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audioguru: I have not read your posted tutorial and therefore I do not know whether I agree with it or not. I do not intend to continue to go head to head with you or anyone else regarding simple op-amp theory. In my original post, I pointed out that you do not need a bias voltage from the power supply added to an op-amp that is run from a bipolar supply. The inputs are referenced to ground, which is the center point of a bipolar supply. Perhaps you disagree with this. So be it. Move on.
In the circuit that you have posted, I know nothing about it, do not have a parts list, nor do I know what the second IC is, but it appears to be an AC or pulsed active circuit. In which case, the pin with the capacitor connected to it is not floating when the field is collapsing and there is capacitive reactance. Perhaps this is what this circuit is meant to do since there is a DC blocking capacitor at the output of the op-amp. Again, I do not intend to spend time in debate with you.
If your purpose here is trying to determine if this circuit works, I must point out that you have spent days in debate when you could breadboard this circuit in minutes to make such a determination.
In reference to your comment, "Why ignore me? You might learn something important". I do not think anyone with your attitude would be able to teach me anything. I now fully understand the prior comments posted by ante.

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It seems simple enough to me. The input can be grounded or grounded through a resistor. There is the case where you want the input bias currents to be the same, for a zero voltage output. This is the resistor which will allow you to affect the input bias current. I think the resistor is for opamps that don't have a null offset pot.

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Wired,
There are many good discussions about electronics at this place.
Please join-in if you have good proof for your arguments, as I have.

Kevin,
The missing resistor provides a VOLTAGE reference of 0V to the non-inverting input of the opamp, not a current source, since the opamp's input current is very, very small.
Offset current equaling or its adjustment is not neccessary for this circuit.

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Wired,
There are many good discussions about electronics at this place.
Please join-in if you have good proof for your arguments, as I have.


audioguru, since when does this forum give you the right to tell people when they can join a discussion? No one has to prove anything in this forum to discuss electronics. >:(

The bottom line is this: I think you just trampled on a new member and sent him away pissed off simply because he did not agree with your views. Then told him he could join in the discussion if he has proof.
This is not allowable.

...as far as I am concerned: you are wrong and, no, I am not going to waste my time to find you the proof. It is an application I use in real life ....and I will join any discussion I please, even with lack of the proof you require.
(as you can see, when you drive people away from the forum, it pisses me off)

MP
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You didn't comment on my University Opamp Tutorial, which explained that BOTH inputs should have a voltage reference.
My University is well respected, and so is National Semiconductor.
In the latter's Application Note #20, An Applications Guide For Opamps, page 2 describes "The Non-inverting Amplifier" (which is what we are discussing):
"The amplifier output will go into saturation if the input is allowed to float".

That is exactly what I say in my quote, above, and have been saying all along.
The Application Note is here:
http://www.national.com/an/AN/AN-20.pdf

The opamp cannot work properly without having a voltage reference (ground, in our application) on its non-inverting input.


You still don't believe my University nor National Semiconductor?
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Who is wrong?
We are talking about range (distance) here, which requires the use of a high-gain LINEAR amplifier in the receiver.
The author probably knew that his project has a range problem when he mentioned using a high-powered IR LED to get more range.
This project has a tone-modulated IR LED as a transmitter, and a phototransistor, amplifier and PLL bandpass filter/detector as a receiver. The tone is used to avoid interference from light, heat and other IR transmitters. The amplitude of the tone received at the detector determines the range.

Put a distance between transmitter and receiver so that the tone level at the collector of the phototransistor is only 200 microvolts. With the opamp having a floating input, its output will be saturated against a power rail and its output level will be ZERO. Only a whopping big input signal will allow the saturated opamp to function.
If the opamp's input is properly biased with a single resistor to ground, then it will be a linear amplifier with a gain of 101, and its output level will be 20.2mV, which is enough to allow the detector to activate.

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Kevin,
No, an opamp will not be a linear amplifier when using positive feedback instead of negative. It will be a Schmitt trigger and produce a square-wave output. Its input will have hysteresis that is determined by the ratio of the feedback resistors.
An opamp with negative feedback has its gain determined by the ratio of its feedback resistors, if it is operating within is passband. In the case of this project, the opamp has a 100K feedback resistor (R4) and a 1K resistor to ground (R3). So its gain (if properly biased) is 100/1 + 1 = 101. The schematic is attached:

post-1706-14279141649121_thumb.gif

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Audioguru, we are not talking about rail to rail operation. We are in fact able to reduce the gain with the feedback resistor because the opamp is a symmetrical device. The positive feeback resistor can lower the gain because it is infact still a collector resistor. You will notice that it works a little different though. Just design it backwards using about a 1k load to -18volts.

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Kevin,
Positive feedback doesn't reduce an opamp's gain, unless its resistance is so low that it overloads the opamps output. With positive feedback, the opamp's output will switch as close to rail-to-rail as it can. An opamp doesn't need a collector resistor because its output is push-pull, usually complimentary emitter-followers. If you load an opamp's output with a 1K resistor to -18V, then you are probably overloading its positive output swing.
Try it.

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The 1Kohm to -18 can simulate a load. The current is even close enough. But more importantly, it shows that this is a collector resistor. Whenever you are able to reduce the voltage away from a rail, such as at 0volts, you are reducing the gain. Just apply the signal and notice the gain is much lower.

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Kevin,
Put a 'scope on your opamp's output and you'll see that the 1k load resistor to -18V is overloading its positive output swing. Its gain is not reduced, it is current-limiting its output to protect itself against your abnormal load.
Which opamp part number? Look at its data sheet to see its maximum positive (sourcing) output current. Then use Ohm's Law to determine how high its output can swing without current-limiting.

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Kevin,
18mA of load current is a lot for most opamps, especially when the load connects to a negative voltage, instead of to ground as is usual.
You don't say which opamp, but a good old 741 is guaranteed to swing its output to only 10V with a 2K load to ground, when using a +, - 15V supply. That's only 5mA. Its output short circuit current is guaranteed to be only 10mA. So your 1K load to -18V will seriously overload the output of a "weak but still guaranteed" 741, and even a typical one that has 25mA of output short circuit current. You can't measure gain when the output is overloaded.
See page 3 of the LM741's data sheet, "Output Voltage Swing" and "Output Short Circuit Current". The data sheet is here:
http://www.national.com/ds/LM/LM741.pdf

You keep saying that your load is a collector resistor. Look at the schematic of the 741 opamp on page 4 of its data sheet and you'll see that the output transistors, Q14 and Q20, are emitter followers that don't have, and don't need, collector resistors.

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You have to investigate the circuit more closely. The opamp is rated for 20mA, which is what the circuit demands. Look at the ouput of an opamp, the emitter follower. You can easily approximate the current because of the diodes in parallel with the resistors. Notice that the load resistor is in parallel with the circuit before the emitter followers. The other resistor is an emitter element because it is a base resistor when compared to the input stage. Look at the input stage. The base resistor directly affects it's gain.

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