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Pimp my pH input

F

Fox one! Fox one!

Jan 1, 1970
0
Hi,

I am trying to debug the first stage of a pH input circuit. The first stage
is just a voltage follower made from a TL084

The pH probe centre electrode is connected directly to the non-inverting
input of one opamp of a TL084 (this connection does not touch the PCB, it's
all in mid air)
The pH probe centre electrode also has a 33pF ceramic capacitor to ground,
close to the TL084
The pH probe outer electrode is connected to ground
There is a 1M feedback resistor from the output of the opamp to the
inverting input.

When I connect a pH probe, the output of this first stage is about 75mV.
According to the TL084 datasheet it has an input offset voltage of 6mV so I
can't understand where the 75mV comes from . I have tried 2 different pH
probes, one old one new, from different manufacturers, one with pointy tip
and one with round bulb, but they both give the same result so I don't think
it's the ph sensor.

The power supply I think is my problem, I decided to use the +10V and -10V
from a MAX232 rather than use a chip like a ICL7660. The MAX232 can only
supply the +V supply of the TL084 with 5.77V and the -V supply with -3.88V.

I am trying to calculate whether this is the problem. With this power
supply, the 'ground' for the opamp is actually at 0.945V relative to circuit
ground. With the pH probe in a pH 7 calibration solution, the output
between electrodes should be close to 0V, so in this case from the opamp's
point of view both inputs are at -0.945V, and the only signal on the output
should be the offset voltage + (common mode voltage * CMRR).

The CMRR on the datasheet is 86dB, or 1 / 20,000. If the common mode
voltage on the inputs is 0.945V the contribution to the output should only
be 47uV i.e. nothing.

So, why is the first stage giving 75mV? Should I use the supply-voltage
rejection ratio instead? It's also 86dB so adding them together still
doesn't explain it.

The output of the first stage is connected via a 510K resistor to the next
stage so it is hard to imagine that that is affecting it.

Any ideas would be much appreciated.
 
T

Tom Biasi

Jan 1, 1970
0
Fox one! Fox one! said:
Hi,

I am trying to debug the first stage of a pH input circuit. The first
stage is just a voltage follower made from a TL084

The pH probe centre electrode is connected directly to the non-inverting
input of one opamp of a TL084 (this connection does not touch the PCB,
it's all in mid air)
The pH probe centre electrode also has a 33pF ceramic capacitor to ground,
close to the TL084
The pH probe outer electrode is connected to ground
There is a 1M feedback resistor from the output of the opamp to the
inverting input.

When I connect a pH probe, the output of this first stage is about 75mV.
According to the TL084 datasheet it has an input offset voltage of 6mV so
I can't understand where the 75mV comes from . I have tried 2 different
pH probes, one old one new, from different manufacturers, one with pointy
tip and one with round bulb, but they both give the same result so I don't
think it's the ph sensor.

The power supply I think is my problem, I decided to use the +10V and -10V
from a MAX232 rather than use a chip like a ICL7660. The MAX232 can only
supply the +V supply of the TL084 with 5.77V and the -V supply
with -3.88V.

I am trying to calculate whether this is the problem. With this power
supply, the 'ground' for the opamp is actually at 0.945V relative to
circuit ground. With the pH probe in a pH 7 calibration solution, the
output between electrodes should be close to 0V, so in this case from the
opamp's point of view both inputs are at -0.945V, and the only signal on
the output should be the offset voltage + (common mode voltage * CMRR).

The CMRR on the datasheet is 86dB, or 1 / 20,000. If the common mode
voltage on the inputs is 0.945V the contribution to the output should only
be 47uV i.e. nothing.

So, why is the first stage giving 75mV? Should I use the supply-voltage
rejection ratio instead? It's also 86dB so adding them together still
doesn't explain it.

The output of the first stage is connected via a 510K resistor to the next
stage so it is hard to imagine that that is affecting it.

Any ideas would be much appreciated.

Can you post the circuit somewhere and the specs on the probe?

Tom
 
J

John Popelish

Jan 1, 1970
0
Fox said:
Hi,

I am trying to debug the first stage of a pH input circuit. The first stage
is just a voltage follower made from a TL084

The pH probe centre electrode is connected directly to the non-inverting
input of one opamp of a TL084 (this connection does not touch the PCB, it's
all in mid air)
The pH probe centre electrode also has a 33pF ceramic capacitor to ground,
close to the TL084
The pH probe outer electrode is connected to ground
There is a 1M feedback resistor from the output of the opamp to the
inverting input.

When I connect a pH probe, the output of this first stage is about 75mV.
According to the TL084 datasheet it has an input offset voltage of 6mV so I
can't understand where the 75mV comes from . I have tried 2 different pH
probes, one old one new, from different manufacturers, one with pointy tip
and one with round bulb, but they both give the same result so I don't think
it's the ph sensor.

The power supply I think is my problem, I decided to use the +10V and -10V
from a MAX232 rather than use a chip like a ICL7660. The MAX232 can only
supply the +V supply of the TL084 with 5.77V and the -V supply with -3.88V.

I am trying to calculate whether this is the problem. With this power
supply, the 'ground' for the opamp is actually at 0.945V relative to circuit
ground. With the pH probe in a pH 7 calibration solution, the output
between electrodes should be close to 0V, so in this case from the opamp's
point of view both inputs are at -0.945V, and the only signal on the output
should be the offset voltage + (common mode voltage * CMRR).

The CMRR on the datasheet is 86dB, or 1 / 20,000. If the common mode
voltage on the inputs is 0.945V the contribution to the output should only
be 47uV i.e. nothing.

So, why is the first stage giving 75mV? Should I use the supply-voltage
rejection ratio instead? It's also 86dB so adding them together still
doesn't explain it.

The output of the first stage is connected via a 510K resistor to the next
stage so it is hard to imagine that that is affecting it.

Any ideas would be much appreciated.

A pH probe set is a kind of electro-chemical cell, with one
fairly low resistance connection to the solution (the
reference probe that makes a connection to the solution with
minimal voltage drop) and one very high resistance
connection that includes the voltage offset produced by the
cell. The hydrogen ion (naked protons that are much smaller
than any atom) concentration of the solution determines the
voltage produced by the cell, as they diffuse through the
glass bulb of the measurement probe and produce a net
voltage across the glass membrane.

You must pass nearly zero current through either of these
connections to get an undistorted measure of the voltage
difference produced by the protons. It sounds like you have
a very high resistance load (very limited current through)
the measurement probe, but how about the reference probe.
If the solution has other ground paths, you may be passing
current through it, and getting some voltage drop through
it. Is the test solution in a glass container (and
electrically isolated from any current path that includes
the reference probe) or is it grounded through other means
than just the reference probe?
 
F

Fox one! Fox one!

Jan 1, 1970
0
Tom Biasi said:
Can you post the circuit somewhere and the specs on the probe?

Tom

Hi Tom,

Here is the circuit.



1M
___
.-----------|___|---------.
| |
| |
| 5.77V |
| |
| | |
| | |
| | |
pH probe | |\| |
.-----. '-----------|-\ |
| G | | >----------o-------------
|_-_-_|--------------------|+/
| | | |/| TL084
'--o--' --- |
| --- |
| | |
|
0V 0V |

-3.88V


(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
 
J

John Popelish

Jan 1, 1970
0
Fox said:
Hi Tom,

Here is the circuit.



1M
___
.-----------|___|---------.
| |
| |
| 5.77V |
| |
| | |
| | |
| | |
pH probe | |\| |
.-----. '-----------|-\ |
| G | | >----------o-------------
|_-_-_|--------------------|+/
| | | |/| TL084
'--o--' --- |
| --- |
| | |
|
0V 0V |

-3.88V
Are the DC supplies batteries or AC derived supplies? If
the latter, you have to include the AC paths through the
inter winding capacitances back to the line and from there
to ground and from there, back to the zero volt node.

Have you checked to see if there is any AC hum on the opamp
output?
 
F

Fox one! Fox one!

Jan 1, 1970
0
Mr.pH said:

Thankyou that is a good page, I have read it before. My circuit is very
similar except it is unity gain. The only reason I have the resistor in the
feedback at all is that I read on this page:
http://www.ibiblio.org/obp/electricCircuits/Semi/SEMI_8.html (search for
"ph") that if the resistor has the same impedance as the probe, the voltage
across it due to the opamp leakage current will match and thus cancel the
voltage error due to the same leakage through the probe.

I forgot to mention to Tom the specs of the probe, but no I don't know what
they are. I am trying to make it work with typical probes, which from what
I've heard are round bulb types and have a impedance of about 400Mohms.
 
J

John Popelish

Jan 1, 1970
0
Fox one! Fox one! wrote:
(snip)
... My circuit is very
similar except it is unity gain. The only reason I have the resistor in the
feedback at all is that I read on this page:
http://www.ibiblio.org/obp/electricCircuits/Semi/SEMI_8.html (search for
"ph") that if the resistor has the same impedance as the probe, the voltage
across it due to the opamp leakage current will match and thus cancel the
voltage error due to the same leakage through the probe.

The probe resistance is not known and very high, compared to
1 meg. You need an opamp that has a bias current so low
that it produces an insignificant drop across the 100 meg or
so probe resistance. Then you can eliminate the feedback
resistor. I would go for an opamp with sub pico ampere bias
current. perhaps the LMC6001 with 25 fempto amps of bias
current.
http://www.national.com/ds/LM/LMC6001.pdf
 
F

Fox one! Fox one!

Jan 1, 1970
0
John Popelish said:
A pH probe set is a kind of electro-chemical cell, with one fairly low
resistance connection to the solution (the reference probe that makes a
connection to the solution with minimal voltage drop) and one very high
resistance connection that includes the voltage offset produced by the
cell. The hydrogen ion (naked protons that are much smaller than any
atom) concentration of the solution determines the voltage produced by the
cell, as they diffuse through the glass bulb of the measurement probe and
produce a net voltage across the glass membrane.

You must pass nearly zero current through either of these connections to
get an undistorted measure of the voltage difference produced by the
protons. It sounds like you have a very high resistance load (very
limited current through) the measurement probe, but how about the
reference probe. If the solution has other ground paths, you may be
passing current through it, and getting some voltage drop through it. Is
the test solution in a glass container (and electrically isolated from any
current path that includes the reference probe) or is it grounded through
other means than just the reference probe?


My understanding of the probes is that the junction and the reference are
all in the one package with only two terminals. In my circuit one terminal
is connected to ground and the other directly to the non-inverting input of
a TL084, so that should be zero current shouldn't it?

It does respond to ph changes, I should have mentioned this:
In a pH 7 buffer the output is 75mV (expecting 0mV)
In a pH 4 buffer the output is 195mV (expecting 180mV, as output should
change roughly 60mV for each pH)

The test solutions are in disposable plastic cups with no other wires in
them.

..
 
J

John Popelish

Jan 1, 1970
0
Fox said:
My understanding of the probes is that the junction and the reference are
all in the one package with only two terminals.

That is just comp[act [packaging. There are still two
distance probes that contact the solution, one through a
direct electrical connection through a fluid bridge, and one
through a membrane.
In my circuit one terminal
is connected to ground and the other directly to the non-inverting input of
a TL084, so that should be zero current shouldn't it?

It does respond to ph changes, I should have mentioned this:
In a pH 7 buffer the output is 75mV (expecting 0mV)
In a pH 4 buffer the output is 195mV (expecting 180mV, as output should
change roughly 60mV for each pH)

The test solutions are in disposable plastic cups with no other wires in
them.

I see. That simplifies things from a DC standpoint. Is the
cup sitting on a metal sink or a non conductive surface? In
a separate post, I have questions for you about the possible
hum being injected through the supply.
 
F

Fox one! Fox one!

Jan 1, 1970
0
Fox one! Fox one! said:
Hi,

I am trying to debug the first stage of a pH input circuit. The first
stage is just a voltage follower made from a TL084

The pH probe centre electrode is connected directly to the non-inverting
input of one opamp of a TL084 (this connection does not touch the PCB,
it's all in mid air)
The pH probe centre electrode also has a 33pF ceramic capacitor to ground,
close to the TL084
The pH probe outer electrode is connected to ground
There is a 1M feedback resistor from the output of the opamp to the
inverting input.

When I connect a pH probe, the output of this first stage is about 75mV.
According to the TL084 datasheet it has an input offset voltage of 6mV so
I can't understand where the 75mV comes from . I have tried 2 different
pH probes, one old one new, from different manufacturers, one with pointy
tip and one with round bulb, but they both give the same result so I don't
think it's the ph sensor.

The power supply I think is my problem, I decided to use the +10V and -10V
from a MAX232 rather than use a chip like a ICL7660. The MAX232 can only
supply the +V supply of the TL084 with 5.77V and the -V supply
with -3.88V.

I am trying to calculate whether this is the problem. With this power
supply, the 'ground' for the opamp is actually at 0.945V relative to
circuit ground. With the pH probe in a pH 7 calibration solution, the
output between electrodes should be close to 0V, so in this case from the
opamp's point of view both inputs are at -0.945V, and the only signal on
the output should be the offset voltage + (common mode voltage * CMRR).

The CMRR on the datasheet is 86dB, or 1 / 20,000. If the common mode
voltage on the inputs is 0.945V the contribution to the output should only
be 47uV i.e. nothing.

So, why is the first stage giving 75mV? Should I use the supply-voltage
rejection ratio instead? It's also 86dB so adding them together still
doesn't explain it.

The output of the first stage is connected via a 510K resistor to the next
stage so it is hard to imagine that that is affecting it.

Any ideas would be much appreciated.

I've noticed that if I ground the ph input the opamp gives 1.7mV!! Maybe
the probe is the problem?
 
F

Fox one! Fox one!

Jan 1, 1970
0
John Popelish said:
Fox one! Fox one! wrote:
(snip)


The probe resistance is not known and very high, compared to 1 meg.

Good point I just thought since the space is there I would put in the
largest value I have in my parts pile. I tried bypassing it (I am beginning
to get superstitious now) and it made no difference (of course I knew it
wouldn't).
You need an opamp that has a bias current so low that it produces an
insignificant drop across the 100 meg or so probe resistance. Then you
can eliminate the feedback resistor. I would go for an opamp with sub
pico ampere bias current. perhaps the LMC6001 with 25 fempto amps of bias
current.
http://www.national.com/ds/LM/LMC6001.pdf

Ah yes I have seen that part but this is a project for a hobby community on
the net, so I am trying to use parts you can get from the local electronics
store in case they ever have to repair it. My local store stocks these
which seem like they are the right type (FET input):

CA3130 5pA bias current
CA3140 10pA
TL084 30pA (what I am using)
TL074 65pA

By my calculations, assuming the probe has an impedance of around 675Mohm
(based on 'standard glass', hemispherical shape on this page:
http://www.asi-sensors.com/learning/lrn_ph_glass.aspx) then a worst case
30pA leakage current would cause a voltage drop of 20mV. That is
surprisingly large but still doesn't explain where 75mV is coming from.
 
F

Fox one! Fox one!

Jan 1, 1970
0
Are the DC supplies batteries or AC derived supplies? If the latter, you
have to include the AC paths through the inter winding capacitances back
to the line and from there to ground and from there, back to the zero volt
node.

Have you checked to see if there is any AC hum on the opamp output?

I've tried a DC supply and an AC supply (my PCB supports both) but both give
the same results. In normal operation I'd expect it to be used with an AC
supply, and for it to be installed in an earthed metal case, with circuit
ground connected to the case by the screw mounting points. i.e. circuit
ground should aways be mains earth

I've probed it with my CRO and no I can't see any mains frequency on the
output at all.

Thanks for your help.
 
F

Fox one! Fox one!

Jan 1, 1970
0
John Popelish said:
Fox said:
My understanding of the probes is that the junction and the reference are
all in the one package with only two terminals.

That is just comp[act [packaging. There are still two distance probes
that contact the solution, one through a direct electrical connection
through a fluid bridge, and one through a membrane.
In my circuit one terminal is connected to ground and the other directly
to the non-inverting input of a TL084, so that should be zero current
shouldn't it?

It does respond to ph changes, I should have mentioned this:
In a pH 7 buffer the output is 75mV (expecting 0mV)
In a pH 4 buffer the output is 195mV (expecting 180mV, as output should
change roughly 60mV for each pH)

The test solutions are in disposable plastic cups with no other wires in
them.

I see. That simplifies things from a DC standpoint. Is the cup sitting
on a metal sink or a non conductive surface? In a separate post, I have
questions for you about the possible hum being injected through the
supply.


They are sitting on top of my computer. I tried putting a book between them
and lifting the cup upwards but that made no difference.
 
J

John Popelish

Jan 1, 1970
0
Fox said:
I've tried a DC supply and an AC supply (my PCB supports both) but both give
the same results. In normal operation I'd expect it to be used with an AC
supply, and for it to be installed in an earthed metal case, with circuit
ground connected to the case by the screw mounting points. i.e. circuit
ground should aways be mains earth

I've probed it with my CRO and no I can't see any mains frequency on the
output at all.

Then I am fresh out of leads to follow. Sorry. Perhaps
your probe set is just contaminated or the reference probe
plug clogged and is producing the 75 mV offset, no matter
what amplifier is used to buffer it. I have seen probe sets
go bad this way. You might try leaving the probe in the pH
7 buffer solution for a few days and see if the offset
drifts in either direction. No need to keep the amplifier
on all the time, though it shouldn't hurt.
 
I've noticed that if I ground the ph input the opamp gives 1.7mV!! Maybe
the probe is the problem?

Or your opamp has a lot higher bias current than it is supposed to
have. Replace the probe with a 10 meg resistor and see how high the
output goes.
 
F

Fox one! Fox one!

Jan 1, 1970
0
Sheesh that was the problem! I don't have a 10meg resistor but I replaced
the TL084 and voila, it works. I could swear I already tried this, god
knows I changed everything else... Thanks a lot for your help.
 
J

John Popelish

Jan 1, 1970
0
Fox said:
Sheesh that was the problem! I don't have a 10meg resistor but I replaced
the TL084 and voila, it works. I could swear I already tried this, god
knows I changed everything else... Thanks a lot for your help.

:)

Live long and prosper.
 
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