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The circuit given below is typical pH meter op-amp circuit. But here LMC6001 is not configured as unity gain.

typical ph meter circuit

My question is this, I want to make this circuit with unity-gain buffer op-amp. What about R2, R4, C1 and R9 in here? What do they do? Are they useful to make signal stable?

What about making LMC6001 unity-gain by removing R1,R2,R3 and moving C1 R9 shunt to LMC6001's feedback loop?

And what I understand yet: R1, R3 are gain resistors. LMC6041's function is adding reference voltage (adjusted by R8 // R6,R7,R8 resistor divider) offset to ADC's input (or pH signal).

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    \$\begingroup\$ Why do you want to make it unity-gain? You do know that the nominal ph-probe output is ~0.06V/ph, right? You're going to need gain somewhere. \$\endgroup\$
    – Connor Wolf
    Commented Jan 27, 2015 at 11:47
  • \$\begingroup\$ Nope. I have got enough offset voltage. No gain is needed. Plus, i don't use negative rail (single supply). \$\endgroup\$
    – user30878
    Commented Jan 27, 2015 at 11:51
  • \$\begingroup\$ You're missing the point. What device are you using that can resolve well below 60 millivolts? Because right now, 60 millivolts change equates to a difference of 1 pH. Unless you need only the coarsest of coarse measurements, a plain MCU ADC probably won't work too well here. \$\endgroup\$
    – Connor Wolf
    Commented Jan 27, 2015 at 11:52
  • \$\begingroup\$ You know there are a lot of ADCs up to 12 bit resolution. \$\endgroup\$
    – user30878
    Commented Jan 27, 2015 at 11:54
  • \$\begingroup\$ I know there are ADCs with 24 bit resolution (or higher!), but no matter what the bit size of your ADC, only using part of it's input range is poor engineering. \$\endgroup\$
    – Connor Wolf
    Commented Jan 28, 2015 at 2:12

1 Answer 1

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It's best to think of this circuit in two sections. Everything before R4 belongs in the first section. R4, and everything after it, belongs in the second section. Now the first section looks like a standard op-amp noninverting voltage amplifier. As you probably know, the gain of such an amplifier is equal to \$1 + Z_f/Z_{in}\$, where \$Z_f\$ is the feedback impedance and \$Z_{in}\$ is the input impedance. The second section is an inverting voltage amplifier, with gain equal to \$-Z_f/Z_{in}\$.

The gain of the first section (with the LMC6001 op-amp) is therefore equal to 1 + (R2+R_V)/R1, where R_V is the variable resistance setting of the R3 potentiometer. The simplest way to change this to a unity-gain buffer is actually just to remove R1 and short the V- terminal to the output. The op-amp forces V- to equal V+, so if V- is shorted to the output, then the output must also equal V+.

The gain of the second section is -R9/R4 = -1. You correctly observe that the network connected to the positive input of the LMC6041 is supposed to add a DC offset. So this section is already at unity gain, but it's inverting.

The capacitor C1, in combination with R9, acts as a low-pass filter with a cutoff frequency of 0.7 Hz. That makes sure that the output voltage changes slowly, so you can read the output with a multimeter.

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  • \$\begingroup\$ Thanks for very clear answer. Three questions now: 1- Why don't remove R1,R2,R3 instead of making them equal? 2- What kind of low pass filtering is this? Its not look like active or passive low pass filters that i learn. What it filters? Is it filtering reference offset voltage or pH signal? 3- With unity gain configuration, just seen in here i.sstatic.net/iZzJl.png adding same valued serial resistor both before Vin and feedback loop is useful for making signal stable? \$\endgroup\$
    – user30878
    Commented Jan 27, 2015 at 12:04
  • \$\begingroup\$ Q1. The whole point of an op-amp is that it has a huge intrinsic voltage gain - generally over 10^6. This gain is "tamed" by the feedback network. If you simply remove the feedback, though, you won't get unity gain. You'll get 10^6*(V+ - V-) where V+ and V- are the voltages at the input pins. Excellent explanations of op-amp basics can be found in The Art of Electronics by Horowitz and Hill (en.wikipedia.org/wiki/The_Art_of_Electronics). \$\endgroup\$
    – Dave Kielpinski
    Commented Jan 27, 2015 at 12:05
  • \$\begingroup\$ Q2. The voltage gain is the feedback impedance divided by the input impedance. The capacitor impedance is equal to 1/(2 pi j f C) where f is the frequency. So the capacitor shorts out the high frequencies. \$\endgroup\$
    – Dave Kielpinski
    Commented Jan 27, 2015 at 12:06
  • \$\begingroup\$ "you won't get unity gain. You'll get 10^6*(V+ - V-).." ok but V- is 0 in my circuit, its grounded. \$\endgroup\$
    – user30878
    Commented Jan 27, 2015 at 12:08
  • \$\begingroup\$ Q3. Adding a series resistor to the output of an op-amp can sometimes make the circuit more stable when high-frequency oscillations are a problem. It can also be useful to avoid overdriving the current of the output stage. In your application, I don't think either of these is a concern. \$\endgroup\$
    – Dave Kielpinski
    Commented Jan 27, 2015 at 12:08

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