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#1 |
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diyAudio Member
Join Date: Mar 2009
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Hi,
I'm trying to build this design: http://www.shine7.com/audio/pa100.htm I've realized that it doesn't have a volume control potentiometer and i want to add it. How can i do this? Is it correct to connect both R2 and R7 to the middle of the potentiometer? Then, one side to ground the other to the RCA In filter (R1-C1) ¿? Thanks a lot ! |
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#2 |
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diyAudio Member
Join Date: Feb 2008
Location: Melbourne, Victoria, Australia
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http://www.minirig.org.au |
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#3 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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J1 Pin1 connects to pot wiper.
J1 Pin2 connects to pot return (bottom). RCA pole connects to pot input (top). RCA barrel connects to pot return.
__________________
regards Andrew T. |
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#4 |
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diyAudio Member
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a very simple way is...the middle pin of the common 3 terminal pots is where u have take the input to your amp..
2 pins remain....to one of them connect your amp and the input signals ground To the one remaining pin..connect the signal from the input source. Only one rule applies, the middle pin is the pin from which you have to take the signal to the amp.. The pot will work even if you exchange the other two pins ( 1 & 3 ) positions but it's direction will change from Clockwise to anticlockwise. |
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#5 |
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diyAudio Member
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When choosing a pot, does it need to be an effect pot that handles xx watts or is a normal pot adequate when used on input signal?
Also what decides whether to use a log or a linear pot and which ohmic value you need? |
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#6 |
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diyAudio Member
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u don't need too much power handling capacity , most pots are rated around 0.25W, generally a log taper is used.
Loudness increases, I mean dB is a logarithmic measure and human ears also perceive loudness change in logarithmic fashion ( in theory...). A linear taper would work far too abruptly..a little turn of the knob would blast the hell out of you.. |
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#7 |
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diyAudio Member
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hehe okay.
What decides the ohmic valu needed then? Trial and error or can you calculate it? |
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#8 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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keep the pot resistance as low as all your sources are able to drive.
As a starter try comparing 10k to the specifications for minimum load resistance, of what you might want to connect to it. If the power amp is also low in input impedance (Zin<20k) you have a problem that a pot alone is unlikely to solve.
__________________
regards Andrew T. |
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#9 |
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diyAudio Member
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AndrewT
My amp says RIN Input Impedance min. 43kΩ nom. 50kΩ max. 57kΩ Does this mean I should take a 50kΩ pot or should I somehow take into account the resistance from the source also? |
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#10 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Hi Havoc,
a 10k pot will work with your Rin=50k, but you must keep the cables short. That places the volume control near the amplifier. If you can operate like this then OK. Now check all the sources that may need to drive the 10k pot. Are they able to do this?
__________________
regards Andrew T. |
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