EQ boost and cut circuit - which end of the pot is which?

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mr coffee

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Aug 1, 2005
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Location
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Hi all,

Here is a discrete transistor circuit to do boost and cut of a (switchable) single band from an old circuit (Acoustic 360 preamp).

<img src="http://216.77.188.54/coDataImages/p/Groups/207/207065/folders/247331/1948945Variampcircuit.gif" width="395" height="240">

Can anyone help me understand which end of the pot is boost and which end is cut? I am not that good at discrete circuitry and I don't understand the concepts around feedback from the collector to the emitter.

Thanks all!

mr coffee
 
you used the wrong syntax when you tried to insert the image in your post. here it is

1948945Variampcircuit.gif


ill take a crack at it. im sure one of the elder-statesmen will correct me if im wrong.

the end of the pot connected to the 680 ohm resistor is boost. the inductor and cap in series has a low impedance at a specific frequency. when the pot wiper is up at the collector end this shunts that frequency to ground (cut) and when its at the emitter end it bypasses the emitter feedback at that frequency (boost).

when the selector switch connects the inductor to ground (with no cap) it acts as a low freq shelf boost-cut.

the 680 ohm resistor limits the amount and 'q' of the boost.
 
Solder_city,

Thanks for your input, man!

I figured that bypassing the emitter resistor to ground would boost at the resonant frequency, but when I got to thinking about the other end, except at the exteme wiper setting, I got to thinking negative feedback to the emitter and got it all screwed up in my head (i.e., if there was less negative feedback at the resonant frequency where it was shunted to ground, wouldn't that BOOST that frequency band - and then I've got BOTH ends boosting which doesn't make any sense).

I agree solder_city's analysis makes the most sense at the extreme rotation points, which means it's bound to be correct, right?

Anyone here who can explain how this can cut other than when it is at the extreme of the 50K pot where it shunts the collector circuit impedance at the resonant frequency?
i.e., How does it cut when the pot wiper is set 5K plus away from the collector end? Or does it?
Or does it NOT cut very well (or not very symmetrically) like the more modern circuit where the pot bridges the input of a differential amp?

Thanks for all input here guys!!
 
> explain the systax for posting an image?

Naked HTML (with angle brackets) is not allowed in posts.

UBB syntax is simplified HTML using square brackets. IMG SRC= is simplified to a pair of tags. Frills like "width" are not allowed.

Code:
[IMG]http://216.77.188.54/coDataImages/p/Groups/207/207065/folders/247331/1948945Variampcircuit.gif[/IMG]

Note: the syntax can be awful fussy about spaces, including automatic line-wraps.
 
My two cents:

There is positive feedback from colector to emiter. Remember
common base amp? Signal input at emiter, otput at colector, and its
noninverting. So if u take signal from colector to emiter you will have
positive feedback.

Now, negative feedback reduces gain. Positive feedback is doing the
oposite. You have two NFBs in circuit: 1Mohm from colector to base
and total AC emiter resistance for emiter degenaration. And there is
positive feedback trough 50k pot. When you push the wiper to colector
you shunt part of the signal in positive FB path to ground, so you reduce
positive FB around resonant frequency => cutting .
At certan point on wiper path you have same impedance toward colector
and emiter. Transistor behaves like phase splitter regarding L-C network
with outputs of phase splitter conected to one point trough same
impedances. So result is that wiper (and thus L-C network) is at AC
ground, making L-C efectively out of circuit. EQ is flat.
When you push wiper to emiter, NFB starts to reduce cus you are
reducing emiter impedance around resonant frequency.
Less NFB=>boosting.
hope this makes sense

cheerz
urosh
 
> explain how this can cut other than when it is at the extreme of the 50K pot where it shunts the collector circuit impedance at the resonant frequency?
i.e., How does it cut when the pot wiper is set 5K plus away from the collector end?


Well, work it out.

Assume the collector impedance is infinite, the coil-cap impedance at resonance is zero.

The collector resistor is 5K. When the coil-cap shunts it, output is zero. Not really, because the coil has resistance. But it can dip pretty good.

When the pot is 5K away, output falls 6dB just due to shunting. That's a non-trivial dip.

You are right, in that the high pot value gives little effect until you get to the end. But the spreads-out the subtle artistic settings, while still allowing unsubtle slam at the ends.

There is also dip action happening at the emitter: the 2.2K is bypassed with 25K, 45K, 50K as you go to dip. That adds another dB at the extreme.

> like the more modern circuit where the pot bridges the input of a differential amp?

Both are very old. Swinging inputs, swinging outputs... either one can work. It is mostly about what fits the rest of your plan. With a few naked transistors, swinging outputs is simpler.
 

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