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I am in first year of Engineering school and I was given an assignment containing this circuit, which drives pressure sensors in a pitot tube :

enter image description here

I am struggling to understand the whole circuit, and more precisely the first op-amp, which output (pin 1) and e- (inverting) input (pin 2) are connected to ground.

What is its use? How can such an op-amp have an influence on the overall circuit, if its output is not used?

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The first OP-amp is actually creating the circuit ground. The 7810 creates a stable 10 volt, which is then divided by the voltage divider R2 and R3, filtered by C3 to make a stable 5 volt level relative to the most negative level.

The OP-amp then buffers this, and the rest of the circuit uses its output as the reference ground. Remember that ground in a circuit like this is just a convenience, a node that is used when referring to other voltages.

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  • \$\begingroup\$ Why does the "ground" need to be buffered? \$\endgroup\$ – Ian Ringrose Apr 28 '16 at 10:23
  • \$\begingroup\$ See @placeholder's answer. \$\endgroup\$ – nekomatic Apr 28 '16 at 11:15
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While I agree with @pipe and in fact upvoted his answer, an additional nuanced reply is that a ground is more than "just" a reference.

What I mean by this is a ground is not just a voltage, but something that can source & sink current and stay at the same potential.

The ground created by that op-amp can both source and sink current and remain roughly half way between the rails of the +12 volt source. If the design were to have just used another regulator like a 7805, that device would have only sourced current and thus would only have put out the right "mid voltage" value when current was flowing out of it's output.

Rather more restricting that the circuit shown.

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    \$\begingroup\$ There is also something a bit dodgy about the circuit- the 2.2uF capacitor is between two op-amp outputs (one being the virtual ground) and there are no other capacitors to the virtual ground (unless there is something in the DPM). Depending on the type of op-amp this is a formula for interesting types of instability. The originator of the circuit apparently got at least one instantiation it to work well enough, but caution is called for. \$\endgroup\$ – Spehro Pefhany Apr 28 '16 at 3:26
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    \$\begingroup\$ @SpehroPefhany that is a valid concern, but take into account that all these devices are in one package, off the same rails, and the pressure sensors draw only ~ 3 mA this circuit should be OK with those concerns. It would be safer to add in some bypass though for sure. However, did you notice the heavy cap on the last stage being driven directly? Yikes \$\endgroup\$ – placeholder Apr 28 '16 at 4:39
  • \$\begingroup\$ @SpehroPefhany And not only that: for several reasons that circuit is a clear example of bad design. \$\endgroup\$ – Massimo Ortolano Apr 28 '16 at 9:40
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    \$\begingroup\$ That's the one I'm talking about. It's between the outputs of amplifiers I and IV. And @MassimoOrtolano is right, there are several other bad things about that circuit- the display will typically change tens of counts per degree C just from the op-amp drift \$\endgroup\$ – Spehro Pefhany Apr 28 '16 at 11:19
  • \$\begingroup\$ We reference all other potentials to ground so how can ground potential change? The only way for it to change potential is to use another ground to reference it too. The designer choses what is ground and if he would have chosen to use a 7805 it would have been a different design, good or bad it doesn't matter it was his choice. \$\endgroup\$ – Daniel P Feb 17 '18 at 10:15
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(TL,DR: see paragraph 5)

The meter module requires the voltage on its GND pin to be between its V+ and V- supply pins. It converts and displays the voltage between its IN+ pin and GND.

The 7810 regulates the input to 10V between nodes labelled +5V and -5V.

R2 and R3 provide a mid-point voltage with Thevenin impedance of 2.5 kOhm (= 10K // 10K) in parallel with 100nF. Thus any (DC) current drawn from or supplied to this node will push the voltage by 2.5V per milliamp.

The GND node will be carrying currents from: Meter VIN+, R9, R11, R15, R16, R13, A2 and C5. These will probably sum to less than a milliamp, but the meter may draw varying current through each measurement cycle.

Amplifier 1 acts as a voltage follower for the R2 R3 chain. It will act to hold its output, the node labelled GND, at the mid-point of nodes labelled +5V and -5V. Looking at it from a different perspective, it acts to pull the mid-point of its supplies towards its output voltage. It will have a closed-loop output impedance of a few ohms, so current drawn on the GND node will have little effect on the voltage between GND and the +5V and -5V lines.

Amplifiers II-IV are all configured as simple differential amplifiers. II and III have a gain of 100V/V and Zin of 10K. IV has a gain of 20 and Zin of 50K.

Having C5 connected directly to the output of amplifier IV is an error. OPAs are not specified to be stable with a large capacitative load. It would probably be better to put it across meter VIN+ and GND, with 10K or so between the wiper of A2 and the VIN+.

The gain of the circuit will directly depend upon the output voltage of the 7810. If the meter has an external reference input it would be best to connect this to a fraction of +5V, this giving a ratiometric reading.

The offset voltage of all four amplifers will contribute to the signal. The amplifiers will need good DC and 1/f noise specifications.

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A simple and direct answer to your main question, is that this is one way of providing the differential voltages that the op-amps need (+ & - 5v). By floating the ground (to +5v) the single 10v source can supply + & - 5v! You should now be able to understand that the output of op-amp I, is being used. It creates a virtual ground (or reference ground).

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