132056133502?hash=item1ebf28af7e:g:QPEAAOSwt5hYb2Nx
I'm basing this on the Eveready datasheet for an Alkaline D-Cell which is rated to 500mA, and with an internal resistance [worst case when fresh] of 400mΩ. So, if the battery arrangement in your geezer-bot is 6 cells in series, then I would expect the current demand to not be much more than a half an amp [might be current surges to 1A to drive the motors, but @ 400mΩ, that's a loss of 400mV at 1A, so probably not -- there is probably sufficient capacitance to even out the current surges].
6 cells, at 1.5V is 9V nominal. So, adjust the DC-to-DC converter for that voltage.
On downside, though. There is the danger of running the batteries down too low, and ruining them, or at least, impacting their longevity. So, either a battery manager is indicated, or because this thing has a nice big voltage readout (defaults to Input Voltage), you can keep an eye on the battery condition. Plus, this will give your robot yet more "tech-vizage"!
6hearts:
Another idea: could I just stick two 5V power banks in each leg (I think they should fit or I can make them fit) and wire them to the 3xD battery holder? Then each leg would have 5V instead of 4.5V and it would be easy to charge via two USB's. Any downsides here?
A fresh Alkaline battery can be as high as 1.62 V so 3 x 1.62 = 4.86 which isn't quite 5V, so it's still a bit of a crap-shoot -- if you wish to play it safe, put a schottky diode in series, like a 1N5818.
ReverseEMF:
So cool that we live in an age where there is such a thing as an "Old Robot" ;D
Haha, indeed 
ReverseEMF:
How about a DC-to-DC converter. It would boost the battery voltage to the nominal voltage, regardless of the battery voltage.
Cool idea, thanks! Thanks also for the link. The robot is a large beast, so I'm sure it's half empty so I can put the circuit inside it somewhere.
Perhaps I can stick a volt-meter it to monitor the voltage. Maybe also a Zener diode that would disconnect once the voltage is too low, but then the question is what voltage is the limit?
6hearts:
Perhaps I can stick a volt-meter it to monitor the voltage.
The board I referenced, includes a voltmeter. So, if you find a way to mount the board on the outside of the bot, the "two birds" would be cool, exposed, robot-ish electronics -- and a voltmeter for monitoring the batteries.
6hearts:
Maybe also a Zener diode that would disconnect once the voltage is too low, but then the question is what voltage is the limit?
If the rechargeable batteries are NiMH, then 1.1V per cell. NiCads can go a bit lower -- I've seen 0.9V, but it's been a long time since I worked with NiCads. Lithium's a different story and I will defer to others, since my experience is limited. One of those 5V USB Battery Packs would shut it self off, so there's that 
Shutting the battery off would be an excellent job for an Arduino (perhaps a Pro Mini, or Trinket). Monitor the battery(s) with a voltage divider and Analog port, and shut off the battery(s) with one or more Digital Outputs and a transistor or two.
High Side switch:
Low Side Switch:
R1 = [100k(VBMAX - VA0MAX)] / VA0MAX
Where:
- VBMAX is the Highest expected Battery Voltage
- VA0MAX is the highest voltage to allow on the A0 pin (4.5V recommended)
The 100nF capacitor is to reduce the impedance to below 10k to satisfy the ADC sample time requirement. Since the voltage divider will be across the battery, we want it to have as little impact on discharge time, as possible. In fact, that 100k resistor could, probably be increased in value [remember to increase it in the formula, as well].
