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how to take 0.5-1.8V signal and change it to 0-4.096V signal. Op-amp difference amplifier maybe?

M

Michael Noone

Jan 1, 1970
0
Hi - I have a signal that ranges from approximately 0.5-1.8V. I want to
convert that into a 0-4.096V signal so that I can feed it into an ADC. So
essentially what I want is an op-amp difference amplifier - but there is a
slight problem. I need to have three of these circuits on the board I'm
making - and they need to be as small as possible. Also - 0.5-1.8V is
approximate - each circuit needs to have these values be adjustable. Just
to keep things complicated - board space is quite limited, so the smaller I
can make this the better. Oh - the input signal is coming from a 3K
potentiometer.

So my best idea was to make a op-amp difference amplifier, something like
the second circuit here: http://hyperphysics.phy-
astr.gsu.edu/hbase/electronic/opampvar6.html. Thus my input signal would be
connected to V+ (through a large resistor), and then somehow I'd need to
get an adjustable voltage to V-. I think I could use a low resistance pot
for this. But then I'd need two potentiometers for adjusting the gain of
the circuit - and they would need to always have an identical value. I
guess I could use a 2 channel potentiometer, though I'm not sure they make
them very small. (I've never seen any, at least)

So - any suggestions? And sorry if my post is a bit hard to understand - I
haven't slept in quite a while...

-Michael
 
J

Joerg

Jan 1, 1970
0
Hello Michael,

Don't know what your supplies are but have a look at ye olde LM324A (not
the non-A). This one has 3mV max offset while the regular version is
7mV. I believe the A also comes in TSSOP which should be small enough.
And it is cheap.

If they must be scattered look for a SOT23-5 opamp such as the TLV2211.

Regards, Joerg
 
M

Michael Noone

Jan 1, 1970
0
Hello Michael,

Don't know what your supplies are but have a look at ye olde LM324A (not
the non-A). This one has 3mV max offset while the regular version is
7mV. I believe the A also comes in TSSOP which should be small enough.
And it is cheap.

If they must be scattered look for a SOT23-5 opamp such as the TLV2211.

Regards, Joerg

All three channels can be on the same chip. Honestly the op-amp is the
least of my worries though - I'm worried about the board being covered in
pots (I think I need something like 9 pots)

-Michael
 
J

Joerg

Jan 1, 1970
0
Hello Michael,
All three channels can be on the same chip. Honestly the op-amp is the
least of my worries though - I'm worried about the board being covered in
pots (I think I need something like 9 pots)

Know what you mean. Too many pots used to be called "rubber
engineering". I guess you'd have to think about an alternative solution
that either doesn't need all this adjustment or where you can do it
digitally. There are lots of multi-channel DACs.

Then there is the measure-and-add-resistor alignment technique.

Regards, Joerg
 
B

Bob Monsen

Jan 1, 1970
0
Michael said:
Hi - I have a signal that ranges from approximately 0.5-1.8V. I want to
convert that into a 0-4.096V signal so that I can feed it into an ADC. So
essentially what I want is an op-amp difference amplifier - but there is a
slight problem. I need to have three of these circuits on the board I'm
making - and they need to be as small as possible. Also - 0.5-1.8V is
approximate - each circuit needs to have these values be adjustable. Just
to keep things complicated - board space is quite limited, so the smaller I
can make this the better. Oh - the input signal is coming from a 3K
potentiometer.

So my best idea was to make a op-amp difference amplifier, something like
the second circuit here: http://hyperphysics.phy-
astr.gsu.edu/hbase/electronic/opampvar6.html. Thus my input signal would be
connected to V+ (through a large resistor), and then somehow I'd need to
get an adjustable voltage to V-. I think I could use a low resistance pot
for this. But then I'd need two potentiometers for adjusting the gain of
the circuit - and they would need to always have an identical value. I
guess I could use a 2 channel potentiometer, though I'm not sure they make
them very small. (I've never seen any, at least)

So - any suggestions? And sorry if my post is a bit hard to understand - I
haven't slept in quite a while...

-Michael

You need a subtractor with a bit of gain. An opamp + 4 resistors.

R1 goes from the 0.5 reference voltage to the inverting input. R2 goes
to the output from there.

R3 goes from the signal input to the non-inverting input, and R4 goes
from there to ground.

R1 = R3, and R2 = R4.

The gain of the thing is R2/R1

This will subtract 0.5V from the 0.5-1.8 volt input, and multiply the
result by R2/R1.

Using a divider with 1.8k and 200, you'll get 0.5V out of a 5V rail.
This is your reference.

Using 24k for R1 and R3, and 100k for R2 and R4 gives you the right
amount of gain.

A rail-to-rail opamp, like the microchip MCP6294, will keep the voltage
output near the rails accurate.

The main issue is input impedance, which is going to be equal to R1.
This can cause inaccuracy if your sensor has high output impedance. If
that's a problem, use a buffer between the sensor and R1 (input to +
input of opamp, - input to output of opamp)

--
Regards,
Bob Monsen

If a little knowledge is dangerous, where is the man who has
so much as to be out of danger?
Thomas Henry Huxley, 1877
 
J

John Popelish

Jan 1, 1970
0
Michael said:
Hi - I have a signal that ranges from approximately 0.5-1.8V. I want to
convert that into a 0-4.096V signal so that I can feed it into an ADC. So
essentially what I want is an op-amp difference amplifier - but there is a
slight problem. I need to have three of these circuits on the board I'm
making - and they need to be as small as possible. Also - 0.5-1.8V is
approximate - each circuit needs to have these values be adjustable. Just
to keep things complicated - board space is quite limited, so the smaller I
can make this the better. Oh - the input signal is coming from a 3K
potentiometer.

So my best idea was to make a op-amp difference amplifier, something like
the second circuit here: http://hyperphysics.phy-
astr.gsu.edu/hbase/electronic/opampvar6.html. Thus my input signal would be
connected to V+ (through a large resistor), and then somehow I'd need to
get an adjustable voltage to V-. I think I could use a low resistance pot
for this. But then I'd need two potentiometers for adjusting the gain of
the circuit - and they would need to always have an identical value. I
guess I could use a 2 channel potentiometer, though I'm not sure they make
them very small. (I've never seen any, at least)

So - any suggestions? And sorry if my post is a bit hard to understand - I
haven't slept in quite a while...

If possible, I would design the amplifier to keep the output between 0
and 4.096 for the worst case tolerance of the input, and digitally
correct the zero and span for all other cases.

I don't think you are going to find any ganged pots, so if you want to
adjust both zero and full scale, you will need two pots per channel.
But remember that opamps are not much larger than resistors, so you
may be able to use a multi opamp amplifier for each channel and reduce
the other parts counts. And as long as you can stand a little
interaction between the zero and span adjustments, you don't need a
full subtracter. A non inverting amplifier (input signal to +, output
to feedback resistor to end of span pot, wiper to -, other end of span
pot to fixed resistor to wiper of zero pot across .6 volts or so and
ground.) will not load your signal pot, and the resistor values can be
chosen to produce only a tiny interaction between the two adjustments.
1 opamp, two fixed resistors and two pots per channel. Use a dual
or quad amp.
 
M

Mook Johnson

Jan 1, 1970
0
Do you really need the full range of resolution from the A2D?

How small of a change are you looking to detect from a pot (noisy)? A 16
bit 0-4.096 A2D can detect down to 62uV of change per bit. With your given
voltage range, you'll have 20K increments to play with (over 14 bits of
resolution).

You can do all the scaling inside the micro.
 
M

Michael Noone

Jan 1, 1970
0
Do you really need the full range of resolution from the A2D?

How small of a change are you looking to detect from a pot (noisy)? A
16 bit 0-4.096 A2D can detect down to 62uV of change per bit. With
your given voltage range, you'll have 20K increments to play with
(over 14 bits of resolution).

You can do all the scaling inside the micro.

Well - it's a 10b ADC with a 4.096V vref (I could probabaly lower that to
2.048 if it would help anything). I want to get about 360 steps between .5
and 1.8v. So - I could do something like just run the input signal through
a non-inverting amplifier which would bring it up to a max voltage of
4.096V, so I'd get ADC readings of about 284 - 1023 (=range of 739) which
would give me about double the range that I need, so it'd probabaly be OK.
But I just think it would be a little more clean if I could use the full
range of the ADC.

-Michael
 
J

Joerg

Jan 1, 1970
0
Hello Michael,
Well - it's a 10b ADC with a 4.096V vref (I could probabaly lower that to
2.048 if it would help anything). I want to get about 360 steps between .5
and 1.8v. So - I could do something like just run the input signal through
a non-inverting amplifier which would bring it up to a max voltage of
4.096V, so I'd get ADC readings of about 284 - 1023 (=range of 739) which
would give me about double the range that I need, so it'd probabaly be OK.
But I just think it would be a little more clean if I could use the full
range of the ADC.

All that effort for a gain of half a bit? If you really need that 1/2
bit why not offset-shift and sample again? Then you'd have 10 1/2 bits.

Regards, Joerg
 
M

Mook Johnson

Jan 1, 1970
0
If you're range is changing like you mentioned earlier, it would me a much
more versitile design if you just used a protion of the A2Ds range and
allowed the input to be whatever range they needed to be. As long as you
have the resolution, you should be fine even with 10 bits. Be sure to
tightly couple and filter the signal so you redice the LSB flicker to a
minimum and you should be fine.

Could you use a multi-input 12 or 16 bit A2D? It will likely take less
space than all those pots and be significantly less trouble as well (noise,
drift, vibration sensitivity...etc). (I don't like using pots to set a
stable voltage reference)
 
W

Wouter van Ooijen

Jan 1, 1970
0
Hi - I have a signal that ranges from approximately 0.5-1.8V. I want to
convert that into a 0-4.096V signal

My first guess would be an A/D, uC, D/A.
so that I can feed it into an ADC.

But when you want to A/D why go to this trouble at all? Just A/D the
original signal and do the rest in software!


Wouter van Ooijen

-- ------------------------------------
http://www.voti.nl
Webshop for PICs and other electronics
http://www.voti.nl/hvu
Teacher electronics and informatics
 
R

Rich Grise

Jan 1, 1970
0
My first guess would be an A/D, uC, D/A.


But when you want to A/D why go to this trouble at all? Just A/D the
original signal and do the rest in software!

Because he'd only be using 26% of the ADC's range, and lose two bits
of resolution.

Cheers!
Rich
 
J

jabara

Jan 1, 1970
0
Wouter van Ooijen (www.voti.nl) said:
My first guess would be an A/D, uC, D/A.


But when you want to A/D why go to this trouble at all? Just A/D the
original signal and do the rest in software!

Op amp. easy fast and cheap. all you need is a gain of about 2. What are
you driving?
 
W

Wouter van Ooijen

Jan 1, 1970
0
But when you want to A/D why go to this trouble at all? Just A/D the
Because he'd only be using 26% of the ADC's range, and lose two bits
of resolution.

- does he realy need those two bits?
- feed the A/D a lower reference
- get an A/D with two more bits
- get two bits less per sample but add some noise and sample more


Wouter van Ooijen

-- ------------------------------------
http://www.voti.nl
Webshop for PICs and other electronics
http://www.voti.nl/hvu
Teacher electronics and informatics
 
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