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January 19, 2020, 05:09:43 PM

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Author Topic: Gallien Kruger GT200  (Read 2951 times)

willpirkle

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Re: Gallien Kruger GT200
« Reply #30 on: December 26, 2019, 09:58:06 PM »
The duty cycle shift is apparent. Is it 50% up to clipping point, then as you push the signal up, the duty cycle keeps getting farther away from 50%?. Or, is it always off like this, even when not clipping? Did you solve the bias issue where the voltage-drop was 10x off?

gbono

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Re: Gallien Kruger GT200
« Reply #31 on: December 26, 2019, 11:11:51 PM »
Bias is still not getting anywhere near .05V. Checked the operation of transistors in the bias network and they appear to be working - all resistors and capacitors check out. Maybe I don't understand the bias network operation?

The waveform is very symmetric at low output levels.

g1

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Re: Gallien Kruger GT200
« Reply #32 on: December 29, 2019, 11:09:38 PM »
What are your base and emitter voltages now for Q9 & Q10 with bias set for max. idle current?
In reply #24, Will posted a 210G schematic.  I like the voltages on that one better.  I notice R11 and R12 are a bit higher values in that one.  I'm wondering if you may have to adjust those values to get your idle current up to spec. with these different replacement output transistors.

willpirkle

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Re: Gallien Kruger GT200
« Reply #33 on: December 30, 2019, 11:03:25 AM »
Agree with g1 — the subbed parts are not identical with respect to the built-in base-emitter shunt resistors (which are slightly larger in the new part) and the abs max Ib, which is more than double the original (0.5A vs. 0.2A). You may need to push the bases a bit harder on the new parts. Following up on g1’s question, are your output Vbe’s healthy at idle?

gbono

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Re: Gallien Kruger GT200
« Reply #34 on: December 30, 2019, 04:01:12 PM »
VBE of the output transistors sits at 1.1 V.
So how are output transistors biased in this design? Point "A" is 12V and R8/9 provide bias to Q7? How is Q9 biased?
« Last Edit: December 30, 2019, 04:11:40 PM by gbono »

g1

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Re: Gallien Kruger GT200
« Reply #35 on: December 31, 2019, 03:29:54 PM »
There is more than R8 &9 in that string, it is R8,9,10, & 11.
Adjusting the trimmer R10 adjusts the Q3 current.  Q3 conducts through D1 and R8.
How hard Q3 is conducting changes the voltage at Q7 base, and in turn, Q9.
The bottom side is the mirror image of the top.

As stated earlier, you will probably have to increase the value of R11 and R12 to get the bias right for the sub. output transistors.

gbono

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Re: Gallien Kruger GT200
« Reply #36 on: January 03, 2020, 03:00:26 PM »
Thanks g1....

So Q3, D1 and R9/10/11 form a VBE multiplier?
« Last Edit: January 03, 2020, 03:01:57 PM by gbono »

g1

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Re: Gallien Kruger GT200
« Reply #37 on: January 04, 2020, 02:25:30 PM »
Yes, that's the VBE multiplier.
Because of the odd setup with the output transformer, there is one for the top, and one for the bottom.

But I think I erred where I said to increase the value of R11 & R12.  You will want to decrease them.  That makes it a bit easier though, you can just tack some other resistors across them to decrease their value.