Krell KSA 100 schematic, resistor value question

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Tube_Dude

Tube_Dude

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Joined 2002
millwood said:


I thought it was settled on that other thread that for such a topology, you cannot calculate VAS and LTP current. Assuming that the current in the LTP legs is evenly split is falwed.

hopefully, the time and effort there were not wasted.
Click to expand...

The time and effort were never wasted in good company!!! ;)

Were your simulation is flawed is:

The bases of the input LTP are not grounded in the real circuits...(i prefer real circuits that simulated ones)...and in the case of the Krell they are connected to ground by a 22 KOhm resistor.

This is reflected in the emmiters of the LTP as as 220 Ohms + the 220 Ohms emmiters resistor ..become 440 Ohms...

So any small diference in the Vbe of the LTP become with no consequence!!!..and the current will be splited almost evenly in the LTP.
:)
 
M

millwood

Disabled Account
Joined 2003
Tube_Dude said:


The time and effort were never wasted in good company!!! ;)

Were your simulation is flawed is:

The bases of the input LTP are not grounded in the real circuits...(i prefer real circuits that simulated ones)...and in the case of the Krell they are connected to ground by a 22 KOhm resistor.

This is reflected in the emmiters of the LTP as as 220 Ohms + the 220 Ohms emmiters resistor ..become 440 Ohms...

So any small diference in the Vbe of the LTP become with no consequence!!!..and the current will be splited almost evenly in the LTP.
:)
Click to expand...

someone had pointed out in the other thread that none of the above will change the nature of the circuitry: it is not determinate, "almost evenly" notwithstanding, :)
 
M

millwood

Disabled Account
Joined 2003
it is not determinate because you cannot calculate it without assuming the current going through at least one leg of the LTP, or vis versa.

that's why simulated results are so widely apart from each other. I was hoping that service manual will have specified the current so we know for sure what it is.
 
J

jcx

Member
Joined 2003
the point of the uastable VAS current thread applies with current mirrors in the complementary diff pair collectors - the high output impedance provides a very large voltage gain at the input to the vas which results in a unstable voltage into the vas and indeterminate vas bais current due to the lack of common mode feedback, differential feedback still holds the current division ratio in the diff pairs close to 1:1


in this circuit the 1KOhm resistive collector load provides a very definite and calcuable voltage into the vas and hence a predictable vas bias current (counting on my thumbs i get ~6 mA)

with a dc bolcking feedback cap for unity dc gain, dc emitter degeneration and offset trim i think the assumption of equal current division in the diff pairs is pretty good


the diff pair current sources could be greatly improved with a couple of bootstrap resistors, the diff pair bypass and feedback dc blocking caps should all be nonpolar/bipolar electrolytics with as high a voltage rating as you are willing to fit
 
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