Hi all
I have a Hoseki Digital LED Alarm Clock Model H-5008 / 230VAC
The primary of the tiny transformer inside has gone open.
I have some similar transformers with me but I do not know what is the output of the original transformer. There is no marking on the pcb about the input voltage.
Is there any way to find out? Start at a low voltage with a variac until the unit starts working?
Any help will be appreciated.
-- Thanks in advance ClueLess
Look at any filter caps near some diodes that the output of the transformer connects to. For example, 6.3 VDC marking on the caps means the supply probably is designed for 5 vdc.
Also look around and see if there is a small 3 pin regulator with markings like 7805 (5 volt) or 7812 (12 volt)
Jeff
-- "Everything from Crackers to Coffins"
Look at the voltage ratings on the capacitors, then go down to about
1/2 of that for the secondary voltage of the transformer. Remember, the peak voltage is 1.414 times the nominal voltage. So, for example, if the capacitors were rated at 50V, I would try a transformer that gave a rectified voltage of about 35 - 40 V DC, Divide the 40V DC by 1.414 to get a transformer that had an AC output of about 25 - 30 V ac."Clueless"
** That is what I would do.Make sure it is working just fine and then add 20 % more to cover low AC supply.
... Phil
Thanks Phil.
I did that and found that it works at 6 volts
Thanks you and others who responded to my query.
PS: I di write to Hoseki twice but the buggers never responded. That is customer relation, I suppose.
-- ClueLess
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