Convert 10v AC to 9v DC

10V AC is 14V peak to peak, so this would work. How did you measure this voltage ? If it is the open circuit voltage, you may have a too large voltage drop under load to be able to use this suply.

You'll need a basic regulated power supply like this :

Bridge ____ rectifier 1000uF 100nF | | 50uF 100nF +-+----+----+----|7809|---+----+--- +9V A A | | I|____|O | | AC ----+ | --- --- | --- --- in ----(-+ --- --- | --- --- A A | | | | | +-+----+----+------+------+----+--- Ground

(created by AACircuit v1.28.6 beta 04/19/05

I'll explain the different parts a bit :

The diode bridge is used to rectify the AC, that is invert all negative portions of the waveform and get only positive half-sines.

Next, the 2 capacitors in parallel filter this rectified waveform, the big 1000uF (use 25V type at least) to filter ripple and the 100nF to filter higher frequency noise. You have about 13V DC at this point.

The 7809 regulates the voltage down to 9V. You need at least 2V higher voltage at input for this to work, so 13V would do, but as I said before under load your power supply may not be able to keep this voltage.

When looking at the regulator with the inscriptions facing you, the three leads are in order : input, ground, output.

After the regulator, add two more capacitors for improved filtering (use

16V models at least).

That's it.

vic

--- View in Courier:

11VAC 16VDC 7809 \\ +----+ \\ +-----+ 120AC>----+ +-----|~ +|------+-----| |--->9V P||S | | |+ +--+--+ R||E | | [1000µF25V] | I||C | | | | 120AC>----+ +-----|~ -|------+--------+------>GND +----+ FWB 50V1A

The assumption is that the 11VAC is coming from a step-down transformer connected to 120VAC mains.

Also, you could make the FWB from 1N400X rectifiers.

-- John Fields Professional Circuit Designer

Thank you both.

Both are good sugestions. I will consider which is best for my the application and build it.

Thanks again.