There are some low cost 1:1 high impedance audio transformers on e-bay that might do the job.
I have no idea if it can handle 12v at 60Hz without core saturation, but for the price it would be worth a try.
**broken link removed**
Not well considered. Saturation would also show at no load. Seeing the "clipped" waveform in loaded state indicates too large transformer windings resistance in combination with the rectifier and filter capacitor, just normal behavior. Measure the transformer parameters and you can predict the waveforms in a simulation.If I load this to about 10ma (that's what I need) voltage drops to about 8V (too low) and the secondary ac waveform is severely clipped, is that due to core saturation?
Not well considered. Saturation would also show at no load. Seeing the "clipped" waveform in loaded state indicates too large transformer windings resistance in combination with the rectifier and filter capacitor, just normal behavior. Measure the transformer parameters and you can predict the waveforms in a simulation.
Calculating transformer windings is easy. Calculate 42 cm² windings/core cross section for 50 Hz or 35 cm² windings/cross section for 60 Hz as a starting point.
Ratio 1:1
12Vrms
0.2VA
60Hz
Is this something I could wind myself in an hour?
What core should I use and roughly how many turns would be required.
I searched for a transformer like this but did not find anything. Perhaps I'm not searching in the right place?
may I ask if your requirements are;
time? 1hr, times up.
cost? how many?
learning how to design magnetics? do some math
- The impedance of the transformer with no load must be high compared to your load to give somewhat ideal transfer and loading characteristics.
- Pmax=ξ*VA =Vrms²/Rload so for ξ=1, Rload= 12² / 0.2 = 72 Ohms
- Meanwhile the magnetizing inductance at 60Hz to be much great than the load means much greater than 10mH.
- This would require a high number of turns or much higher permeability core.
purpose?
If using to rectify into DC the load becomes the ESR of the diodes and Cap during charge and then open circuit during discharge.
Since %ripple can be equated to Iavg/Ipk = % duty cycle... it is equal to Impedance ratio % {source / load}.
In order to provide low conduction losses I²* R, the winding source resistance must be very low to achieve desired % ripple, and % load regulation.
make sense?
try again. This time with purpose and constraints.
The numbers do not add up. You only want to pay $2 to buy each transformer but you are willing to spend 1 hour of labor to make one.
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