Powering an old two box metal detector

Question

I recently bought an old two-box metal detector that uses radio tubes and runs on the old radio batteries. There are two boxes, One is a transmitter box, the other is the receiver box. one with 6 tubes and one 45V battery (but has a third, center + connection, so is - + + the center plus terminal is 22.5V ) Then, there are two seperate 1.5V batteries also. The other box has two tubes and a 45V battery (no center 22.5V connection) and one seperate 1.5V battery also.

My question is this: I am trying to change the battery system to rechargable batteries. I made a battery pack out of 5 nine volt batteries, but the rechargables charge up to 10 volts 50V total) initially each, then I made a separate battery pack for the 22.5V out of 1.2 AA batteries, which initally charge up to 1.364 volts each (16 come up to about 23.188V 15 to 20.46V) how can I run down the batteries to get to the correct voltage(s)?

Or, does anyone know if these old detectors will take those over voltages? all batteries are connected "in series".

Also, do you think the rechargables probable higher mHa ratings(250 for 9V and 2,000 for 1.2V) could create any problems? This is the first time I have done this, so kind of new at it!

I was going to use alkalines first, because the voltages are closer, but this machine eats batteries, so I am told and rechargable batteries are so much cheaper, in the long run.

Answer 1

You will probably find it eats 1.5V batteries but the 45V current draw is quite low...

As for approximating the 22.5/45V voltages; I recommend erring on the low side rather than going over-voltage. The glassfets will take the over-voltage quite happily, but 50-year old capacitors, probably not...

And HT batteries were commonly used down to a fraction of their nominal voltage, so a stiff supply from rechargeables at 90% of rated voltage should be fine.

An alternative to rechargeables for the 45V supply would be a stack of cheap eBay lithium coin cells.

Also be aware that old battery equipment often used a quite alien design pattern to modern designers : there was no direct DC path from HT to earth; therefore the HT battery was permanently connected, not switched. The only current path from HT was through the glassfets, so switching their heaters off reduced the HT current to 0. (Check this with the heaters off; any leakage may indicate a faulty capacitor.)

Answer 2

To run this relic from a modern battery pack, I'd probably use a single off the shelf pack with boost and buck converters to make the separate voltages. The higher voltages will probably require little current. The 1.5 V is probably for the fillaments so will have substantial current, probably a few 100 mA.

One issue may be whether it is OK to connect the 1.5 V supply somehow with the other supply. Since this was a separate battery, the design may have made use of the fact that it was completely floating with respect to the other voltages. For efficiency, these tubes probably don't have separate heaters. They probably have a integrated cathode/heater. Sometimes these are just two connections with one serving both purposes, and sometimes the fillament is center-tapped for the cathode connection. Either way, it may be that neither side of the 1.5 V supply can be tied to the ground of the high voltage supply.

If the supplies do need to be floating, then I'd probably put the battery on the high curren one and make the 1.5 V from a straight buck converter. The higher voltage can be produced by a flyback converted.

Real current measurements and finding whether the supplies need to be isolated would help a lot.