Acurite Digital Readout Power Supply
Hope someone can help.
I could not find a listed device even close.
The primary windings of the transformer have gone (open circuit). I am trying to find the secondary voltage so I can purchase a new transformer. There are no markings on the transformer. One of the first components after the rectifier on the pcb is an LM340T5780P+ voltage regulator (wired as fixed output). Is it possible to work out it’s supply voltage and hence input voltage from other components from the pcb?
Spec sheet says Vout - 5,12,15. Vin - 10,19,23. Are these just test voltages for the specs or are they the only 3 operating voltages?
It is connected to 2 capacitors with markings - K5R 334K & 105 +25K.
The transformer connects to the two pins.
There is no model number - Acurite Millmate.
The only numbers on the board are; 387575-6033 (printed) and 7-19 570702 1-29-90 (hand written) and ACU-RITE 387575-6034 ASSM. REV A (printed).
There does not appear to be a thermal fuse within the transformer. It has 4 primary windings and a single secondary winding.
The transformer is connected to the two pins shown in photos 1 and 3.
Is it possible to workout the supply voltage from the voltage regulator and the components around it - capacitors and resistors?
Do you mean the tiny capacitors near the regulator that I gave the numbers for, or the electrolytic capacitors?
The main transformer is connected directly to a RMS401L rectifier (Max 35V). The rectifier is connected directly to some electrolytic capacitors, the LM340T5 voltage regulator, an IC - TL494 and the small capacitors associated with the LM340T5 as listed above. The electrolytic capacitors are rated at 16V, the IC is rated at 7-40Vand the regulator is rated at 5V out and 5 to 18V in.
If I had to take a guess, it would be 12V supply to the board. Would that be correct? Could I damage anything with 12V - Would 10V be better?
Can anyone help