Phototransistor with virtual ground opamp design

I am setting up a high sensitivity light detection circuit and came across this thread just a bit ago:

Phototransistor transimpedance amplifier

Within it was a response with this image:

My question is this: Should the phototransistor in the diagram above be connected to the true ground coming from the power supply or to the virtual ground?

Thank you and I know this may seem like a newbie question and perhaps similarly asked in other threads, but it doesnt seem to have been explaind in my opinion.

First of all the phototransistor seems to be drawn upside down, the emitter should be at the negative side.

In your circuit, the inverting input is held at the virtual ground potential by the feedback loop. In order for the output of the opamp to be made to go positive, current must be drawn from the circuit node to which the inverting input is connected. This will cause the inverting input to tend to go negative and the output voltage will rise, causing current to flow (to the left) through the feedback components thereby restoring the voltage at the inverting input terminal. In order for this to happen the phototransistor must be connected to a more negative potential than the virtual earth. The "true ground" would be a suitable point. If it were returned to the virtual ground, then there would be no potential difference across the transistor and the circuit would not work.

It has to be tied to your true ground, not virtual ground. The virtual ground is used by your opamp

COMPLETE REWRITE:

I kinda cheated last time by using floating voltage sources as the opamp inputs.

Here's something a little closer to the truth, using the phototransistors in optocouplers.

Both ways work, (grounded and pseudogrounded) but the grounded version has a lot more gain as evidenced by the difference in the LED ballast resistors.

Version 4
SHEET 1 3416 680
WIRE 1696 -176 1584 -176
WIRE 1920 -176 1760 -176
WIRE 2672 -176 2560 -176
WIRE 2896 -176 2736 -176
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WIRE 1696 -64 1584 -64
WIRE 1920 -64 1920 -176
WIRE 1920 -64 1776 -64
WIRE 2560 -64 2560 -176
WIRE 2672 -64 2560 -64
WIRE 2896 -64 2896 -176
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WIRE 1712 80 1680 80
WIRE 1824 80 1824 64
WIRE 1824 80 1792 80
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WIRE 2800 80 2800 64
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TEXT 1048 472 Left 2 !.tran 7

• How does this work when the phototransistor is referenced to virtual earth? The only way I can see that working is if the phototransistor exhibits significant photovoltaic effect. Is this the mechanism you are thinking of for this circuit configuration? – user1582568 Feb 24 '16 at 16:45
• No. All I was pointing out was that if a voltage source is connected between an opamp input (- in this case) and either a ground or a pseudoground related to the opamp, there will exist a potential difference across the opamp inputs which the opamp's output will servo to cancel out and thereby provide an output. – EM Fields Feb 24 '16 at 20:32
• Yes, if it is a voltage source that's true, but this is a phototransistor – user1582568 Feb 24 '16 at 20:34
• Yes, but it's acting like a voltage-variable resistor in a voltage divider, (with a potential across it) the other resistance being the feedback resistor from the opamp's output to its inveting input. Check out my edited/new answer. :) – EM Fields Feb 24 '16 at 23:12
• Ah, but it would not work if all the components were ideal. I suspect the tiny amount of response that you are getting in your simulation is due to the opamp bias current providing a little offset and so biasing the phototransistor. In the ideal circuit the opamp output would be at the virtual earth potential and there would not be any voltage across the feedback resistor. – user1582568 Mar 1 '16 at 13:28