Current Mode Amplifiers, how do they work?
Oct 25, 2014 at 4:56 AM Post #16 of 28
oh forgot to add, the high gain (RL/R = 100M/10) would effectively be negated enough by the current divider
 

 
Oct 25, 2014 at 11:18 AM Post #17 of 28
I see the difference, but still wondering does this difference make damping factor not an issue for the current amp? How?


It's not an issue when the driver has sufficient mechanical damping, like orthos do.

If you look at frequency response plots in this "Transient and frequency response of the moving-coil loudspeaker current-drive and voltage-drive method" paper mentioned above (which deals with full-size traditional dynamic drivers), you will see that current drive causes a hump around the driver's resonant frequency. The improvements they obtained thanks to current drive happened either far away from the resonant frequency or with use of external driver motion sensors which allowed the amplifier to reduce driving current in response to speaker resonance.

Using current drive with traditional dynamic speakers without any compensating circuity does affect their frequency response.
 
Oct 25, 2014 at 11:42 AM Post #18 of 28
  Looks like the output impedance is R thevnin so RL || output impedance.  Rsmall || Rbig, which is Rsmall.  So basically, the output impedance is RL from what I'm seeing.  

Found something related to this.  Firstwatt paper explains in details.  http://www.firstwatt.com/pdf/art_cs_amps.pdf
 
 
"The adjustments for damping at the low end and the mid/high peaking are easily
accomplished using R, RC or RLC (resistor/inductor/capacitor) networks wired in parallel with
the output of the amp. You occasionally see this sort of equalization with full-range highefficiency
drivers using voltage-source amplifiers, but there the elements are paralleled and
then placed in series with the driver. With a current source, the equivalent circuit becomes
the elements placed in series and then paralleled with the driver."
 
Bakoon has the RL which is parallel with the output of the Mega Ohm output impedace circuit which affectively reduces the output impedance.
 
2-1_satri_circuit.jpeg
 
 
Oct 25, 2014 at 12:54 PM Post #19 of 28
Measurements of the HD600 into the Bakoon indicate that Current mode drive will change the frequency response. This is only suitable for Orthos. When they claim DF will not be issue, it means that the Damping Factor will not affect the amp's non-linear distortion(THD+N) which is true. There is a nice lengthy and digestable thread on the Bakoon HPA-01 and how it works on another site, just google.
 
Oct 25, 2014 at 1:00 PM Post #20 of 28
  Measurements of the HD600 into the Bakoon indicate that Current mode drive will change the frequency response. This is only suitable for Orthos. When they claim DF will not be issue, it means that the Damping Factor will not affect the amp's non-linear distortion(THD+N) which is true. There is a nice lengthy and digestable thread on the Bakoon HPA-01 and how it works on another site, just google.

Do you have those or can you link them?
 
Oct 25, 2014 at 5:22 PM Post #21 of 28
  Do you have those or can you link them?

 
They are on a site - known for its CSD plots - that cannot be named here, but it was created by those who started this thread, if that helps finding out what site it is. Check the forum there, the Bakoon HPA-01 thread is a recent one and should be easy to find.
 
Oct 25, 2014 at 5:28 PM Post #22 of 28
  This is only suitable for Orthos. When they claim DF will not be issue, it means that the Damping Factor will not affect the amp's non-linear distortion(THD+N) which is true.

 
It could affect the distortion of the HD600, though, as I have shown above. Unfortunately, those measurements at the site that cannot be named used the HD600 only for frequency response testing (which produced the results expected from the impedance vs. frequency plots for the HD600), but the distortion tests were performed with resistors instead.
 
Oct 26, 2014 at 1:59 AM Post #23 of 28
   
It could affect the distortion of the HD600, though, as I have shown above. Unfortunately, those measurements at the site that cannot be named used the HD600 only for frequency response testing (which produced the results expected from the impedance vs. frequency plots for the HD600), but the distortion tests were performed with resistors instead.

I looked at the measurements, and it looks similar to how the HD800 reacted to added significant resistance at the output with boosted bass around the resonance.  I'd expect the the distortion to rise around the resonance with the reactive driver as load than the resistor they have used.  
 
Regarding the Bakoon, I don't know how the voltage and current mode are done, but given they have specs for input impedance for voltage mode and separate outputs, probably has separate circuits.  I guess having only current mode would not be that much useful beyond orthos. Looking at their Bakoon diagram for the current mode circuit, it's really just a current conversion in between and back to voltage again with RL setting the output voltage.  If RL is volume control it would increase the output impedance as you turned the volume up.  
 
Oct 26, 2014 at 5:08 PM Post #24 of 28
Efficiency is 106dB/mW.  It's quite efficient.  
That's the article that confused me.  It's not a good explanation.  
I don't think he understands the workings, but regurgitated what he was told.  
Headfonia?  I seriously doubt it's anything, but hearsay.  


project86 is a nice guy, but his understandings of the workings of that particular amp are rather poor, to put it nicely.

from the satri circuit link


if their circuit has an output impedance of 100 Mohms (RL is what they call it I believe) then another load of nominal impedance say 32 ohms is will not affect the output impedance. 100M +32 = 100M

http://en.wikipedia.org/wiki/Current_divider


specifically check the loading effect section

I might be wrong but that is how I understand it.


In a true current mode amplifier, the output impedance of the amp is IN PARALLEL with the load impedance, which in our case is the headphone.
 
Oct 26, 2014 at 5:35 PM Post #26 of 28
In a true current mode amplifier, the output impedance of the amp is IN PARALLEL with the load impedance, which in our case is the headphone.

Are you talking about current source?  My understanding of current source is ideally has infinite output impedance which is at the output of the source.  And then you have the Norton resistance that closes the circuit that give the voltage out.  Voltage out is the product of Norton Current and Resistance.  Then Norton resistance is the like RL at the output of the SATRI-IC.  Which is assentially outputting voltage like a voltage source.  You can make a Thevenin equilvilent that gives you the output impedance.  In the case of the Bakoon it's the RL.   Norton resistance parallels the ideal infinate output impedance of the current source.  The headphone is connected parallel to the RL which iike being parallel to a Norton Resistor.  So it can be modeled equivelant to a voltage souce like the Thevenin circuit with the output impedance in series to the headphone load. 
Thevenin_to_Norton2.PNG
 
 
Oct 26, 2014 at 5:39 PM Post #27 of 28
Are you talking about current source?  My understanding of current source is that it ideally has infinite output impedance which is at the output of the source.  And then you have the Norton resistance that closes the circuit that give the voltage out.  Voltage out is the product of Norton Current and Resistance.  Then Norton resistance is the like RL at the output of the SATRI-IC.  Which is assentially outputting voltage like a voltage source.  You can make a Thevenin equilvilent that gives you the output impedance.  In the case of the Bakoon it's the RL.   Norton resistance parallels ideal infinate output impedance of the current source.  The headphone is connected parallel to the RL which iike being parallel to a Norton Resistor.  So it can be modeled equivelant to a voltage souce like the Thevenin circuit with the output impedance in series to the headphone load. 
Thevenin_to_Norton2.PNG


Yes.
A conventional voltage amplifier is a voltage source.
You want the output impedance to be as low as possible so virtually all the output voltage is applied to the load.

A current amplifier would be a current source, the output impedance would be extremely high, hence virtually all the output current would be applied to the load.
 
Apr 2, 2022 at 1:56 PM Post #28 of 28
Are current mode amplifiers class A? Or A/B? Or something else entirely?

Also, a related question - are THX and/or NFCA Topping amps a subtype of class D amps? Or, again their own category?
 

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