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Old 25th October 2010, 10:52 PM   #1
dmx2020 is offline dmx2020  New Zealand
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Default Calculating gain of loudness control

Hi,

As the title says i need to calculate the gain of my loudness circuit that im about to build. (circuit diag attached)

Itís borrowed from a JVC amp schematic and the specs say it gives a gain of +6dB at 100Hz and +4dB at 10kHz.

I need to change the circuit, so itíll give say +10dB at 100/10k. How would change the resistors/caps (22k and 0.033u) to do this.

And what if I wanted to change the frequencies, say get +10dB at 60Hz instead of 100Hz.

Much appreciate some help on this. Thanks.
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Old 25th October 2010, 10:58 PM   #2
dmx2020 is offline dmx2020  New Zealand
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Old 25th October 2010, 11:35 PM   #3
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What is the 4-terminal device in the diagram that looks like a pot? What schematic does this little section come from?

If it's a pot then the circuit has no gain...

w
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Old 25th October 2010, 11:51 PM   #4
dmx2020 is offline dmx2020  New Zealand
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Hi, the 4 terminal device is a volume pot (used to control the volume of a amp).

It is from a amp schematic, IN is the input from source preamp (tape/tuner/phone) and OUT is output to amp (as in power amplifier used to amplify music).Thanks

Last edited by dmx2020; 26th October 2010 at 12:01 AM.
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Old 26th October 2010, 12:01 AM   #5
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Right... resistive divider formula :

Vo = (R1*R2)/(R1+R2)*Vin,

thus by superimposition -> Vo=(Vin*R2)/(R1+R2)
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Old 26th October 2010, 12:03 AM   #6
godfrey is offline godfrey  South Africa
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Quote:
Originally Posted by wakibaki View Post
What is the 4-terminal device in the diagram that looks like a pot? ...
Looks like one of those pots with a tap part-way along the track i.e. non-standard and hard to find. I'd recommend looking for a circuit that uses normal off-the-shelf parts.
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Old 26th October 2010, 12:05 AM   #7
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Quote:
Originally Posted by wakibaki View Post
What is the 4-terminal device in the diagram that looks like a pot? What schematic does this little section come from?

If it's a pot then the circuit has no gain...

w
It's a pot - a log pot with a "loudness" tap. Almost every amplifier with a "loudness" switch has such a pot.

The section is from the schematic of a JVC amp.

The gains talked about are relative to the gain (loss) of the circuit at 1 KHz.

The 180 pf capacitor provides HF boost.
The 1M ohm resistor sets the amount of HF boost.
The 0.0033 microfarad capacitor provides HF cut (LF boost).
The 22K ohm resistor sets the amount of HF cut (LF boost).

All of the components interact, so changing any one will alter the response over the whole range. You will need to find the resistance values of the two sections of the pot each side of the tap. Once you have that, you can work out the response using Ohm's law. Or you can build the circuit in a simulator such as SPICE and measure the results of component changes.
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Old 26th October 2010, 12:13 AM   #8
dmx2020 is offline dmx2020  New Zealand
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Thank you for the quick responses..

I take it that there is some misunderstanding, the schematic claims the loudness circuit produces the specified gain values (6db at 100Hz and 10khz at 10kHz).

Apologies if this post is in the wrong category but the loudness filter is something that provides attenuation at a particular band of frequencies in the audible range so that some frequencies (i.e.:100Hz and 10khz in this case) appear to be amplified more than others. I was just wondering if someone knew how the resistor/caps values affected this so that i could change the values to get the desired gains/attenuation. Many thanks
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Old 26th October 2010, 12:16 AM   #9
dmx2020 is offline dmx2020  New Zealand
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@Don Hills: Thanks for your help. Just the information i was looking for...Thank you so much
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Old 26th October 2010, 12:55 AM   #10
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Quote:
Originally Posted by Don Hills View Post
It's a pot - a log pot with a "loudness" tap. Almost every amplifier with a "loudness" switch has such a pot.
Thanks. Never seen one before...

w
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