AC coupled offset and gain problem

D

Douglas Beeson

Guest
I am having trouble understanding the behavior of a circuit I am using to apply
gain and offset to a triangle wave.

The circuit in question is here:

http://gallery.me.com/doug.beeson#100069/offset%252Bgain

My triangle wave output is 4 +/- 1.6 V, and I want 2.55 +/- 2.45 V out of the
gain stage. (0.1 - 5 Vpp)

To find the topology and calculate the resistor values I used the TI offset and
gain calculator located at:
http://www-k.ext.ti.com/srvs/cgi-bin/webcgi.exe?
Company={5761bcd8-11f5-4e08-84e0-8167176a4ed9},kb=analog,case=obj%2835457%29,new

My circuit is AC-coupled, but I can only get it to work properly if I shunt the
input signal to ground, between the cap and gain resistor Rg=1370, with a low-
value resistor (60 or so). This effectively creates a voltage divider with Rg
and sets the DC part of Vi to 0.04 V or so. But it also loads the previous
output stage...which I don't want.

If I remove the shunt resistor, the output is centered at about 1.5 V instead of
2.5 V, so my output signal dips a volt or so below ground.

What I don't understand is why I need the shunt resistor in this inverting
topology. All the op amp gain/offset diagrams I have ever seen show the input
from the cap straight to the gain resistor, with no shunt. How could that ever
work? The input's DC level, after the cap, seems to be set by V- (and hence by V
+), instead of being 0 as the design tool assumes.

I'm confused! Am I making a mistake with the gain calculator or am I wrong to
assume that an AC-coupled signal into an inverting amp should not need a shunt
resistor to force its DC to 0?

Thanks,

doug
 
"Douglas Beeson" <unkown@xnntp> wrote in message
news:9485231D-446A-4F2D-A638-B2AF2BDB8457%Xnntp@beta07.com...
I am having trouble understanding the behavior of a circuit I am using to
apply
gain and offset to a triangle wave.

The circuit in question is here:

http://gallery.me.com/doug.beeson#100069/offset%252Bgain

My triangle wave output is 4 +/- 1.6 V, and I want 2.55 +/- 2.45 V out of
the
gain stage. (0.1 - 5 Vpp)

To find the topology and calculate the resistor values I used the TI
offset and
gain calculator located at:
http://www-k.ext.ti.com/srvs/cgi-bin/webcgi.exe?
Company={5761bcd8-11f5-4e08-84e0-8167176a4ed9},kb=analog,case=obj%2835457%29,new

My circuit is AC-coupled, but I can only get it to work properly if I
shunt the
input signal to ground, between the cap and gain resistor Rg=1370, with a
low-
value resistor (60 or so). This effectively creates a voltage divider with
Rg
and sets the DC part of Vi to 0.04 V or so. But it also loads the previous
output stage...which I don't want.

If I remove the shunt resistor, the output is centered at about 1.5 V
instead of
2.5 V, so my output signal dips a volt or so below ground.

What I don't understand is why I need the shunt resistor in this inverting
topology. All the op amp gain/offset diagrams I have ever seen show the
input
from the cap straight to the gain resistor, with no shunt. How could that
ever
work? The input's DC level, after the cap, seems to be set by V- (and
hence by V
+), instead of being 0 as the design tool assumes.

I'm confused! Am I making a mistake with the gain calculator or am I wrong
to
assume that an AC-coupled signal into an inverting amp should not need a
shunt
resistor to force its DC to 0?

Thanks,

doug
Click to expand...
You DC offset is as it should be with those values of resistors, 1.5 volts.
You need to change the values of the resistors to the non - inverting input
of the OP-AMP. It's a simple voltage divider then multiplied by the gain of
1.5. You have 1.0 volts going to the non - inverting input, you need 1.7
volts, so change the resistor to ground to a higher value using R1 / (R1 +
R2) = 1.7 R1 being the resistor to ground.

Shaun
 
"Shaun" <rowl@nomail.com> wrote:

"Douglas Beeson" <unkown@xnntp> wrote in message
news:9485231D-446A-4F2D-A638-B2AF2BDB8457%Xnntp@beta07.com...
I am having trouble understanding the behavior of a circuit I am using to
apply
gain and offset to a triangle wave.

The circuit in question is here:

http://gallery.me.com/doug.beeson#100069/offset%252Bgain

My triangle wave output is 4 +/- 1.6 V, and I want 2.55 +/- 2.45 V out of
the
gain stage. (0.1 - 5 Vpp)

To find the topology and calculate the resistor values I used the TI
offset and
gain calculator located at:
http://www-k.ext.ti.com/srvs/cgi-bin/webcgi.exe?
Company={5761bcd8-11f5-4e08-84e0-8167176a4ed9},kb=analog,case=obj
%2835457%29,new

My circuit is AC-coupled, but I can only get it to work properly if I
shunt the
input signal to ground, between the cap and gain resistor Rg=1370, with a
low-
value resistor (60 or so). This effectively creates a voltage divider with
Rg
and sets the DC part of Vi to 0.04 V or so. But it also loads the previous
output stage...which I don't want.

If I remove the shunt resistor, the output is centered at about 1.5 V
instead of
2.5 V, so my output signal dips a volt or so below ground.

What I don't understand is why I need the shunt resistor in this inverting
topology. All the op amp gain/offset diagrams I have ever seen show the
input
from the cap straight to the gain resistor, with no shunt. How could that
ever
work? The input's DC level, after the cap, seems to be set by V- (and
hence by V
+), instead of being 0 as the design tool assumes.

I'm confused! Am I making a mistake with the gain calculator or am I wrong
to
assume that an AC-coupled signal into an inverting amp should not need a
shunt
resistor to force its DC to 0?

Thanks,

doug


You DC offset is as it should be with those values of resistors, 1.5 volts.
You need to change the values of the resistors to the non - inverting input
of the OP-AMP. It's a simple voltage divider then multiplied by the gain of
1.5. You have 1.0 volts going to the non - inverting input, you need 1.7
volts, so change the resistor to ground to a higher value using R1 / (R1 +
R2) = 1.7 R1 being the resistor to ground.

Shaun
Click to expand...
Thank you, Shaun.

I assume you got the 1.7 by dividing the output DC level (2.55) by the gain
(1.53), giving 1.67. Makes sense.

What is the approach, in general, to these problems? Calculate gain first, then
set offset from Vo=Vi(R1/R1+R2)(Rg/Rf+Rg) ?

When I used the TI calculator I used +1.6 and -1.6 as the full scale and zero
scale input values, and 0.1 and 5 as the output values, respectively. That
assumes the DC input is 0. The tool calculated the R values I gave (which were
wrong).

What did I input incorrectly there

--
______________
Douglas Beeson
Java Software Design and Development
Montreal, QC
dbeeson99@videotron.ca - remove the nines
 
You DC offset is as it should be with those values of resistors, 1.5
volts. You need to change the values of the resistors to the non -
inverting input of the OP-AMP. It's a simple voltage divider then
multiplied by the gain of
1.5. You have 1.0 volts going to the non - inverting input, you
need 1.7 volts, so change the resistor to ground to a higher value
using R1 / (R1 + R2) = 1.7 R1 being the resistor to ground.

Shaun




Thank you, Shaun.

I assume you got the 1.7 by dividing the output DC level (2.55) by
the gain (1.53), giving 1.67. Makes sense.

What is the approach, in general, to these problems? Calculate gain
first, then set offset from Vo=Vi(R1/R1+R2)(Rg/Rf+Rg) ?

When I used the TI calculator I used +1.6 and -1.6 as the full scale
and zero scale input values, and 0.1 and 5 as the output values,
respectively. That assumes the DC input is 0. The tool calculated the
R values I gave (which were wrong).

What did I input incorrectly there?
Click to expand...

An op amp is a device that attemps to keep it's two inputs at the same
voltage reference. The inputs also do not draw any current.


Hence, your positive side is simply a voltage divider

V+ = Vref*R2/(R1+R2)

But then V- = V+

So the current through 1370 res, lets call it R3, is

I = (Vi - V-)/R3

This current is the same current that is going through the 2100 resistor,
call it R4.

Hence (V- - Vo) = I*R4


So

V- = (Vi - V-)*R4/R3 + Vo

==> Vo = V-*(1 + R4/R3) - Vi*R4/R3
==> Vo = Vref*R2*(1+R4/R3)/(R1+R2) - Vi*R4/R3

So R3 and R3 not only affect the gain in such a circuit but also the offset.
You must pick R1 and R2 to "undo" the effect but this also changes your
offset. It will not be easy to guess at the values of R1,..,R4 because of
the way they all depend on each other. But note that R3 and R4 is also the
gain of the input just like a normally configured op amp.


Note that Vref*R2*(1+R4/R3)/(R1+R2) = Vref*(1 + Gain)/(1 + R1/R2) = Offset
and R4/R3 = Gain

I believe you said your gain is 2.45/1.6

Hence, choose R4/R3 = 2.45/1.6. (you could simply choose R4 = 2.45k and R3 =
1.6k)

This will give you the appropriate gain.

Now

The offset is then

8*(1 + 2.45/1.6)/(1 + R1/R2) = Offset

You simply need to solve for R1/R2(the ratio, not R1 and R2 separately as
that is not important... only the ratio matters). Once you do this then you
know what the ratio of R1 and R2 needs to be to get the appropriate offset.
Do not choose R1 and R2 too small nor too large. On the order of 10^4 is ok.

Pick the offset carefully since your original signal is already offset... at
least I think you mentioned that.
 
"Douglas Beeson" <dbeeson@videotron.ca> wrote in message
news:DFF72036-0C64-46BA-B812-04A3886FE7A0%dbeeson@videotron.ca...
"Shaun" <rowl@nomail.com> wrote:


"Douglas Beeson" <unkown@xnntp> wrote in message
news:9485231D-446A-4F2D-A638-B2AF2BDB8457%Xnntp@beta07.com...
I am having trouble understanding the behavior of a circuit I am using to
apply
gain and offset to a triangle wave.

The circuit in question is here:

http://gallery.me.com/doug.beeson#100069/offset%252Bgain

My triangle wave output is 4 +/- 1.6 V, and I want 2.55 +/- 2.45 V out
of
the
gain stage. (0.1 - 5 Vpp)

To find the topology and calculate the resistor values I used the TI
offset and
gain calculator located at:
http://www-k.ext.ti.com/srvs/cgi-bin/webcgi.exe?
Company={5761bcd8-11f5-4e08-84e0-8167176a4ed9},kb=analog,case=obj
%2835457%29,new

My circuit is AC-coupled, but I can only get it to work properly if I
shunt the
input signal to ground, between the cap and gain resistor Rg=1370, with
a
low-
value resistor (60 or so). This effectively creates a voltage divider
with
Rg
and sets the DC part of Vi to 0.04 V or so. But it also loads the
previous
output stage...which I don't want.

If I remove the shunt resistor, the output is centered at about 1.5 V
instead of
2.5 V, so my output signal dips a volt or so below ground.

What I don't understand is why I need the shunt resistor in this
inverting
topology. All the op amp gain/offset diagrams I have ever seen show the
input
from the cap straight to the gain resistor, with no shunt. How could
that
ever
work? The input's DC level, after the cap, seems to be set by V- (and
hence by V
+), instead of being 0 as the design tool assumes.

I'm confused! Am I making a mistake with the gain calculator or am I
wrong
to
assume that an AC-coupled signal into an inverting amp should not need a
shunt
resistor to force its DC to 0?

Thanks,

doug


You DC offset is as it should be with those values of resistors, 1.5
volts.
You need to change the values of the resistors to the non - inverting
input
of the OP-AMP. It's a simple voltage divider then multiplied by the gain
of
1.5. You have 1.0 volts going to the non - inverting input, you need
1.7
volts, so change the resistor to ground to a higher value using R1 / (R1
+
R2) = 1.7 R1 being the resistor to ground.

Shaun




Thank you, Shaun.

I assume you got the 1.7 by dividing the output DC level (2.55) by the
gain
(1.53), giving 1.67. Makes sense.

What is the approach, in general, to these problems? Calculate gain first,
then
set offset from Vo=Vi(R1/R1+R2)(Rg/Rf+Rg) ?

When I used the TI calculator I used +1.6 and -1.6 as the full scale and
zero
scale input values, and 0.1 and 5 as the output values, respectively. That
assumes the DC input is 0. The tool calculated the R values I gave (which
were
wrong).

What did I input incorrectly there?

--
______________
Douglas Beeson
Java Software Design and Development
Montreal, QC
dbeeson99@videotron.ca - remove the nines
Click to expand...

1 - Rf
Your gain is Av = ------ = 1.532
R1

The voltage the you want on the output is 2.55 volts; so divide that by your
gain.

Vout(divider) = 2.55/1.532 = 1.6645 volts

Vin (divider) = 8 volts
R4
Vout / Vin = 1.6645 / 8 = 0.2086 = ---------
R3 + R4

we use R3 as 12.1 Kohms

we need to solve for R4

do some algebra and solve

R4 = 3181 ohms
This is your bottom resistor that was 1740 ohms on your diagram.


I hope this helps;

Shaun
 
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