Hi everyone.
Is it possible to make a bridge amplifier coexist with a current sensor, like husband and wife after many years?.
the purpose is to create an intelligent network for feedback and correcting the current when the speaker impedance changes with frequency.
I read up on current sensors but none of them satisfied me completely then one day Mr ELvee (member of this forum) was kindand he pointed out his circuit to me:
https://www.diyaudio.com/community/...ransformer-based-circuit.356221/#post-6247241
the heart of the sensor is the clamp but for a feedback network a small toroid whose enamelled wire winding has the task of detecting is more appropriatethe signal
generated by the speaker current which is amplified by the two transimpedance opamps , everything else is useless. what I like about this system is the absence of contact and no
resistors in series with the speaker.
I subsequently thought of an alternative which you can see in the photograph. to do a quick test and verify its functioning I used 15 (15 is not a fixed number, it varies according to needs) small plastic toroids (therefore magnetic permeability equivalent to air, no ferromagnetic core) arranged in a row to capture the maximum intensity of the magnetic field.
the windings are connected to each other to form a larger one, a single layer of turns for each toroid. you can see it in action in the first attached video, the terminals are connected directly to the oscilloscope probe so no amplification with opamp.I used a 20-20khz sweep,
I can't tell you what the volume was but I'm sure it didn't exceed 50%. the amplitude of the signal seen on the oscilloscope is adjusted on the volt scale.
to create this type of amplifier I thought of the scheme shown in the second attached video. in the simulator, I used a transformer as a current sensor because I couldn't use anything else and instead of the speaker I used a variable trimmer. apparently if the circuit is applied in reality, it doesn't work and this is a real shame.
this was stated by people accustomed to chewing electronics every day. Mr ELvee also advised me against going down this road, so I'll stop here.
however it seems right to me to share what I have done and the information collected, in this forum there are many brilliant minds who can express an appreciated opinion.
Is it possible to make a bridge amplifier coexist with a current sensor, like husband and wife after many years?.
the purpose is to create an intelligent network for feedback and correcting the current when the speaker impedance changes with frequency.
I read up on current sensors but none of them satisfied me completely then one day Mr ELvee (member of this forum) was kindand he pointed out his circuit to me:
https://www.diyaudio.com/community/...ransformer-based-circuit.356221/#post-6247241
the heart of the sensor is the clamp but for a feedback network a small toroid whose enamelled wire winding has the task of detecting is more appropriatethe signal
generated by the speaker current which is amplified by the two transimpedance opamps , everything else is useless. what I like about this system is the absence of contact and no
resistors in series with the speaker.
I subsequently thought of an alternative which you can see in the photograph. to do a quick test and verify its functioning I used 15 (15 is not a fixed number, it varies according to needs) small plastic toroids (therefore magnetic permeability equivalent to air, no ferromagnetic core) arranged in a row to capture the maximum intensity of the magnetic field.
the windings are connected to each other to form a larger one, a single layer of turns for each toroid. you can see it in action in the first attached video, the terminals are connected directly to the oscilloscope probe so no amplification with opamp.I used a 20-20khz sweep,
I can't tell you what the volume was but I'm sure it didn't exceed 50%. the amplitude of the signal seen on the oscilloscope is adjusted on the volt scale.
to create this type of amplifier I thought of the scheme shown in the second attached video. in the simulator, I used a transformer as a current sensor because I couldn't use anything else and instead of the speaker I used a variable trimmer. apparently if the circuit is applied in reality, it doesn't work and this is a real shame.
this was stated by people accustomed to chewing electronics every day. Mr ELvee also advised me against going down this road, so I'll stop here.
however it seems right to me to share what I have done and the information collected, in this forum there are many brilliant minds who can express an appreciated opinion.
Attachments
You can make an op-amp like amplifier circuit that provides a programable output impedance, without resorting to a current transformer nor a return path sensing resistors, therefore useful for bridging amplifiers. But the math is tricky, so I advise using spice to check the results. The circuit is similar to a "Howland current source", but with more negative feedback than positive feedback so that the impedance is not "infinite" ie not a current source, but is a multiple of the sensing resistor.
https://www.ti.com/lit/an/snoa474a/snoa474a.pdf
https://en.wikipedia.org/wiki/Current_transformer
https://www.ti.com/lit/eb/slyy154b/slyy154b.pdf?ts=1697806873268
The behavior of loudspeakers is complicated but there is a basic relationship between the applied voltage and cone movement, so especially for lower frequencies, a voltage drive is the accepted ideal, but of course, some people will not agree.
https://www.ti.com/lit/an/snoa474a/snoa474a.pdf
https://en.wikipedia.org/wiki/Current_transformer
https://www.ti.com/lit/eb/slyy154b/slyy154b.pdf?ts=1697806873268
The behavior of loudspeakers is complicated but there is a basic relationship between the applied voltage and cone movement, so especially for lower frequencies, a voltage drive is the accepted ideal, but of course, some people will not agree.
What you have implemented is a Rogowski sensor variant. They do work, and are useful for metrology, but they have a low output and they linearly depend on frequency: at the low-end of the audio spectrum, the output is tiny, meaning the conditioning circuit has to be an integrator having a very high gain.
This is problematic, especially at low frequencies. At higher frequencies, this means difficult tradeoffs, corrections, etc.
Including such a thing inside a FB loop is fraught with difficulties; every problem is solvable, look at the MFB system made by Philips for example: it is an even harder challenge, but it has been solved (sort of).
The correction was relatively weak and limited to low frequencies.
In your case, the solution proposed by Steve, or the one I discussed with you in PM is much more realistic.
Here it is, outlined graphically:
U2 is a floating amplifier boosting the shunt voltage, and U4 is a differential or instrumentation amplifier delivering a GND-referenced image of the output current.
Such an arrangement simplifies the offset and accuracy requirements of the differential amplifier, and the bandwidth and phaseshift are compatible with an inclusion into a feedback loop
This is problematic, especially at low frequencies. At higher frequencies, this means difficult tradeoffs, corrections, etc.
Including such a thing inside a FB loop is fraught with difficulties; every problem is solvable, look at the MFB system made by Philips for example: it is an even harder challenge, but it has been solved (sort of).
The correction was relatively weak and limited to low frequencies.
In your case, the solution proposed by Steve, or the one I discussed with you in PM is much more realistic.
Here it is, outlined graphically:
U2 is a floating amplifier boosting the shunt voltage, and U4 is a differential or instrumentation amplifier delivering a GND-referenced image of the output current.
Such an arrangement simplifies the offset and accuracy requirements of the differential amplifier, and the bandwidth and phaseshift are compatible with an inclusion into a feedback loop
Look at Lem hall effect current sensors.
Another way for a fast high bandwith galvanic isolation current sense is using a HCNR20x plus a bunch of external parts.
Broadcom makes the hcpl7800 series of isolation amplifiers, but these have a bandwith sub 100Khz
The lem sensors are generaly very linear with good BW, their output is differential
Another way for a fast high bandwith galvanic isolation current sense is using a HCNR20x plus a bunch of external parts.
Broadcom makes the hcpl7800 series of isolation amplifiers, but these have a bandwith sub 100Khz
The lem sensors are generaly very linear with good BW, their output is differential
Years ago, I thought I had the math down to a formula, but I seem to have lost it and today I just hack it in spice. This (attached) is a bridged amp with an output impedance of about 8 Ohms. The bridge complicates it because each side is actually driving half the impedance, ie 4 Ohms.
Attachments
Methods to increase or control the output impedance are useful for current-drive or intermediate drive, but if you want a more complex behaviour, you need an image of the current, to be used as an input of whatever servo system you want to use.
I don't know what Arivel wants to achieve exactly
I don't know what Arivel wants to achieve exactly
Then, using Steve's solution is the simplest option, but a problem remains: how to interlink the two sides so that each one takes a fair share of the voltage.
Without an explicit balancing mechanism, one side will always be a little stronger, and its output will remain stuck at the + or - rail
Without an explicit balancing mechanism, one side will always be a little stronger, and its output will remain stuck at the + or - rail
With a little help from Wolfram Alpha, I think I have the non-inverting circuit math under control.
In post #5, I simply slaved the inverting side as a voltage inverter which doubles the voltage and therefor doubles the artificial impedance. The non-inverting side will clip a bit sooner under load than the slave. I plan to solve the same formulas for an inverting version, which may be a better way to bridge.
I pasted the CSV spreadsheet data for this topology as text in the schematic.
If you want an ~infinite impedance, you can set the Z value to some high number, but I'd be worried about instabilities and any voltage by an infinite impedance is zero current, so it will probably do strange things. Wolfram had a long list of conditions where it doesn't work.
In post #5, I simply slaved the inverting side as a voltage inverter which doubles the voltage and therefor doubles the artificial impedance. The non-inverting side will clip a bit sooner under load than the slave. I plan to solve the same formulas for an inverting version, which may be a better way to bridge.
I pasted the CSV spreadsheet data for this topology as text in the schematic.
If you want an ~infinite impedance, you can set the Z value to some high number, but I'd be worried about instabilities and any voltage by an infinite impedance is zero current, so it will probably do strange things. Wolfram had a long list of conditions where it doesn't work.
Attachments
I tried to make measurements with the following tones: 50, 100, 150 Hz. current approximately 10 mA with multimeter.
If I haven't done something wrong, what the oscilloscope (probe x1) shows me is not a tiny signal.
The theory is merciless: the output is strictly proportional to the frequency, and it is shifted by 90°.
This means that the output at 20Hz is 1,000x times smaller than for 20kHz.
This requires a huge dynamic range from the conditioning circuit, with worries about the S/N ratio, the flatness (because parasitics inevitably play a role) and unwanted phaseshifts.
It is already complicated enough when the circuit is used open-loop, for metrology but the difficulties become insurmountable if you try to include it in a closed-loop.
If in your experiments you see levels larger than those predicted by the theory, it means that there are parasitic couplings of some sort
This means that the output at 20Hz is 1,000x times smaller than for 20kHz.
This requires a huge dynamic range from the conditioning circuit, with worries about the S/N ratio, the flatness (because parasitics inevitably play a role) and unwanted phaseshifts.
It is already complicated enough when the circuit is used open-loop, for metrology but the difficulties become insurmountable if you try to include it in a closed-loop.
If in your experiments you see levels larger than those predicted by the theory, it means that there are parasitic couplings of some sort
After about 3 attempts, given my limited math skills, I am victorious. The attached circuit is a bridged pair with a total output impedance of 8 Ohms, ie 4 Ohms each side and a gain of x23.5 (make it what you will now that I have the spread sheet working). I put the spread sheet CSV data on the schematic but again it is:
A,R,Z,N,P,Na,Pa,Nb,Pb
23.5,0.39,4,=1+(A2-1)B2/C2,=((A2-1)B2+C2)/(C2-B2),10000,10000,=F2(D2-1),=G2(E2-1)
,,,,,,,,
inverting,,,=1+A2*B2/C2,=(A2*B2+C2)/(C2-B2),10000,10000,=F4*(D4-1),=G4*(E4-1)
Note that you cannot stray far from the calculated values because the difference between the feedback ratios "P" and "N" is critical. So, I simulate using close values that are relatively practical. For simulation, the loaded voltage should be half the open circuit voltage when the load value matches. This means that the circuit may clip unloaded while not clip when loaded. An infinite impedance, ie a "Howland current source" will clip (or worse) when unloaded because any current into an open is an infinite voltage.
Also note that the input buffer is not optional because the input source resistance must not add to the feedback networks.
Oops: There is a minor error with resistor designators Na1 and Pa1 but both are 10K here, so it doesn't matter.
A,R,Z,N,P,Na,Pa,Nb,Pb
23.5,0.39,4,=1+(A2-1)B2/C2,=((A2-1)B2+C2)/(C2-B2),10000,10000,=F2(D2-1),=G2(E2-1)
,,,,,,,,
inverting,,,=1+A2*B2/C2,=(A2*B2+C2)/(C2-B2),10000,10000,=F4*(D4-1),=G4*(E4-1)
Note that you cannot stray far from the calculated values because the difference between the feedback ratios "P" and "N" is critical. So, I simulate using close values that are relatively practical. For simulation, the loaded voltage should be half the open circuit voltage when the load value matches. This means that the circuit may clip unloaded while not clip when loaded. An infinite impedance, ie a "Howland current source" will clip (or worse) when unloaded because any current into an open is an infinite voltage.
Also note that the input buffer is not optional because the input source resistance must not add to the feedback networks.
Oops: There is a minor error with resistor designators Na1 and Pa1 but both are 10K here, so it doesn't matter.
Attachments
Last edited:
It occurs to me that if you actually want a "Howland current source", the gain A becomes ~infinite but it's actually simpler. For a current source, the feedback ratio's N and P are the same and the transconductance is N/Ri.
PS. Bridging makes the problem of two current sources in series, ie doesn't work. So, I suggest you just invert the voltage as I did in the first suggestion of post #5.
PS. Bridging makes the problem of two current sources in series, ie doesn't work. So, I suggest you just invert the voltage as I did in the first suggestion of post #5.
Attachments
Last edited:
Well done Steve.
I was just waiting for someone to talk about this:
"two current sources in series".
do we want to talk about this problem? of the causes and implications of this system? it is useful to know more even for those who know little about these things, it is not for nothing that we are in a forum which by definition means "discussion".
Thank you Steve for your kind efforts.
where I come from there is a popular saying: unity is strength. it means that if many brains come together to solve a problem, there is a high probability of success.
is there anyone else who wants to join and take up this challenge?.
I was just waiting for someone to talk about this:
"two current sources in series".
do we want to talk about this problem? of the causes and implications of this system? it is useful to know more even for those who know little about these things, it is not for nothing that we are in a forum which by definition means "discussion".
Thank you Steve for your kind efforts.
where I come from there is a popular saying: unity is strength. it means that if many brains come together to solve a problem, there is a high probability of success.
is there anyone else who wants to join and take up this challenge?.
Oops: If you stick to the feedback ratio definition I used before, the transconductance of a Howland current source is actually (N-1)/Ri, ie ((10k+10K)/10K -1)/1. So the output of post #13 is 1V*(2-1)/1 = 1 Amp, ie 10 V across 10 Ohms. Note that the feedback networks N and P are the same for inverting and non-inverting Howland current sources. Way simple!
It occurs to me that if Howland can publish a paper about his current source, I could/should do the same for the much more complicated artificial impedance circuits in post #12. It would not be the first important bit of technology that I have released without any patents or copy-right. I don't think the idea is completely new. But if you use it in any publication, I appreciate you mentioning where it came from.
I just looked that up:
Copyright for your IP remains with you in any case, as per forum rules, see "User Contribution to our website."
HI.
for my personal whim I wanted to make other measurements on the sensor.
this time the user is a 10 kohm resistor and as a source I used a function generator with an output of 5Vpp.
I measured the output voltage for some frequencies and the results are as follows:
50Hz - 4.4Vpp
100Hz - 4.5Vpp
150Hz - 4.5Vpp
200Hz - 4.5Vpp
then I measured the phase shift of the current.
begins to become visible at 1Mhz and reaches 90° at 15Mhz.
for my personal whim I wanted to make other measurements on the sensor.
this time the user is a 10 kohm resistor and as a source I used a function generator with an output of 5Vpp.
I measured the output voltage for some frequencies and the results are as follows:
50Hz - 4.4Vpp
100Hz - 4.5Vpp
150Hz - 4.5Vpp
200Hz - 4.5Vpp
then I measured the phase shift of the current.
begins to become visible at 1Mhz and reaches 90° at 15Mhz.