This is what Arty and I have done, stacked the buffers as well using 100ohm output isolation resistors it's good, very very good. (thanks must go to Arty for putting us onto this)
And you can use the dc null pins 1 and 8 with a 20kohm trimpot on just one of the AD844's to null out any dc offset. Mine stays at just +- 1mV dc offset!!!! I'm now direct (dc) coupled from D to A convertor output all the way to the speakers, and the bass I'm getting is just stunning, and the transparency is really spooky, speakers just are not in the room at all.
Cheers George
@Calvin
Thanks for writing such a clear explanation of the AD844 and going on to give me excellent advice regards the gain resistor function.
The idea of developing a gain volume control at this early small signal stage really appeals. Do you have any suggestions on the best design?
Just a series 100ohm resistor off each pin 5 and then joined together for the output.
Cheers George
Come on George, you can't expect me to read back a few pages to find out what you actually mean. Yes in hindsight it is clear, but everything is clear in hindsight.
Jan
Come on Jan, the " is way to high to drive anything external direct" part of that statement should have made you of all people tweak, that I meant "output IMPEDANCE"
Cheers George
I am still a Little lost concerning the "output impedance" issue. Is the 3M ohm figure refered to in previous post, and on the data sheet as open loop transresistance, the Rt referred to in the equlvalent Circuit diagram? Is this the same resistor you are paralleling with a 2.7k ohm resistor to get your output voltage from Tz? Is this still a "high output impedance?
The output impedance is not a physical resistor. If you don't use any resistor on Tz it appears as if it has a series resistor on the output of nominal 3 megohms. You can simply short Tz to ground with no bad effects; it's a current output.
The output voltage from Tz, for load resistors appreciably below 3 megohms, is simply Iout * Rt where Rt is the external load resistor on pin 5.
And, when you use an external load resistor Rt, the output impedance becomes Rt//3 megohms.
Jan
Last edited:
Whats the main function of the 100ohm resistors on Pin6 - are they to prevent an individual AD844 over-driving relative to the others?
Would a lower value resistor be a good idea to keep the output impedance nice and low?
Any idea how to estimate the lowest resistor value for the Pin6 output?
Would a lower value resistor be a good idea to keep the output impedance nice and low?
Any idea how to estimate the lowest resistor value for the Pin6 output?
Art is the man to ask this, as he has done more measuring with this part. I'll get him to respond.
Cheers George
Hi,
if I may just chime in ...
The high value of -as it was named in this thread- ´output impedance´ related to the impedance of the TZ-node, that is the output of the I-V part of the AD844.
It´s the parameter ´Open-Loop Transresistance´ specced on page 3 of the AD844´s Datasheet, Rev.F, at typical 3MOhm.
The circuitry between TZ (Pin5) and Output(Pin6) is the voltage buffer part that features a low output impedance.
That is specced at 15Ohm as ´Output Resistance´on page 4 of the DS.
As Jan already wrote creates the (I-V)resistor connected from TZ to gnd almost exclusively the output voltage and it defines the output impedance of this node.
The function of the 100Ohm resistors in the Buffer outputs of the combined AD844 is due to the fact, that without the resistors, each output would ´see´ not only the attached load but also the paralleled other buffer outputs.
As the single outputs are low-ohmic in value and the combined ouputs even lower ohmic, large compensation currents would flow between the outputs due to the ever occuring tolerances between the devices.
The 100Ohm resistances reduce these compensation currents to suffciently low values
Say You´ve paralleled 4x AD844 with 15R output impedance each, no external load attached and one of the AD844´s a bit off, while the three others behave identical.
Then this ´off-AD844´ would see the remaining three AD844 paralleled as load, hence 15R/3=5R.
A output voltage difference due to device tolerances of just 5mV would create a compensation current of 1mA in the ´off-AD844´.
Luckily the output voltage differences remain small and(!) the open-loop resistance of the AD844 is so high already, that the compensation currents remain small enough to be a real problem.
But think about what happened if R were just a few milliohms, as it is typical for feedback OPAmp designs.
Then the compensation current may easily even surpass the OPAmps current drive capabilities, leaving to too less or no current drive reserve for the signal swing --> clipping and heat power loss.
In our case with the 100R included, the output resistance of the single AD844 becomes 115R and it would ´see´ 115R/3=38R3 as paralleled load.
Hence the compensation current would sink from 1mA to 130µA for a 5mV difference.
As the 100R resistors are in series with the buffers outputs they reduce voltage swing slightly as they form a voltage divider with the attached external load.
They should be chosen high enough in value to prevent large compensation currents but low enough to keep the voltage losses low.
jauu
Calvin
if I may just chime in ...
The high value of -as it was named in this thread- ´output impedance´ related to the impedance of the TZ-node, that is the output of the I-V part of the AD844.
It´s the parameter ´Open-Loop Transresistance´ specced on page 3 of the AD844´s Datasheet, Rev.F, at typical 3MOhm.
The circuitry between TZ (Pin5) and Output(Pin6) is the voltage buffer part that features a low output impedance.
That is specced at 15Ohm as ´Output Resistance´on page 4 of the DS.
As Jan already wrote creates the (I-V)resistor connected from TZ to gnd almost exclusively the output voltage and it defines the output impedance of this node.
The function of the 100Ohm resistors in the Buffer outputs of the combined AD844 is due to the fact, that without the resistors, each output would ´see´ not only the attached load but also the paralleled other buffer outputs.
As the single outputs are low-ohmic in value and the combined ouputs even lower ohmic, large compensation currents would flow between the outputs due to the ever occuring tolerances between the devices.
The 100Ohm resistances reduce these compensation currents to suffciently low values
Say You´ve paralleled 4x AD844 with 15R output impedance each, no external load attached and one of the AD844´s a bit off, while the three others behave identical.
Then this ´off-AD844´ would see the remaining three AD844 paralleled as load, hence 15R/3=5R.
A output voltage difference due to device tolerances of just 5mV would create a compensation current of 1mA in the ´off-AD844´.
Luckily the output voltage differences remain small and(!) the open-loop resistance of the AD844 is so high already, that the compensation currents remain small enough to be a real problem.
But think about what happened if R were just a few milliohms, as it is typical for feedback OPAmp designs.
Then the compensation current may easily even surpass the OPAmps current drive capabilities, leaving to too less or no current drive reserve for the signal swing --> clipping and heat power loss.
In our case with the 100R included, the output resistance of the single AD844 becomes 115R and it would ´see´ 115R/3=38R3 as paralleled load.
Hence the compensation current would sink from 1mA to 130µA for a 5mV difference.
As the 100R resistors are in series with the buffers outputs they reduce voltage swing slightly as they form a voltage divider with the attached external load.
They should be chosen high enough in value to prevent large compensation currents but low enough to keep the voltage losses low.
jauu
Calvin
Last edited:
Calvin was on the right track.
It's expected that there's a slight difference between dc offsets and possibly TZ signal currents too, so the 100 ohms prevents potential loading effects.
You could select a lower value but I figured with 2 or more in parallel that the output impedance is sufficiently low enough to drive a long interconnect without introducing measurable HF droop.
Interesting...but how to work out Iout from the TZ stage eg using PCM63 with 0-2ma current out and a TZ resistor= 2.3kohms....
Thanks for another great explanation. Much appreciated.
I wonder if the output impedance from Pin6 is much lower then the 15ohm open loop spec, in the audio band frequency range? On the Data Sheet Pg7, Figure 11. Output Impedance vs. Frequency, it looks like output impedance is only 0.3 Ohm under 100kHz. Does that make the 100ohm resistor even more important?
Thanks Art. Im curious just how much the AD844 will differ in this stacked configuration without a global feedback resistor I think is used in the datasheet to give only +/- 0.03% differential gain error. I wonder if its possible to measure any mV differences between individual Pin 6's of your AD844 stack before the 100ohm resistors? Can a scope do that somehow?
Last edited:
They mustn't oscillate at all. Vcc / Vee shouldn't be more than + - 11V. They are fine under these conditions.
As mentioned before, dynamic impedance of less than 1 ohm can be achieved + the voltage can be amplified at the same time. They are very fast, settle very quickly, provide reasonably low input capacitance. The static impedance is 14 ohms (!)
I'd like to try something else here, but neither AD nor BB are keen on producing better, more suitable OP for I/V purpose, so 20-year old IC will have to do.
Nick