You found that too?
Yeah, I had this Philips CD player and it must have been early 90's and it had them running 15V x 2 and max is supposed to be 18V x 2. Wound them back to 12V and even put some small heatsinks on them, just to be sure - never ever had to do that before. Come to think of it, never again.
Cheers, Joe
Hi,
The DAC´s output current is mirrored into the TZ-node and through the I-V Resistor.
Mirroring means that a 1:1 copy of the input current shows at the TZ-node.
Hence the +-2mA a PCM63 can source into the input will also flow through the I-V resistor.
As Jan noted, the exact value for the I-V resistor would include the internal 3MOhm part of the AD844 in parallal to the external resistor.
Since the internal value is very much larger than the external resistor, it plays a negligable role and may be safely and sanely neglected in the calculations.
+-2mA of p-p signal current translates to 1.4142mA rms (the rms value beeing the typical notation for audio voltage or current levels)
A 2k3 I-V resistor would yield +-2mApp x 2300Ohm = +-4.6Vpp or 1.442mArms x 2300Ohm = 3.253Vrms.
The PCM itself features a internal 1k5 resistor (named Rf).
That would yield +-3Vpp or 2.12Vrms.
Yes, its lower when the AD844 is used in a closed loop configuration like the one shown in Fig.13, on p8 (Diagram Fig.11 refers to the circuit of Fig.13)
No, if the AD844 runs open-loop.
And that is the way its used in our application here.
Both the I-V part as well as the buffer part run open-loop.
Hence the 15Ohm open-loop output impedance for the buffer applies.
As demonstrated before this is a sufficiently high value that You wouldn´t need the 100R compensation resistors when paralleling multiple AD844s.... but they don´t hurt either.
jauu
Calvin
Read #1055 and 1069 again
The DAC´s output current is mirrored into the TZ-node and through the I-V Resistor.
Mirroring means that a 1:1 copy of the input current shows at the TZ-node.
Hence the +-2mA a PCM63 can source into the input will also flow through the I-V resistor.
As Jan noted, the exact value for the I-V resistor would include the internal 3MOhm part of the AD844 in parallal to the external resistor.
Since the internal value is very much larger than the external resistor, it plays a negligable role and may be safely and sanely neglected in the calculations.
+-2mA of p-p signal current translates to 1.4142mA rms (the rms value beeing the typical notation for audio voltage or current levels)
A 2k3 I-V resistor would yield +-2mApp x 2300Ohm = +-4.6Vpp or 1.442mArms x 2300Ohm = 3.253Vrms.
The PCM itself features a internal 1k5 resistor (named Rf).
That would yield +-3Vpp or 2.12Vrms.
Well, yes and no.
Yes, its lower when the AD844 is used in a closed loop configuration like the one shown in Fig.13, on p8 (Diagram Fig.11 refers to the circuit of Fig.13)
No, if the AD844 runs open-loop.
And that is the way its used in our application here.
Both the I-V part as well as the buffer part run open-loop.
Hence the 15Ohm open-loop output impedance for the buffer applies.
As demonstrated before this is a sufficiently high value that You wouldn´t need the 100R compensation resistors when paralleling multiple AD844s.... but they don´t hurt either.
jauu
Calvin
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It's a current conveyor - the current going into pin 2 is coming out from pin 5, unchanged. The voltage at pin 2 sets the DC output offset.
This part is really much easier to use than an opamp!
BTW for paralleling, you should also // pin 5 on the devices as that will lower the offset between the buffered outputs so lowering chance for equalizing currents. You can then use a single load resistor on the // pin 5's and don't need the 100 ohms as Calvin noted.
BTW Anybody here ever experimented with using pin 5 as an input? Opens up all kinds of interesting circuits...
Edit: for the not faint at heart: check out http://linearaudio.nl/sites/linearaudio.net/files/UK-1 2008040241.pdf (fig 7) and http://linearaudio.nl/sites/linearaudio.net/files/UK-2 2008050441.pdf (fig 4) ;-)
Jan
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The smart thing to do is // at pin 5 and use a single load resistor, that makes any output offset even smaller.
Jan
Hi,
Well, yes and no.
Hi Calvin
Been following what you are saying and pretty much in agreement. Peculiar thing this topic about Tz and current send in general. When it comes to figuring out the source impedance, something interesting happens. Lets say we use 1K5 as the example value (the Pedja value I recall). And let's say that we take the Tz output and happen to feed it to a stage that also happens to be 1K5 (rather low, but this is a thought experiment), then we see the output halved.
This is interesting because it now seems as if the the 1K5 values are in series and that effectively this confirms that the 1K5 original value is the source Z. The two 1K5 value acts as if they were in series and a voltage divider. Yet we know they are connected in parallel.
But in reality, using 1K5 on Tz and the fact that it will likely see 10K or more in the following stage/preamp, means that whatever value we start out with, does indeed become the source Z.
The beauty is of course that normally if we had a conventional source Z of 1K5 would be considered high and unlikely to be able to drive another 1K5 as load. But in this case, no problem other than -6dB because of the voltage divider effect.
But the above example shows how weird things become when we look at things from a current model rather than voltage. The above two 1K5 resistors are physically connected in parallel, but as this is current and not voltage, they act electrically as being in series. Does it come much weirder than that?
Cheers, Joe
Joe I don't think that is true, unless I misunderstand what you are doing. The input current into pin 2 is the same as what comes out from pin 5. So if you have a source signal and feed it to pin 2 through 1.5k, and load pin 5 wih 1.5k, you find the exact source signal value on pin 5.
Jan
Hi Jan
I am talking about how it behaves - I was referring to Pedja's circuit and taking it from there. There the 1K5 resistor value fitted does become the source impedance because it behaves like it. George is taking the output from Pin 5 Tz rather than the buffered Pin 6. Any added load seen by Tz will now reduce in level as if you hade a series source Z of 1K5. So the output does look like the two values, the 1K5 resistor and the load, whatever value that might be, are in series.
Of course Pin 2 sets/affects the current at Tz, but this has no place in the above point, the current value x at Tz is not the issue.
Is that a bit clearer?
Cheers, Joe
PS: Working on understanding loudspeaker behaviour from a current perspective/model and coming across some strange things that are the complete opposite and very non-intuitive. Keep in mind that dynamic drivers, loudspeakers, are current devices and that is what make the coil in the gap move, the current through the coil rather than the voltage across it. Note the juxtaposition of through and across. Definitely through. For example, we know that we can now drive a loudspeaker from any source impedance, even in the bass, provided a technique is found to control motional EMF. Imagine a 2nd order Butterworth (sealed) that can be current driven - it is possible.
Hi,
I agree with Jan.
I can´t really follow that thought and I think You err.
The matter follows without doubt 3 basic laws:
- Kirchhoff´s law telling us the sum of all currents flowing into and out of a circuit node equals 0 .... or in other words: what goes in must come out.
- the voltage level of each branch connected to a common node is the same ... or in other words: the node voltage is the node voltage is the node voltage.
- and finally Ohms law stating R=V/I or V=RxI (applying also to complex figures) .... or in other words: current values and resiastances behave inverse proportinal to each other.
If a source feeds to 2mA into a node to which a single 1.5k resistor is connected all of the 2mA flow out of the node and through the resistor, thereby generating a voltage drop of 2mAx1k5 = 3V.
Adding a second resistor of 1.5k and the current splits up in two, each of 1mA value, through each of the resistors, which then drop 1.5V.
If we add a third resistor now of say 750Ohms, the current splits into three, 2x 0.5mA through the 1.5k resistors and 1mA through the 750R following Ohm´s law.
Hence the node voltage becomes 0.75V.
Following Your logic Joe of a somehow series connection voltage divider, results in a false.
I also don´t follow his thoughts about the current drive.
Indeed must a current flow through the voicecoil to create a motion of the coil.
But this current must not be constant over frequency if a linear amplitude response of the speaker is the aim.
Supplied by a constant current over frequency a dynamic driver´s amplitude response plot is modified to a shape following the impedance plot.
Instead a current inversly following the impedance curve (Ohm again) provides for a linear amplitude response.
A voltage source with a linear amplitude response over frequency can supply such currents.
So its rather that the dynamic speaker should be viewed as a voltage controlled device.
Current drive is possible, no question about that, but it´d rather make sense for drivers with a very linear impedance curve and at very low impedance values, like ribbons.
jauu
Calvin
I agree with Jan.
I can´t really follow that thought and I think You err.
The matter follows without doubt 3 basic laws:
- Kirchhoff´s law telling us the sum of all currents flowing into and out of a circuit node equals 0 .... or in other words: what goes in must come out.
- the voltage level of each branch connected to a common node is the same ... or in other words: the node voltage is the node voltage is the node voltage.
- and finally Ohms law stating R=V/I or V=RxI (applying also to complex figures) .... or in other words: current values and resiastances behave inverse proportinal to each other.
If a source feeds to 2mA into a node to which a single 1.5k resistor is connected all of the 2mA flow out of the node and through the resistor, thereby generating a voltage drop of 2mAx1k5 = 3V.
Adding a second resistor of 1.5k and the current splits up in two, each of 1mA value, through each of the resistors, which then drop 1.5V.
If we add a third resistor now of say 750Ohms, the current splits into three, 2x 0.5mA through the 1.5k resistors and 1mA through the 750R following Ohm´s law.
Hence the node voltage becomes 0.75V.
Following Your logic Joe of a somehow series connection voltage divider, results in a false.
I also don´t follow his thoughts about the current drive.
Indeed must a current flow through the voicecoil to create a motion of the coil.
But this current must not be constant over frequency if a linear amplitude response of the speaker is the aim.
Supplied by a constant current over frequency a dynamic driver´s amplitude response plot is modified to a shape following the impedance plot.
Instead a current inversly following the impedance curve (Ohm again) provides for a linear amplitude response.
A voltage source with a linear amplitude response over frequency can supply such currents.
So its rather that the dynamic speaker should be viewed as a voltage controlled device.
Current drive is possible, no question about that, but it´d rather make sense for drivers with a very linear impedance curve and at very low impedance values, like ribbons.
jauu
Calvin
Sorry Joe, that's wrong also, read before quoting. Arty and I have stacked the buffers as well, and taking the output from pin 6!!!
http://www.diyaudio.com/forums/digital-source/227677-using-ad844-i-v-54.html#post4409559
Cheers George
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Big thank you to Calvin and Jan for teaching me so generously and patiently. True experts can make complex things appear so easy. I read your article Jan but its way beyond my grasp but still fascinating to see what can be done.
I have another beginner query. Id like to run the AD844 on just +/- 5v supplies as that's what the PCM63 needs. The AD844 DS says AD844 can be operated from ±4.5 V to ±18 V (page 1) and then has graphs of
TYPICAL PERFORMANCE CHARACTERISTICS (page 6)
at +/-5V (compared to +/- 18V) and it seems to me the effects are
MHz bandwidth falls
Input voltage max falls to <2V
Ouput voltage falls to <4V
Does that mean the only disadvantage running the AD844 at +/-5V is the final output voltage will be limited to about 4Vp-p = 1.4Vrms which is fine for my preamp anyway. Ive probably missed something important?
I have another beginner query. Id like to run the AD844 on just +/- 5v supplies as that's what the PCM63 needs. The AD844 DS says AD844 can be operated from ±4.5 V to ±18 V (page 1) and then has graphs of
TYPICAL PERFORMANCE CHARACTERISTICS (page 6)
at +/-5V (compared to +/- 18V) and it seems to me the effects are
MHz bandwidth falls
Input voltage max falls to <2V
Ouput voltage falls to <4V
Does that mean the only disadvantage running the AD844 at +/-5V is the final output voltage will be limited to about 4Vp-p = 1.4Vrms which is fine for my preamp anyway. Ive probably missed something important?
Guys, it is NOT !!!
But not surprised by the reaction - I did say it was weird, but get your head around the subject and you guys will see it. Think a bit more Einsteinian, OK?
But in reality it is quite simple and yet this is so typical when discussing things when things behave in a way that seems contradictory, then we dismiss it, too quickly.
Remember, George, what I said when addressing ASoN last month, that thinking in 'current mode' is often completely non-intuitve and you often loose your audience even trying to explain it. Morris even said not even to try because he and I have had that discussion - before ASoN meeting. While that topic was about driving loudspeakers, current driving them is very hard to explain to novices when even more experienced people grapple with it.
Same goes here, but this is pretty airtight argument, so read on:
Example 1. We have a speaker, let's say it is = to a 10 Ohm perfect resistor. We feed 10V RMS into it. Then we remove the load and we now see that it has jumped to 11V RMS. The source impedance is now obviously 1 Ohm. That is the output Z of this basic example.
We all know that the source Z and the load is in series - that is inarguable !!!
So we are agreed upon that, right?
Same applies here:
Example 2. Let us make the Tz Resistor 1K. Now add a 10K load and adjust the input signal to 844 to give us 1V RMS. Now take away the 10K load and we will get 1.1VRMS.
Bingo!
Do the basic maths, the output impedance is 1K.
We all know that the source Z and the load is in series - that is inarguable !!!
Hence, from a voltage source perspective, the 1K and the 10K are in series.
Einsteinian conclusion: While the 1K and 10K are physically in parallel, one acts as the output impedance of the load of the other, hence in series, either 10K into 1K or 1K into 10K, they are are reversible depending on use or the way you look at it.
It is an airtight argument.
Cheers, Joe
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Right
Indeed - it just looks like more snake oil to me - no evidence and no reasonable explanation.
Nick, we are not talking about followers here, at least not emitter followers. We are talking about how current send behaves.
Nice PDF and a good summary - but really, other than pointing to Thevenin and nothing I have said in any way disgards it, in fact Thevenin proves the point.
Maybe I have to ask a series of questions.
1. When 1K is fitted to the Tz to ground, it defines the output Z, correct?
Are we all in agreement - other than mOhm value already in parallel - which Calvin said we can ignore and we will, alright?
When I have that answer I will proceed to the next one.
Awaiting....
Au contraire, I explained it well. If something behaves as a voltage divider, then that is what it is.
Is it snake oil to say that the output impedance of an amplifier is in fact in parallel with the load - assuming single-ended operation?
Thevenin tells us that it is - it is a voltage divider.
Nothing I have said here is earth-shattering!
Cheers, Joe
.
"Originally Posted by Joe Rasmussen View Post
George is taking the output from Pin 5 Tz rather than the buffered Pin 6."
Joe!!!!! you really need to read carefully. I'll say one more time.
Myself and Arty are NOT!!! I repeat NOT!!! taking the output from pin 5!!! As of page 52 we taking it from pin 6!!!!! and are also using the stacked internal AD844 buffers with 100ohm series resistors!!!!!!!
And I wouldn't use last months ASON audio meeting as any sort of "example", it was probably the worst meeting we've had in ages.
OVER and OUT Cheers George