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voltage output

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bogdanfirst

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ok, so i have a transformer wich has an output of AC of 11.8V with no load. the resistance of the secondary is aproximately 0.6Ohms.
i connect it to a rectifier bridge and a filtering cap, 1000uF or more if necessary. what output oc D.C. do i get with no load and with 1.5A load?
is it :
1. 11.8*2/sqrt(2)-1.2V (no load) or
2. (11.8-1.2)*2/sqrt(2) (no load)?
the 1.2 volts i consider from the voltage drop of the 2 diodes that the current passes trough.
well, i actually want to get 12V out of this transformer and i thought of using a 7812 votage regulator. the fact is that the regulator requires a drop of at least 3V on it. so i was curious if when i have the maximum load of 1.5A(but i actually don't think that my circuit will draw more than 1A-1.2A) the regulator can still suply me with 12V.
hope someone can clarify this to me.
 

Chippie

Member
I generally use the ac*1.414 product as a rule of thumb, so at 11.8v ac that will yield somewhere around the 16.5v mark......off load!!!loose about 1.5v for the diodes and you are around the 15.v mark.

As your reg needs Vout+3.......you are pretty close to drop out...Try it and see..... 8)


As for the 1.5 amp load.....the ac output of the transformer will drop, by how much depends on the regulation....this is where the quality of the transfo comes in.....if its built to a price then it will not be too good......usually the case.


Have fun :)
 

stevez

Active Member
The rectifier output is pulsed DC with a peak value equal to the AC peak value minus a little for loss across the rectifier. Without a load the capacitor(s) will be pumped up to the peak voltage (1.414 X the RMS voltage, as suggested). As the load is increased the voltage will drop to approximately the RMS value minus some losses across the rectifier. There will be some ripple which will increase as the load increases. There is a formula that will help to calculate precisely where the voltage will settle but it appears that many designers just take the RMS voltage because it's easier and yeilds conservative results. Chapter 6 of the 1994 Radio Amateur's Handbook does a great job explaining all this and includes the formulas to help get the info you need. If I had an electronic file version I'd send it - but I don't. I had to dig a little further into my textbooks to get the formulas for estimating where the output voltage would settle. Somehow, somewhere on the internet there must be an on-line version of what I have in the handbook.
 

bogdanfirst

New Member
thanks Chippie!
wel my question was actually do i multiply with 1.41(or 2/sqrt(2)) and then remove the 1.5V from the dioades or remove the 1.5V and then multiply with 1.41.
but i got all sorted aout now.
i'll try it and see how it reacts.
thanks.
 
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