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Old 23rd September 2009, 07:36 PM   #1
Artie is offline Artie  United States
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Default Help understanding this DC offset circuit?

Hi all. Even though I've been a tech for several decades, this particular DC offset circuit has me perplexed. This is the schematic for the amp section of a Yamaha CR-1040 receiver:

Click the image to open in full size.

Note the placement of the "DC offset" trim-pot, which is wired as a 2-terminal rheostat, within a bizarre T-network, on the differential input circuit. Assume for a moment that the DC offset is actually right on zero. (As measured to ground.) That makes two legs of that T-network at zero volts. Where does the voltage come from to vary the offset? The only leg left is the base of the differential input transistor, and that doesn't make sense to me.

On a side note: If you look at the 4-cap totem-pole thats close to dead center in that schematic, most everything to the left is the pre-driver hybrid IC, and everything to the right is the output IC. Only the caps and resistors with values assigned are discrete external components. Everything else is within one of the IC's.

Can anyone help me to understand how this DC offset circuit works?
Thanks.

Last edited by Artie; 23rd September 2009 at 07:57 PM.
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Old 23rd September 2009, 07:46 PM   #2
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It equalizes the impedances to ground, that bases of the differetial stage see. 33k+10k trimpot decoupled by 2u2 are acting as a constant voltage source of a value og base curret times DC resistance. The passives at the right hand side of that is a typical divider for a feedback path.
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Old 23rd September 2009, 07:46 PM   #3
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well, the resistance of it, and the 33k, is complementary to the 39k resistor from base to ground, of the input. since it carries base current, there is a slight voltage drop, which can adjusted to get the output on 0. in short.
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Old 23rd September 2009, 07:49 PM   #4
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Since they are ICs with no component values shown, one cannot assume that the schematic is other than representative.

However, assuming that the input diff pair is monolithic, the matching will be extremely good, so that effectively the base currents each side will be identical.

Any R with series C can be discounted for DC offset, of course, so the input Rs is 39K on the left, and 33K - 43K on the right.

Does that give you a clue?

This probably would not work with unmatched discrete transistors.
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Old 23rd September 2009, 07:53 PM   #5
Artie is offline Artie  United States
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Quote:
Originally Posted by cliff View Post
However, assuming that the input diff pair is monolithic, the matching will be extremely good, so that effectively the base currents each side will be identical.
Cliff; thanks for pointing that out. I've been looking at my drawing for a couple of weeks, and thats the one "mistake" I didn't catch. Drawing those input tranny's as "monolithic" is my doing. They aren't shown that way in that actual schematic. I'm not sure to what degree that changes things.

I'll fix my drawing ASAP.

Edit: Drawing fixed.

Last edited by Artie; 23rd September 2009 at 07:58 PM.
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Old 23rd September 2009, 09:27 PM   #6
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I think one can say that the circuit WILL work if the i/p pair is very closely matched. This is easily obtained if they are part of a hybrid or mono IC.

Someone can do the math on a difference of Vbe's of, say, 10mV, and Ib's of, say, 10%.

But not me, I'm off to bed!

Last edited by cliffforrest; 23rd September 2009 at 09:27 PM. Reason: c**p spelling
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Old 24th September 2009, 09:05 AM   #7
Artie is offline Artie  United States
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Ok, thanks guys. As soon as I get the output IC replaced, I'll take some measurements, which I'm sure will help to clarify whats going on.
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Old 24th September 2009, 02:46 PM   #8
mannycc is offline mannycc  Singapore
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Hi,
There seems to be missing on the posted diagram. For 1, correct me if i'm wrong but the Vbe multiplier is lacking some resistors and the VAS buffer connection seems dubious?
Regards,
mannycc
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Old 24th September 2009, 02:54 PM   #9
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Quote:
Originally Posted by cliff View Post
Since they are ICs with no component values shown, one cannot assume that the schematic is other than representative.

However, assuming that the input diff pair is monolithic, the matching will be extremely good, so that effectively the base currents each side will be identical.

Any R with series C can be discounted for DC offset, of course, so the input Rs is 39K on the left, and 33K - 43K on the right.

Does that give you a clue?

This probably would not work with unmatched discrete transistors.

it will work as an ofset adjustment ...the thing is that yes the trimer will vary ofset in the output but if we suppose that ltp is not a monolithic ic with matched units inside but they are simple transitors there is a number of things that can go wrong if the ltp is not matched properly ...though the above network will introduce no ofset in the out


i would mod the circuit for a trimer between the two transitors and make sure that the bases of my ltp are biased from each side with precision .
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Old 24th September 2009, 07:39 PM   #10
Artie is offline Artie  United States
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Quote:
Originally Posted by sakis View Post
i would mod the circuit for a trimer between the two transitors and make sure that the bases of my ltp are biased from each side with precision .
I'm not sure I could get to those two points because they're inside a hybrid IC. Moreover, this is a classic old Yamaha design that sounds great, works good. I'm not sure a "mod" is in order. I was just trying to figure out how that DC offset worked.

Btw: Here's the relevant page from the service manual:

Click the image to open in full size.

Just because of the way that Yamaha lays out their manuals, I'm trusting that the internal diagrams of those IC's are fairly accurate.
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