How to drop voltage for a TT motor?

Status
Not open for further replies.
Hi,

vintage Dual turntables are driven by synchronous (or asynchronous, depending on the unit) motors with dual windings that can be series connected for 220-240 Vac use or in parallel for 110-120 Vac - at least unless the TT comes from a Dual integrated console, containing the amplifier and sometimes a radio. These TT's are fed from a 150 V tap of the unit's mains transformer - for whatever reason. So the motor's winding is designed for this supply.

Sometimes we want to use such TT's as stand-alone units, so the voltage has to be adapted. We could use the console's mains transformer, which were a bad solution, due to it's weight and inevitable magnetic stray. Now voltage dropping comes into mind.

I'm aware of the complex nature of the load that a motor presents, consisting of the winding's inductance as the reactive part and the resistance, in series (or in parallel?) with the mechanical loading, as the real (ohmic) part. How can I discern between both in order to calculate an appropriate dropping resistor? Is even a dropping capacitor (in order to minimize loss) possible? How would I calculate this one?

To give you some information: In the datasheets of their turntables, Dual usually tells 10 watts as the motor's power consumption.

Best regards!
 
Last edited:
Jon's capacitor may work fine.

Depends a lot on the motor details.

I have seen a series cap (on a non-sync induction motor) RAISE the speed and power consumption. The cap resonates with the motor inductance. A Synchronous motor "can" be "tuned" to act inductive, capacitive, or resisty- but that requires adjusting the Field current, which is not an option on tiny sync motors.

Actually a "sync" motor this small may not be a full Synchronous machine, but a salient-pole induction motor. Which will be inductive. In this size, probably a large resistive component. Which will tend to damp any L-C resonance, but in my work with a 43W motor, not enough.

I have had fine results (every time my furnace runs) with a simple resistor.

10W @ 150V is 0.066A
230V-150V = 80V drop
80V @ 0.066A is 1,200 Ohms 5W (use 10+W)

Certainly a handy 1K 10W is a quick and safe thing to try.

Finally a split-primary 115V+115V:36V 10VA transformer can be cobbled-up as a 230V:151V auto-transformer.
 
I'm aware that reactances of capacitive and inductive natures are of opposite phase. So the voltage over a dumping capacitor will exceed the mains voltage by the value of the voltage the motor's inductance sees. To do relilable math, I'll have to determine/measure the motor's inductance and it's ohmic resistance under load. How to do this?


Even calculating the exact value of a dropping resistor needs knowledge of the inductance's value.


And as said before, I'd want to avoid any transformer.


Best regards!
 
> calculating the exact value of a dropping resistor

The main claim of a synchronous motor is that it is NOT critical about its voltage.

If voltage is low, it will run weak. If voltage is high it may run warm.

It is likely a well-stocked junk-box can get you a trial value faster than thinking.
 
Status
Not open for further replies.