@Mark Tillotson this is true at higher frequencies and large signal. Using the formula 2 * pie * f * V, a 20kHz sine at 40V peak, I get about 5V/us.
Most musical content is lower frequency and frequencies say 10kHz and above are pretty small. I think for the most part slew current requirement are small.
I was thinking more of the Cob in an uncompensated circuit.
I think all this unnecessarily complicates Stuat's question, which is where is the current going.
Most musical content is lower frequency and frequencies say 10kHz and above are pretty small. I think for the most part slew current requirement are small.
I was thinking more of the Cob in an uncompensated circuit.
I think all this unnecessarily complicates Stuat's question, which is where is the current going.
Thanks for trying to explain it. I guess I'm still a little confused as to how it's actually operating. I understand the differential amplifier concept and the balance of the currents etc. But what don't make sense to me is if the is 1mA shared between each collector and one of these collectors is connected to the input of the current mirror. Then if the input of the current mirror sees 0.4mA won't it mirror the current to the output of the current mirror?
The explanation lies in the fact that although the current mirror has equal currents in the two branches, it does not mean that they have to sum to 1mA, and does not mean that the two collector currents are equal.
For example the current mirror sourcing 0.45 + 0.45 mA in each branch and one collector drawing 0.45 and the other collector drawing 0.55mA, total 1mA. So 0.1mA coming from the load.
Jan
My numbers add up just fine.
I intentionally (as clearly drawn/shown in the simplified diagram) omitted the parallel signal ground return path through the opposing channel "hum-blocking" resistor simply to convey in a first order analysis the wrongheadedness of separating the so-called "dirty" speaker ground return from the "clean" signal ground with said resistor and so as to not confuse/compound the issue further with:
I see nothing here to argue about, unless someone insists on being deliberately obtuse.
Sorry to jump into a discussion without really being able to find the preceding post with a diagram, so I may miss hitting the plank, and I don't necessarily disagree with you, we might even be in violent agreement. So here it goes.
When I hear the notion of 'dirty speaker ground' that has to be isolated, it gives me shivers in tropical temperatures. The reason is that the error signal to be used by feedback should be as close as possible exactly the (voltage difference between output and ground) minus ((voltage difference between input and ground) times (a constant)). Any differences between the two grounds mentioned here will introduce an error that cannot be corrected by feedback because it occurs outside the loop.
So, speaker ground an input ground should be tied together as closely as possible.
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Post 9307: to miss hitting the plank-> to find a mare´s nest
Edmond - good to see you back.
Assume that the tail current is 2 mA and that the current mirror loading the input LTP is perfect. If the amplifier is slew rate limiting, virtually all of the tail current, here 2 mA, will flow into the Miller compensation capacitor if the current gain of the VAS is high, as with a 2T VAS. If Cm is 30 pF, slew rate will be approximately I/Cm = 2mA/30pF = 66 V/us.
This assumes that there is no other slew-rate limiting mechanism at work, such as the VAS having inadequate current to support that slew rate at its collector node. If the VAS is single-ended with a current source load of 10 mA, and total capacitance at the VAS collector node is 30pF, and output stage current gain is high (as with a Triple), that secondary limit on slew rate would be on the order of 10mA/30pF = 333 V/us.
Cheers,
Bob
I think so. And the ground of the signal input socket should not be isolated, assuring the continuity of the shielding between the input cable and the chassis.