Author: eljaincDate: 22:34 01-04-08
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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
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Author: Phil AllisonDate: 23:08 01-04-08
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"eljainc"
> I have an AC signal that varies between 0-15VAC
** What frequency, what wave shape, how does it vary and how fast ?
Why did you leave all this out ?
Reckon it don't matter ?
> that I would like to
> rectify/convert to a standard DC signal (0-15VDC).
** 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 ?
> 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?
** Yeah - add a DC filter choke in series and a second filter cap.
LOL !
> The current involved is low (on the order of
> 20mA to perhaps as much as 100mA).
>
> Any thoughts or ideas welcome.
** 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
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Author: ZedDate: 23:13 01-04-08
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Author: Phil AllisonDate: 23:16 01-04-08
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"Zed" <lizongyin@gmail.com>
( snip totally asinine advice )
** Zed = Zilch.
Which equals how much this trolling FUCKWIT knows about anything.
Piss off Zed head.
...... Phil
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Author: ZedDate: 23:43 01-04-08
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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.
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Author: Phil AllisonDate: 23:47 01-04-08
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"Zed = Zilch "
** You are a fucking piece of shit
- Zed head
> if you have anything to say to the topic,
** Anencephalic cunts like YOU do not get to tell anyone anything.
FUCK OFF ASSHOLE !!
..... Phil
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Author: Bill BowdenDate: 01:21 02-04-08
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On Apr 1, 6:34=A0pm, eljainc <elja...@sbcglobal.net> wrote:
> 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? =A0The current involved is low (on the order of
> 20mA to perhaps as much as 100mA).
>
> Any thoughts or ideas welcome.
>
> Thanks.
> Mike
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
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Author: Bob WoodwardDate: 03:03 02-04-08
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Bill Bowden wrote:
> On Apr 1, 6:34 pm, eljainc <elja...@sbcglobal.net> wrote:
>> 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
>
> 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
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
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Author: Bill BowdenDate: 19:42 03-04-08
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On Apr 1, 11:03=A0pm, Bob Woodward <"Bob Woodward"@no.org> wrote:
> Bill Bowden wrote:
> > On Apr 1, 6:34 pm, eljainc <elja...@sbcglobal.net> wrote:
> >> 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? =A0The current involved is low (on the order of=
> >> 20mA to perhaps as much as 100mA).
>
> >> Any thoughts or ideas welcome.
>
> >> Thanks.
> >> Mike
>
> > 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
>
> 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
> =A0 =A0ouput 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- Hide quoted text -
>
> - Show quoted text -
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
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