Dividing and Multiplying Voltages

Guest
Hi mates.

Albert's here. I'm new here and i'm seeking your helps regarding the
following.

I'm working on a project which involves the arithmetic operations (+,
-, X, /) on voltages. Using ADC conversion is not applicable in my
case. The project is to capture 4 voltages from 4 testpoints from a
PCBA, ie A, B, C and D. After capturing the 4 voltages, we need to do
the following calculations:

Result = ((A+B)-(C+D))/(A+B+C+D)

For the addition and subtraction parts, i've got the idea of using
summing amplifier and inverting amplifier to perform the functions. If
you have other suggestion, kindly share with me. Fyi, the voltage
values are about 0.3V.

For the multiplying & division's parts, please give suggestion for the
circuit.
Thanks and hope to get your response soon.

Cheers,
Albert Leng
 
<albertleng@gmail.com> wrote in message
news:1103102166.687320.223670@f14g2000cwb.googlegroups.com...
For the multiplying & division's parts, please give suggestion for the
circuit.
Click to expand...
Is this for a test/homework or do you actually just want to build it? The
answers would be different :)

Homework answer: Investigate Logarithims.

If I remember correctly (it's been a while)....

LOG(A*B) = LOG(A) + LOG(B)

so A*B = LOG^-1 (LOG(A) + LOG(B))

Also for division..

LOG(A/B) = LOG(A) - LOG(B)

so

A/B = LOG^-1(LOG(A) - LOG(B))

If you wnt to build something...

Here is another approach that avoids logs...

http://www.edn.com/archives/1996/101096/21di_01.htm

There are commercial chips to do the job. Perhaps one of these black
boxes...

http://www.analog.com/en/subCat/0,2879,773%255F862%255F0%255F%255F0%255F,00.html
 
On Wed, 15 Dec 2004 17:00:56 GMT, "CWatters"
<colin.watters@pandoraBOX.be> wrote:

albertleng@gmail.com> wrote in message
news:1103102166.687320.223670@f14g2000cwb.googlegroups.com...
For the multiplying & division's parts, please give suggestion for the
circuit.

Is this for a test/homework or do you actually just want to build it? The
answers would be different :)

Homework answer: Investigate Logarithims.

If I remember correctly (it's been a while)....

LOG(A*B) = LOG(A) + LOG(B)

so A*B = LOG^-1 (LOG(A) + LOG(B))

Also for division..

LOG(A/B) = LOG(A) - LOG(B)

so

A/B = LOG^-1(LOG(A) - LOG(B))

If you wnt to build something...

Here is another approach that avoids logs...

http://www.edn.com/archives/1996/101096/21di_01.htm

There are commercial chips to do the job. Perhaps one of these black
boxes...

http://www.analog.com/en/subCat/0,2879,773%255F862%255F0%255F%255F0%255F,00.html
Click to expand...
And then there are slide rules ;-)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
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