 ### Network # summing amplifier with linear offset

?

#### ?ine Canby

Jan 1, 1970
0
hi all,

I need to add two signals, a DC and an AC. The AC signal comes from a
DAC, and I wish to supply the DC signal using a varible resistor pot
on the circuit board. The thing is, I'd like the DC output to be
linear with respect to the pot position. In other words, if I turn the
pot 1 turn and get an output increase of 0.1 Volts, then I'd like to
get a further increase of 0.1Volts when I again turn the pot by one
rev.

I need to attenuate the DAC output by a factor of 2 to obtain the
correct scaled AC value, and I was thinking of using the -15 (to gain
a positive output) rail to obtain an output offset voltage of between
0 Volts and 1.6 Volts. In otherwords, I need to add a DC Voltage of
between 0 Volts and 1.6Volts to my scaled AC value when producing my
output voltage.

If we ignore the AC section of the summing amp for a second and
concentrate on the DC section, we basically hav an inverting amp. The
problem is, how would I introduce a variable resistor into an
inverting amplifier circuit to obtain the linear output I desire?
Obviously the resitor beween the differential input and the output
must remain at a fixed value as it would be shared by the AC section
of the summing circuit. I've tried voltage dividing the -15Volt rail
with two resitors in series(between 0 and -15), and connecting their
intersection to the input resistor of the a typical inverting amp
circuit. Making any of these resistor varing, will not result in a
linear output.

Any suggestions?

Thanks.

J

#### John Popelish

Jan 1, 1970
0
?ine Canby said:
hi all,

I need to add two signals, a DC and an AC. The AC signal comes from a
DAC, and I wish to supply the DC signal using a varible resistor pot
on the circuit board. The thing is, I'd like the DC output to be
linear with respect to the pot position. In other words, if I turn the
pot 1 turn and get an output increase of 0.1 Volts, then I'd like to
get a further increase of 0.1Volts when I again turn the pot by one
rev.

I need to attenuate the DAC output by a factor of 2 to obtain the
correct scaled AC value, and I was thinking of using the -15 (to gain
a positive output) rail to obtain an output offset voltage of between
0 Volts and 1.6 Volts. In otherwords, I need to add a DC Voltage of
between 0 Volts and 1.6Volts to my scaled AC value when producing my
output voltage.

If we ignore the AC section of the summing amp for a second and
concentrate on the DC section, we basically hav an inverting amp. The
problem is, how would I introduce a variable resistor into an
inverting amplifier circuit to obtain the linear output I desire?
Obviously the resitor beween the differential input and the output
must remain at a fixed value as it would be shared by the AC section
of the summing circuit. I've tried voltage dividing the -15Volt rail
with two resitors in series(between 0 and -15), and connecting their
intersection to the input resistor of the a typical inverting amp
circuit. Making any of these resistor varing, will not result in a
linear output.

Any suggestions?

Thanks.

I would try connecting the variable resistor as a voltage divider
across a well regulated voltage, and combine the output (possibly
buffered by a follower amplifier) with the dac output (also possibly
buffered) with a subtractor configured opamp. The subtractor could be
configured for an overall gain of 1/2 to correct the DAC output to the
right level.

D

#### dB

Jan 1, 1970
0
The
problem is, how would I introduce a variable resistor into an
inverting amplifier circuit to obtain the linear output I desire?
Obviously the resitor beween the differential input and the output
must remain at a fixed value as it would be shared by the AC section
of the summing circuit. I've tried voltage dividing the -15Volt rail
with two resitors in series(between 0 and -15), and connecting their
intersection to the input resistor of the a typical inverting amp
circuit. Making any of these resistor varing, will not result in a
linear output.

Any suggestions?

Use a potential divider and pot to establish your 0V to 1.6V. Buffer
it with a unity gain follower. Connect the output of the follower to
the summing amp through an R of the same value as the feedback R.

B

#### Ban

Jan 1, 1970
0
?ine Canby wrote:
|| hi all,
||
|| I need to add two signals, a DC and an AC. The AC signal comes from a
|| DAC, and I wish to supply the DC signal using a varible resistor pot
|| on the circuit board. The thing is, I'd like the DC output to be
|| linear with respect to the pot position. In other words, if I turn
|| the pot 1 turn and get an output increase of 0.1 Volts, then I'd
|| like to get a further increase of 0.1Volts when I again turn the pot
|| by one rev.
||
|| I need to attenuate the DAC output by a factor of 2 to obtain the
|| correct scaled AC value, and I was thinking of using the -15 (to gain
|| a positive output) rail to obtain an output offset voltage of between
|| 0 Volts and 1.6 Volts. In otherwords, I need to add a DC Voltage of
|| between 0 Volts and 1.6Volts to my scaled AC value when producing my
|| output voltage.
||
|| If we ignore the AC section of the summing amp for a second and
|| concentrate on the DC section, we basically hav an inverting amp. The
|| problem is, how would I introduce a variable resistor into an
|| inverting amplifier circuit to obtain the linear output I desire?
|| Obviously the resitor beween the differential input and the output
|| must remain at a fixed value as it would be shared by the AC section
|| of the summing circuit. I've tried voltage dividing the -15Volt rail
|| with two resitors in series(between 0 and -15), and connecting their
|| intersection to the input resistor of the a typical inverting amp
|| circuit. Making any of these resistor varing, will not result in a
|| linear output.
||
|| Any suggestions?
||
|| Thanks.

You have to buffer your DC with a non-inverting opamp to get a linear range.
Here is a circuit, you can extend easily to also a negativ range. Maybe your
AC can be coupled in by a capacitor, but also DC is possible as shown. I
added an offset trimmer to give exactly 0.0V at the lower end of the pot.
The upper end can be adjusted by adding a 5k rheostat with a 33k in series.

-15V
VEE ||
| +-----||--+
..-. | 47p|| |
| | ___ | ___ |
| |36k AC o--------|___|-+---|___|-+
'-' 10k | 5k1 |
| 1/2 NE5532 | |
..-. |\ | |
| |<---+-------|+\ ___ | |\ |
| |10k | | >-+--|___|-+----|-\ | ___
'-' --- +-|-/ | 10k___ | >-+-|___|-o
| --- | |/ | +-|___|-+-|+/ 50R out
| |1u +------+ | 1M | |/
| | | |1/2 NE5532
=== === VCC | .-.
GND GND | | | |
.-. | 1k| |
| |<-+ '-'
100k| | |
'-'0-trim ===
| GND
VEE
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

B

Jan 1, 1970
0
?

#### ?ine Canby

Jan 1, 1970
0
Thanks very much for your advice, I've a better idea of what I'm doing
now. This is the circuit diagram of my design -
http://www.geocities.com/aine_canby/atten.gif

I'm now getting a linear DC offset at the output, ranging from 11mV to
1.411, which I am happy with. The POT is a 1K. I wasn't sure what
resistance value to give Remit so I chose 1k. Does that sound
reasonable? I also didn't buffer the DAC output - under what
conditions might I need to buffer the DAC output signal?

The 15 and -15 Volt rails are produced by a regulated supply.

The output voltage of my circuit will be used to drive a constant
current driver. Up until now I've simply been using a signal generator
to produce the driver input voltage. The driver converts this input
voltage to a proportional output current. I need to ensure that I
supply the driver with voltages between 0V and 1.6Volts. If I supply
the driver with a voltage greater than 1.6Volts I will produce a
current at the driver output which _will_ damage the equipmant I'm
using. Is there such a thing as a 1.5 or 1.6 Volt zener diode which I
could place accross my output? If not, then what might I do?

Thank you.

Aine.

D

#### dB

Jan 1, 1970
0
In other words, I need to add a DC Voltage of
between 0 Volts and 1.6Volts to my scaled AC value when producing my
output voltage.

If we ignore the AC section of the summing amp for a second and
concentrate on the DC section, we basically hav an inverting amp. The
problem is, how would I introduce a variable resistor into an
inverting amplifier circuit to obtain the linear output I desire?
Obviously the resitor beween the differential input and the output
must remain at a fixed value as it would be shared by the AC section
of the summing circuit. I've tried voltage dividing the -15Volt rail
with two resitors in series(between 0 and -15), and connecting their
intersection to the input resistor of the a typical inverting amp
circuit. Making any of these resistor varing, will not result in a
linear output.

Any suggestions?

Establish your offset range of 0 to 1.6V with a potential divider and
pot.

Buffer the pot's output with a unity gain follower, e.g. an OP07.
Connect the follower's output to the summing amp input via a resistor
of the same value as the summing amp's feedback R.

G

#### Genome

Jan 1, 1970
0
?ine Canby said:
hi all,

Any suggestions?

Thanks.

Try the schematics newsgroup for one

AND

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