was Re:Ir-remote Now is: Op Amp discussion
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audioguru2
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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.
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.
Kevin Weddle
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One might get a better feel for opamps if you connect the negative feedback from ouput to noninverting input instead of the inverting input. This is called positive feedback, but the idea of gain is relatively the same. In other words, connect the opamp backwards and realize the similarities.
audioguru2
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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:
View attachment 35501
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:
View attachment 35501
Kevin Weddle
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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.
audioguru2
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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.
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.
Kevin Weddle
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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.
audioguru2
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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.
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.
Kevin Weddle
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The input signal as well as the output signal and the DC bias looks correct. What is wrong with 18ma load current? The gain is a little higher than RF over RI but the gain is much lower than open loop.
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audioguru2
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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.
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.
Kevin Weddle
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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.
audioguru2
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Kevin,
I don't understand you.
You didn't attach a sketch of whatever you are talking about.
In my last reply, I posted the data sheet for a 741 opamp. It doesn't have the diodes that you are talking about, and it uses transistor current sources instead of the collector resistors that you are talking about. Its input stage doesn't have a base resistor.
Did you see it? Would you like to discuss it?
I don't understand you.
You didn't attach a sketch of whatever you are talking about.
In my last reply, I posted the data sheet for a 741 opamp. It doesn't have the diodes that you are talking about, and it uses transistor current sources instead of the collector resistors that you are talking about. Its input stage doesn't have a base resistor.
Did you see it? Would you like to discuss it?
Kevin Weddle
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RI is your base resistor. The emitter resistance of the other transistor sees this base resistor. The impedance, looking at the emitter of a transistor, is resistance of the base divided by beta. RF is a collector resistor as seen by the output stage. You are right, there are no diodes. But there are PN junctions in parallel with the resistors that can be used to calculate the current.
You may not believe me when I tell you there is a different version of that circuit you posted. It's missing 2 diodes and a connection point. Anyways, the currents subtract with the upper current being higher and the load gets the rest of the current.
You may not believe me when I tell you there is a different version of that circuit you posted. It's missing 2 diodes and a connection point. Anyways, the currents subtract with the upper current being higher and the load gets the rest of the current.
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audioguru2
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Kevin,
You keep thinking that an opamp is a transistor. They are not the same.
An opamp is a complete amplifier with a voltage gain of about 200,000, a low-impedance symmetrical push-pull output and a high-impedance differential input. It doesn't need a resistor at its output to a supply voltage, like a transistor needs for its collector current. An opamp with + and - supply voltages usually uses 0V as ground. Its load is also usually connected to ground since its output is symmetrical.
The voltage gain of a complete opamp circuit is determined by the ratio of its external negative feedback resistors. Since the voltage gain of an opamp without feedback is so high, its output will swing to a voltage level that is controlled by its input voltage and feedback resistors, so that the voltages at its inputs are practically the same.
You keep thinking that an opamp is a transistor. They are not the same.
An opamp is a complete amplifier with a voltage gain of about 200,000, a low-impedance symmetrical push-pull output and a high-impedance differential input. It doesn't need a resistor at its output to a supply voltage, like a transistor needs for its collector current. An opamp with + and - supply voltages usually uses 0V as ground. Its load is also usually connected to ground since its output is symmetrical.
The voltage gain of a complete opamp circuit is determined by the ratio of its external negative feedback resistors. Since the voltage gain of an opamp without feedback is so high, its output will swing to a voltage level that is controlled by its input voltage and feedback resistors, so that the voltages at its inputs are practically the same.
