Full wave bridge rectifier circuit recommendations needed

Hello,

I have an AC signal that varies between 0-15VAC that I would like to rectify/convert to a standard DC signal (0-15VDC). I realize that a full wave rectifier along with a capacitor filter circuit can produce a more DC like signal. Is there an IC or circuit that can remove the ripple almost entirely? The current involved is low (on the order of

20mA to perhaps as much as 100mA).

Any thoughts or ideas welcome.

Thanks. Mike

Reply to
eljainc
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"eljainc"

** What frequency, what wave shape, how does it vary and how fast ?

Why did you leave all this out ?

Reckon it don't matter ?

** How fast does this DC level need to rise and fall in response to changes in the AC one ?

Why did you leave this out ?

Reckon it don't matter ?

** Yeah - add a DC filter choke in series and a second filter cap.

LOL !

** Explain just what it is YOU are trying to do !!!

Seeing as you have ZERO ability to write a technical specification - maybe we can decipher it from the nature of the task.

...... Phil

Reply to
Phil Allison

Hi, there

KA7815 may help.

You may dowload KA7815 from this address:

formatting link

You could check the application circuit on page 21. hope this will help

Reply to
Zed

"Zed"

( snip totally asinine advice )

** Zed = Zilch.

Which equals how much this trolling FUCKWIT knows about anything.

Piss off Zed head.

...... Phil

Reply to
Phil Allison

Hi, phil,

pay attention to your mouth, ok?

if you have anything to say to the topic, you may say it; if not, just leave. this group is for serious discussion.

have a good night.

Reply to
Zed

Well, you could use an DRC circuit (Diode / resistor / capacitor in series) so the voltage on the capacitor represents the peak of the AC signal minus the diode drop. If the AC signal is 15 volts RMS, sine wave, the peak will be about 1.414 times higher or 21.21 volts and the diode will reduce that by about 700mV, so you get 20.51 on the capacitor. But you can't put any load on it (no current) or the voltage will fall. You could use an opamp to buffer the voltage so the external load doesn't affect it much.

-Bill

Reply to
Bill Bowden

Bill

In your proposal the capacitor will get charged via the diode but will not become decharged except by its own leakage. So the voltage over the capacitor will not follow the input signal. But as soon as you connect a load to the output the capacitor will be decharged and the ripple will appear.

Therefore you have to split the functions needed to separate circuits.

You need:

a "precision rectifier" ( google ) mostly 2 op-amps + diodes ouput voltage can be adjusted by resistor-value an integrator ( as simple as an RC up to an n.th order low-pass ) an output buffer that can supply the current needed ( LT1010 )

Robert

Reply to
Bob Woodward

Yes, you are right. I was thinking of a peak detector, but in that case the resistor is not needed, just the diode and cap should do. But without knowing the frequency or rate of amplitude change, how can you make the capacitor voltage follow the input? What is the response time that wasn't specified? If it's 1MHz and the amplitude changes at 1 Hz, that's 1 problem, but if it's 10 Hz, and the amplitude changes at

5 Hz, that's another problem.

-Bill

Reply to
Bill Bowden

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