Voltage levels and VU meters...
Harry Bissell
harrybissell at prodigy.net
Thu Jul 29 07:50:29 CEST 1999
DANGER Will Robinson....
watch out, don't call an LM324 a "quad 741". They are very different. The LM324
is specialized for single supply operation. And its current source/sink
capability are not symmetric... which could be a problem here. It also has a
nasty crossover notch if used on a bipolar supply as the input swings through
zero. Someone suggested a load resistor to the negative rail might cure this,
but I never tried..
If you use an active rectifier and a very small cap, followed by a buffer you
don't have to worry about how much current is available to charge the hold
cap... Is peak hold really a VU response, or do you really want some rolloff on
the leading edge (i.e. you really need R1.)
The Volume Unit was supposed to give a quasi-average reading... it would allow
peaks that were too short to be obvious as distortion when recording to analog
tape. A good VU meter lets you drive a good hot signal to tape without
(audible) clipping.
:^) Harry
Jaroslav Lukesh wrote:
> > On Tue, 27 Jul 1999, Chris Crosskey wrote:
> >
> > > If it needs DC then what do I do, if I rectify it with signal diodes,
> how do
> > > I go about smoothing it when the signal might be anywhere between 20Hz
> and
> > > 20kHz?....I know that I'm unlikely to need the bottom octave of that or
> the
> > > top two but I want it to be capable of the full range....
> >
> > You might want to use active rectifier to get rid of the 0.6V diode drop.
> > For smoothing, make an RC lowpass filter with cutoff set low enough (say,
>
> > 20Hz).
> > Disclaimer: I don't have any experience using VU meters so this advice
> > might be wrong.
> >
> > Antti
>
> Hi,
>
> Vu meter shoud have following charecteristic:
>
> 1. FAST response (<1 msec)
>
> 2. SLOW decay (0.05 to 0.2 sec)
>
> How to do it:
>
> in o--R1---+------+--------o out
> | |
> C R2
> | |
> gnd gnd
>
> decay: t=R2*C
> response (rise time): t=R1*C
>
> R1 may be omit, but output of prev. stage must have good current limiter
> (should be used 741 type opamp or LM324 - quadruple 741, max. output
> current is 25mA)
>
> BUT Dont forget for equation
>
> C*U=I*t
>
> if max detected voltage is for example 5 volts ie., C*5=0.025*0.001 ;
> C=0.025*0.001/5=0.0052
> ie., C must be less than 0.0052F, if we use for example 10uF, condition is
> OK, rise time is OK too.
>
> Best Regards
>
> Jerry LSH
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