AC Frequency Conversion – How to Convert 60V AC 50Hz to 60V AC 25Hz

I may be a year to late, but yes you may use it, and in fact your "new" wooden phone are more advanced than the plastic one.

Somehow you need to configure the hook switch with 2 sets of contacts or 3 contacts opening when going on hook. The impedance of your receiver may not match the one in the other phone, in that case we must try your induction coil/transformer, but you will probably make it good without, just copying the diagram you showed. The old transmitter may be another thing, the carbon granulates may be stuck in there, and some shaking, or even a little knocking on the rim of the transmitter element may help. The ringer should be fine as it is. You will find dedicated expertise on the Rotary phone forum.

http://www.classicrotaryphones.com/forum

What are you trying to do? You can step down voltages (90 to 3.3), and you can rectify alternating current to direct current using a bridge rectifier. Mentioning a current capacity like "1000 mAh" suggests you are using a battery or trying to replace a battery.

Batteries are typically rated with current capacities to show how much energy they can deliver over time. A 1000 mAh (or 1 amp-hour) battery would in theory be able to deliver 1A for one hour, or 0.5A for two hours, and so on. Typically they are derated as the current being drawn increases. Thus, 1 Ah battery may supply 100 mA for ten hours, but can only sustain 1A for a few minutes. Check the battery datasheet.

If you've simply meant milliamperes (mA) instead of milliamp-hours (mAh), then this means you need to be able to supply 1000 mA at 3.3V DC.

You can use a step-down regulator, also known as a buck converter, to convert 90V to 3.3V. No converter is 100% efficient, so you will need to have some overhead.

A bridge rectifier can convert AC to DC. There is an associated voltage drop, so you will need to account for that.

Assuming that you need 1A at 3.3V, that is 3.3 watts. Therefore you will need a bit more on the input side to account for inefficiencies. Using Ohm's Law, the current on the input side (\$I = \frac{P}{E}\$) will need to be more than 37 mA.