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From: Tony Williams
Subject: Re: Interface Conversion
Date: Mon, 04 Nov 2002 14:48:40 +0000 (GMT)
References: <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
NNTP-Posting-Date: Mon, 4 Nov 2002 14:48:36 +0000 (UTC)
User-Agent: Pluto/1.14i (RISC-OS/3.60)
In article ,
Grahame Kelly wrote:
> Tony Williams wrote:
> > (4.94 + 0.4) says that AVcc is 5.34V.
> > Is that correct?
> AVcc is 4.94 measured.
There's still a typo somewhere in your original
You give V-adc(max) as (AVcc-0.4), which is
(4.94 - 0.4), equals 4.54V.... not the 4.90V
stated later on in the post.
Whatever it is, maybe the circuit below would be
Vin+--/\/\--+ +-------/\/\-----+--> Vout
0-6.2v | | AVcc | (50k load)
| +---|- \| |
| | | >--/\/\--+
+--|---|+ /| R5
| | |
\ \ |
R2/ /R4 |
\ \ |
| | |
Use a CMOS opamp (eg, Texas TLC2201) which has
a MOSFET rail-rail output stage and a common-mode
input capability that includes 0v (0-2.7V) for
The output resistance of that sketch is R5 plus
the 400ohm (approx) internal o/p resistance of
the TLC2201. That is loaded by (R3+R4) in series,
in parallel your ADC's 50k input impedance. So you
can juggle the values of R3/4/5 so that the voltage
presented to the ADC cannot exceed whatever is req'd.
Vout = Vin * R2/(R1+R2) * (R3+R4)/R4.
But to keep within the CMV range, [6.2 * R2/(R1+R2)]
must be less than 2.7V (and you want R1+R2 = 50k).
R1=27k and R2=20k would give Rin=47k and V(R2)=2.64V
Do similar sums on R3/4/5 to get the req'd Vout(max).
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