Post History
That circuit doesn't do what you think it does. Here is the circuit in question: The impedance seen at IN is definitely higher at high frequencies, but that is only due to it being artificially l...
Answer
#2: Post edited
by
Olin Lathrop
·
2020-09-14T18:26:20Z (about 4 years ago)
- That circuit doesn't do what you think it does. Here is the circuit in question:
- 
- The impedance seen at IN is definitely higher at high frequencies, but that is only due to it being artificially low at DC.
- At DC, there is a load of 2R on IN, in parallel with the opamp input impedance. At high frequencies, but within the opamp's range, the voltage on the left side of C1 is the same as IN. That means there is no voltage across R1, which means there is no current thru it. At high frequencies, therefore, the load on IN is only the opamp input impedance.
- Put another way, the resistors add load on IN, which the circuit removes at high frequencies. However, the load due to the opamp input is always there.
- A similar circuit can offset input impedance, but there has to be some positive gain from IN to OUT. Then at high enough frequencies, the left end of C1 is the same signal as IN, but a little larger. This then has the effect of adding a negative resistance load to IN. If that negative resistance exactly offsets the positive resistance of the opamp input, the net effect is apparent infinite input impedance. However, again, that requires a gain above 1 from IN to OUT. Your circuit is a voltage follower, which has a gain just under 1.
- The technique above is very difficult to get right, mostly because the opamp input impedance needs to be known accurately. For example, if the opamp input impedance happens to already be 0, then trying subtract a fixed impedance from it will make things worse.
- Nowadays, opamps with high input impedance are cheap and available. There is no point trying to play games like this, and the cost is minimal compared to the whole system.
You say you are trying to build a field mill. The cost of the motor and battery alone will swamp any reasonable electronics for detecting the field.
- That circuit doesn't do what you think it does. Here is the circuit in question:
- 
- The impedance seen at IN is definitely higher at high frequencies, but that is only due to it being artificially low at DC.
- At DC, there is a load of 2R on IN, in parallel with the opamp input impedance. At high frequencies, but within the opamp's range, the voltage on the left side of C1 is the same as IN. That means there is no voltage across R1, which means there is no current thru it. At high frequencies, therefore, the load on IN is only the opamp input impedance.
- Put another way, the resistors add load on IN, which the circuit removes at high frequencies. However, the load due to the opamp input is always there.
- A similar circuit can offset input impedance, but there has to be some positive gain from IN to OUT. Then at high enough frequencies, the left end of C1 is the same signal as IN, but a little larger. This then has the effect of adding a negative resistance load to IN. If that negative resistance exactly offsets the positive resistance of the opamp input, the net effect is apparent infinite input impedance. However, again, that requires a gain above 1 from IN to OUT. Your circuit is a voltage follower, which has a gain just under 1.
- The technique above is very difficult to get right, mostly because the opamp input impedance needs to be known accurately. For example, if the opamp input impedance happens to already be 0, then trying subtract a fixed impedance from it will make things worse.
- Nowadays, opamps with high input impedance are cheap and available. There is no point trying to play games like this, and the cost is minimal compared to the whole system.
- You say you are trying to build a field mill. The cost of the motor and battery alone will swamp any reasonable electronics for detecting the field.
- <hr>
- <blockquote>Without cap C1, the current through the sine generator (and hence through C2) is the same as previously</blockquote>
- That means that the impedance presented to the sine generator is the same in both cases. Since the voltage is the same, and you say the current is the same, the impedance must also be the same.
- However, that doesn't make much sense. At DC, the load should be R1+R2, which is 200 kΩ. There will be some load due to the opamp input, but that should be small relative to 200 kΩ.
- The opamp is a voltage follower, so OUT should be driven to the sine generator output. For sufficient frequency, that will also appear on the left end of C1. That results in 0 V across R1, which means 0 current thru it, which means R1 presents infinite impedance to its top end.
- This means there should have been a difference in sine generator output current with and without C1.
- What is the amplitude of the sine generator output? That divided by 200 kΩ should be the current out of the sine generator with C1 removed.
- I noticed that the two voltage sources for powering the opamp are labled AC1 and AC2. Those need to be DC sources.
#1: Initial revision
by
Olin Lathrop
·
2020-09-14T12:41:12Z (about 4 years ago)
That circuit doesn't do what you think it does. Here is the circuit in question:  The impedance seen at IN is definitely higher at high frequencies, but that is only due to it being artificially low at DC. At DC, there is a load of 2R on IN, in parallel with the opamp input impedance. At high frequencies, but within the opamp's range, the voltage on the left side of C1 is the same as IN. That means there is no voltage across R1, which means there is no current thru it. At high frequencies, therefore, the load on IN is only the opamp input impedance. Put another way, the resistors add load on IN, which the circuit removes at high frequencies. However, the load due to the opamp input is always there. A similar circuit can offset input impedance, but there has to be some positive gain from IN to OUT. Then at high enough frequencies, the left end of C1 is the same signal as IN, but a little larger. This then has the effect of adding a negative resistance load to IN. If that negative resistance exactly offsets the positive resistance of the opamp input, the net effect is apparent infinite input impedance. However, again, that requires a gain above 1 from IN to OUT. Your circuit is a voltage follower, which has a gain just under 1. The technique above is very difficult to get right, mostly because the opamp input impedance needs to be known accurately. For example, if the opamp input impedance happens to already be 0, then trying subtract a fixed impedance from it will make things worse. Nowadays, opamps with high input impedance are cheap and available. There is no point trying to play games like this, and the cost is minimal compared to the whole system. You say you are trying to build a field mill. The cost of the motor and battery alone will swamp any reasonable electronics for detecting the field.