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To maintain CMRR (common mode rejection ratio) of op amp

R

raf

Jan 1, 1970
0
Hi,
i am working on an Electronic control unit. I have a current sensing
circuit for motor (high side sensing) in which a differential amplifier
used to sense the current from shunt resistors and to amplify it. In
this ckt two transistors are added to the input terminals of the op amp
(one pnp and anothet npn) whose emitters are connected. I am sure that
these transistors are to maintain the CMRR of the ckt by decreasing the
common mode voltage effect.
Could anyboday have idea how these transistors help to increase the
CMRR of the op amp in case of change in resistor values(tolerances)
Thanks in advance
 
K

kell

Jan 1, 1970
0
raf said:
Hi,
i am working on an Electronic control unit. I have a current sensing
circuit for motor (high side sensing) in which a differential amplifier
used to sense the current from shunt resistors and to amplify it. In
this ckt two transistors are added to the input terminals of the op amp
(one pnp and anothet npn) whose emitters are connected. I am sure that
these transistors are to maintain the CMRR of the ckt by decreasing the
common mode voltage effect.
Could anyboday have idea how these transistors help to increase the
CMRR of the op amp in case of change in resistor values(tolerances)
Thanks in advance

"two transistors are added to the input terminals of the op amp...
whose emitters are connected"
can you give a more thorough description of the circuit config or
better yet an ASCII diagram
 
A

Ancient_Hacker

Jan 1, 1970
0
raf said:
Hi,
i am working on an Electronic control unit. I have a current sensing
circuit for motor (high side sensing) in which a differential amplifier
used to sense the current from shunt resistors and to amplify it. In
this ckt two transistors are added to the input terminals of the op amp
(one pnp and anothet npn) whose emitters are connected. I am sure that
these transistors are to maintain the CMRR of the ckt by decreasing the
common mode voltage effect.

What is the high-side voltage? Does it vary? What are the power
supply voltages for the op-amp?

I can't think of anything useful you can add to an op-amp to improve
its CMRR in the way of a pair of transistors. Maybe the transistors
are being used as Zeners to limit the input voltages?
 
T

TuT

Jan 1, 1970
0
Ancient_Hacker said:
What is the high-side voltage? Does it vary? What are the power
supply voltages for the op-amp?

I can't think of anything useful you can add to an op-amp to improve
its CMRR in the way of a pair of transistors. Maybe the transistors
are being used as Zeners to limit the input voltages?

You can use a matched pair of PNPs as a long-tailed pair to minimise the
common-mode voltage, provided you have a suitable +ve supply to power the
tail of the LTP.

But a PNP and an NPN . . . ?
 
R

raf

Jan 1, 1970
0
Hi,
thanks for u r reply
http://rapidshare.com/files/4191857/currentsensor.doc.html
here is the link for the circuit. Its high side sensing for the motor
current. In PSPICE i tried to see the effect of common mode voltage
using parametric simulation( giving different values of common mode
voltages) but the PSPICE is taking ideal conditions as its not showing
any difference in output voltage ( common mode voltage does not appear
in the output for ideal opamp) . Also tried to see the effect of these
transistors on the gain of op amp i tried without transistors ckt and
with. The gain is same in the PSPICE. So these transistors are placed
for controlling the CMRR i guess. The highside voltage is 14V battery
supply. Rshunt=2.5 mOhm, Ipeak=100A. So the voltage across the shunt is
250mV. Now whats the purpose of these transistors?
Any suggestions are appreciated.
 
R

raf

Jan 1, 1970
0
Hi,
thanks for u r reply
http://rapidshare.com/files/4191857/currentsensor.doc.html
here is the link for the circuit. Its high side sensing for the motor
current. In PSPICE i tried to see the effect of common mode voltage
using parametric simulation( giving different values of common mode
voltages) but the PSPICE is taking ideal conditions as its not showing
any difference in output voltage ( common mode voltage does not appear
in the output for ideal opamp) . Also tried to see the effect of these
transistors on the gain of op amp i tried without transistors ckt and
with. The gain is same in the PSPICE. So these transistors are placed
for controlling the CMRR i guess. The highside voltage is 14V battery
supply. Power supply voltage of op amp is +/-18v. CMRR is 90db.
Rshunt=2.5 mOhm, Ipeak=100A. So the voltage across the shunt is 250mV.
Now whats the purpose of these transistors?
Any suggestions are appreciated.
 
J

Jim Thompson

Jan 1, 1970
0
Hi,
thanks for u r reply
http://rapidshare.com/files/4191857/currentsensor.doc.html
here is the link for the circuit. Its high side sensing for the motor
current. In PSPICE i tried to see the effect of common mode voltage
using parametric simulation( giving different values of common mode
voltages) but the PSPICE is taking ideal conditions as its not showing
any difference in output voltage ( common mode voltage does not appear
in the output for ideal opamp) . Also tried to see the effect of these
transistors on the gain of op amp i tried without transistors ckt and
with. The gain is same in the PSPICE. So these transistors are placed
for controlling the CMRR i guess. The highside voltage is 14V battery
supply. Rshunt=2.5 mOhm, Ipeak=100A. So the voltage across the shunt is
250mV. Now whats the purpose of these transistors?
Any suggestions are appreciated.

To simulate the effects of common mode input voltages, you need an
OpAmp model that includes these effects... or roll your own:

Using an EVALUE behavioral element, create an output voltage that is
(for example) 100K times the input differential voltage PLUS (again
for example) ONE times the input common mode voltage [(VP + VN)/2]

That's RIGHT, ONE!

OpAmps are always specified as so many dB of CMRR, because, if they
specified common mode GAIN, you would be appalled.

Be appalled ;-)

...Jim Thompson
 
L

legg

Jan 1, 1970
0
Hi,
thanks for u r reply
http://rapidshare.com/files/4191857/currentsensor.doc.html
here is the link for the circuit. Its high side sensing for the motor
current. In PSPICE i tried to see the effect of common mode voltage
using parametric simulation( giving different values of common mode
voltages) but the PSPICE is taking ideal conditions as its not showing
any difference in output voltage ( common mode voltage does not appear
in the output for ideal opamp) . Also tried to see the effect of these
transistors on the gain of op amp i tried without transistors ckt and
with. The gain is same in the PSPICE. So these transistors are placed
for controlling the CMRR i guess. The highside voltage is 14V battery
supply. Rshunt=2.5 mOhm, Ipeak=100A. So the voltage across the shunt is
250mV. Now whats the purpose of these transistors?
Any suggestions are appreciated.

One interesting feature of this circuit that suggests that it may not
occur in the real world, is the choice of diode in D1 location.

Nobody in their right mind would apply a 120Amp schottky to establish
bias in a low power sensing circuit, unless it was present and in use
to do something else. If the latter is the case, it might be relevent
to the disection of the circuit, as to proper function.

You need to show us connections to this diode and to the sensor and to
the battery, without just resorting to VCC5 or VCC12 tags in the
drawing.

The full nature of high current shunts can be pretty elusive, and
physically hard to represent on paper accurately. Try harder.

RL
 
R

raf

Jan 1, 1970
0
Hi Mr.Legg,
The diode i placed could be a zener diode. As i dint able to recognize
it and i jsut want to see the effect of the circuit focussing on CMRR
so i just took it as a simple diode in PSPICE. But i would like to know
what the transistors do actually? what could be their function in this
HIGH SIDE CURRENR SENSING. If we remove the transistor ckt then its a
simple differential op amp which aplifies the differential input.
Expecting some suggestions....
 
L

legg

Jan 1, 1970
0
Hi Mr.Legg,
The diode i placed could be a zener diode. As i dint able to recognize
it and i jsut want to see the effect of the circuit focussing on CMRR
so i just took it as a simple diode in PSPICE. But i would like to know
what the transistors do actually? what could be their function in this
HIGH SIDE CURRENR SENSING. If we remove the transistor ckt then its a
simple differential op amp which aplifies the differential input.
Expecting some suggestions....

Your shunt currently has no reference to system ground, except through
the sensor.

The correct way to insert common-mode interference would be simply to
drive one end of the shunt with a ground-referenced signal.

Figure out what D1 is supposed to do, pick an accurate representation
of this device and stick it in your model. The same goes with any
other device in the bising network.

R4 and R5 dominate CM input network gain and sets the output to report
with respect to the emitter jn.. R6 offsets this output by an EB drop.
There is currently no reason for zero input to produce zero volts, as
reported.

s.e.d. tries to discourage top-posting. Please respect this
convention.

RL
 
A

Ancient_Hacker

Jan 1, 1970
0
this circuit doesnt make any sense to me either.

The choice of parts is very curious, the 4XXXXX is a very old RCA
series, from circa 1970, I doubt if they're still in production.

The other transistor and diode are rather unusual parts too.

Also any attempt at improving CMMR is going to require all those
resistors hanging on the op-amp inputs to be of very high precision and
stability. For 60dB CMMR you'll need resistors of considerably better
than 0.1% precision and stability. That's assuming the transistors do
anything.

One way of improving CMMR *might* be to somehow pull down the voltages,
precisely and exactly by the same amount. Unfortunately the circuit
as-is has no way of measuring the original input voltages, so it can't
subtract anything precisely.

A MUCH more sane way to improve CMMR would be to slave the op-amp's
power supplies to follow the input signal. A pair of resistors and
zeners would do the trick, assuming you have enough voltage headroom
on the op-amp supplies, or you can tie their common side to the shunt
bar.
 

neon

Oct 21, 2006
1,325
Joined
Oct 21, 2006
Messages
1,325
i asume the 2 transistors that you are refering is related to the amp input circuit not some external tranx. Actualy the emitters are connected usualy to a current source. And in an amplifier of good quality$ the emitters diodes are matched for voltage and current drop. therefore low mv offset and current offset. the CMMR is a function of the current source we all know that current sources don't care about voltage. In reality the input does't need 2 tranx only one the other can be a diode matched to the emitter Vbe drop of the transistor.
 
T

TuT

Jan 1, 1970
0
[snip]
Also any attempt at improving CMMR is going to require all those
resistors hanging on the op-amp inputs to be of very high precision and
stability. For 60dB CMMR you'll need resistors of considerably better
than 0.1% precision and stability. That's assuming the transistors do
anything.

[snip]

That's where the long-tailed pair can come into its own, with most of the
common-mode voltage supported across the transistors' c-e.

Say you have a tail resistor R, with bias voltage Vt across it and two
matched collector resistors to ground of 2R, then you have a common-mode
output voltage of Vt to deal with in a subsequent diff amp, which could be
much less than the common-mode at the supply. Errors then reduce to
transistor pair offset performance and mismatch between the two collector
resistors as a function of Vt, not Vcm.

Unfortunately, I'm blocked from accessing the link to the circuit here at
work, so I'm unable to add anything further in that respect.
 
R

raf

Jan 1, 1970
0
Hi all,
Great to recieve different ideas. In the ckt due to the unavailability
of data sheets (transistors) i chose available transistors in the
PSPICE, to run the simulation. Actually they are SMD(surface mounted
devices) one is named as SC and another one LD. And the diode which is
connected to Vcc (5V in this case) could be a ZENER diode.
And op amp is a dual opamp named 513AI. The second op amp (which is not
in the ckt) is sisilarly connected to the transistors but the diode is
COMMON(its connecetd to the base of the transistor). Now i hope the
things are bit clear.

Any way thanks and awaiting for suggestions................

RAF
 
R

raf

Jan 1, 1970
0
Hi,
after conneceting to groung ( the shunt ) with common mode voltage
source the simulation is running. Initially i ran the simulation by
removing the transistors and diode (R4 and R5 aswell) and resistor
R6(200k) to ground making a simple differential op amp.
By applying parametric simulation( different values to common mode
voltage) with input voltage as 250mV. The output is with different
curves with different values . Means common mode voltage effect at the
output.
Next i included the transistor circuit , resistors and ran same
simulation, Now the output is different. All common mode voltages are
separated from the output voltage and the amplitude between is alos
reduced.
From this its clear that the transistors reducing the common mode gain
and reducing the offst output voltages.

How they are performing (trnasistors) means whats happening
theoretically i am unable to understand. As u have suggested (Mr .Legg)
up to some level i can understand. som more datailed expanation i am
requesting u.
 
L

legg

Jan 1, 1970
0
Hi,
after conneceting to groung ( the shunt ) with common mode voltage
source the simulation is running. Initially i ran the simulation by
removing the transistors and diode (R4 and R5 aswell) and resistor
R6(200k) to ground making a simple differential op amp.
By applying parametric simulation( different values to common mode
voltage) with input voltage as 250mV. The output is with different
curves with different values . Means common mode voltage effect at the
output.
Next i included the transistor circuit , resistors and ran same
simulation, Now the output is different. All common mode voltages are
separated from the output voltage and the amplitude between is alos
reduced.
and reducing the offst output voltages.

How they are performing (trnasistors) means whats happening
theoretically i am unable to understand. As u have suggested (Mr .Legg)
up to some level i can understand. som more datailed expanation i am
requesting u.

Your circuit uses a single supply for the op amp.

Any differential measurement that reports with respect to ground will
suffer from the devices inability to produce outputs (and possibly
receive inputs) that approach the 0V supply limit.

Reporting with respect to a diode drop above ground, providing this
information is intelligable downstream, will improve linearity at the
extremes of the circuit's compliance.

The output of your circuit is correctly taken between the Op amp
output and the emitter node, not ground.

The accuracy of your model will depend on the model's ability to mimic
real physical limitations in the op amp.

RL
 
L

legg

Jan 1, 1970
0
Hi,
after conneceting to groung ( the shunt ) with common mode voltage
source the simulation is running. Initially i ran the simulation by
removing the transistors and diode (R4 and R5 aswell) and resistor
R6(200k) to ground making a simple differential op amp.
By applying parametric simulation( different values to common mode
voltage) with input voltage as 250mV. The output is with different
curves with different values . Means common mode voltage effect at the
output.
Next i included the transistor circuit , resistors and ran same
simulation, Now the output is different. All common mode voltages are
separated from the output voltage and the amplitude between is alos
reduced.
and reducing the offst output voltages.

How they are performing (trnasistors) means whats happening
theoretically i am unable to understand. As u have suggested (Mr .Legg)
up to some level i can understand. som more datailed expanation i am
requesting u.

Your circuit uses a single supply for the op amp.

Any differential measurement that reports with respect to ground will
suffer from the devices inability to produce outputs (and possibly
receive inputs) that approach the 0V supply limit.

Reporting with respect to a diode drop above ground, providing this
information is intelligable downstream, will improve linearity at the
extremes of the circuit's compliance.

The output of your circuit is correctly taken between the Op amp
output and the emitter node, not ground.

The accuracy of your model will depend on the model's ability to mimic
real physical limitations in the op amp.

RL
 
R

raf

Jan 1, 1970
0
Hi Mr.Legg,
I have gone through the material "single supply amplifiers" (rail-rail)
from Texas instruments. In that it has clearly mentioned about input
and output stages of the op amp. In the output stage the out put should
swing between positive supply rail and ground ( for battery
applications for example). Using BJTs the out put cannot swing
complterly to the rails.For that two complementary transistors ( one
PNP and NPN) are used . As u said the two resistors (r4 and r5)
overtake common mode gain and sets the output to report w.r.t emitter
junction. And u mentioned to take output between emitter junction and
output resistor.

I have tried using PSPICE by taking the outputs as u mentioned. I am
sending theoutput link here. The input i applied 250 mV ac with dc
common mode voltage rane from 1-100 with 10 V increment.
http://rapidshare.com/files/5013917/Output_plots.doc.html
I am still not getiing exact picture. Do u get some idea please do
inform me.
Regards
 
L

legg

Jan 1, 1970
0
Hi Mr.Legg,
I have gone through the material "single supply amplifiers" (rail-rail)
from Texas instruments. In that it has clearly mentioned about input
and output stages of the op amp. In the output stage the out put should
swing between positive supply rail and ground ( for battery
applications for example). Using BJTs the out put cannot swing
complterly to the rails.For that two complementary transistors ( one
PNP and NPN) are used . As u said the two resistors (r4 and r5)
overtake common mode gain and sets the output to report w.r.t emitter
junction. And u mentioned to take output between emitter junction and
output resistor.

I have tried using PSPICE by taking the outputs as u mentioned. I am
sending theoutput link here. The input i applied 250 mV ac with dc
common mode voltage rane from 1-100 with 10 V increment.
http://rapidshare.com/files/5013917/Output_plots.doc.html
I am still not getiing exact picture. Do u get some idea please do
inform me.
Regards

The pictures are nice, but have little meaning without carefull
labeling.

Just examining the schematic and assuming that the biasing network is
successfull in establishing an output reference that is within a diode
drop of the 5V supply, you can expect your circuit to function so long
as the common mode voltage does not exceed the limits of

12v-Vref x (15/5) + Vref positive (the situation where the
input pins approach the positive supply rail)

Vref x (15/5) - Vref negative. (the situation where the
input pins approch the negative supply rail - in this case 0V)

......if the input common mode range of your op amp model includes both
supply rails.

If the emitter junction is the intended output reference, then R6
should also connect to it and not to the base of Q1. If you need to
bias Q1, add an appropriate part that doesn't function in the gain
network.

AH has mentioned this circuit's sensitivity to tolerance. You can
model this by altering R1, R4, R3, or R5 by a fracction of a percent.
This error is amplified by R6, R2, the same as the differential
signal.

RL
 
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