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#1 |
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diyAudio Member
Join Date: Jul 2007
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Hi does anybody know how to read the circuit attached, I'm trying to find out how to work out the max and min Gain for each circuit,
can anyone explain what is going on in this circuit the part that is comfusing me the most is the capacitor. i'm trying to find out how the circuits reacts to various frequecies If anybody can help i would be very greatful |
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#2 |
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diyAudio Member
Join Date: Oct 2003
Location: Ottawa, Canada
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You obviously forgot to attach whatever circuit you are asking about. Try again.
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#3 |
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diyAudio Member
Join Date: Jul 2007
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Hi does anybody know how to read the circuit attached, I'm trying to find out how to work out the max and min Gain for each circuit,
can anyone explain what is going on in this circuit the part that is comfusing me the most is the capacitor. i'm trying to find out how the circuits reacts to various frequecies If anybody can help i would be very greatful i cant seem to attach the picture of the circuit, if anyone can help could i email it to you |
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#4 |
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diyAudio Member
Join Date: Jan 2005
Location: Phoenix, Az.
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You can easily simulate the circuits using switcherCAD- a free spice simulator that will show you exactly what is going on.
The first circuit is a treble control that provides +/- 17 dB or so of boost/cut in the audible frequency range. The other circuit is supposed to be a bass control circuit that can be simulated the same way, but it doesn't seem to work. Better check that one. I_F |
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#5 |
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diyAudio Member
Join Date: Oct 2003
Location: Ottawa, Canada
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I'll try to explain the circuits without a lot of math.
First you need to understand that a capacitor has a high impedance at low frequencies, and a low impedance at high frequencies. So, to simplify the circuit to help to understand it, we will replace the capacitor by a open circuit (no connection) at low frequencies, and a short circuit (a piece of wire) at high frequencies. We won't worry about what happens in the middle for now. Let's look at the treble circuit first. The opamp is a basic inverting buffer, with gain equal to -Rf/Ri (feedback resistance / input resistance). We'll ignore the negative sign for this. But just remember that the output will be out of phase with the input. At low frequencies, replace C1 by an open circuit. Then Rf = R2 and Ri = R1, so gain is R2/R1 or 1. There is also a 500+10+10 k resistance between the output and input, but it doesn't affect the gain of the circuit, because it doesn't connect to the input pins of the opamp. So the gain for bass frequencies is 1. Bass is unaffected by this circuit, no matter where the pot is set. At high frequencies, think of replacing C1 by a wire. Now the pot, R4, is part of the feedback circuit. So let's look at what happens when the pot is centered. There will be 250k on each side, and both sides are exactly the same. So then Rf = Ri and the treble gain is 1. But when turned fully CW, there is 0 on the left side and 500 k on the right, so Rf = 100k//(500k+10k) = 83.6k and Ri = 100k//10k = 9.1k. So treble gain = 83.6/9.1 = 9.2 or +19.3 dB. This is the maximum treble boost. [The // means "in parallel with", and dB = 20*log(voltage Gain) ]. When turned fully CCW, the treble gain can be calculated to be 9.1/83.6 = 0.109 or -19.3 dB. This is maximum treble cut. The frequency at which the boost/cut begins to have effect is determined by the value of the capacitor. You can figure out the bass circuit yourself the same way. |
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#6 |
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diyAudio Member
Join Date: Jul 2007
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How would i work out the gain of the circuit if a number of frequencies where inputted e.g. 3khz,
do you add the capacitor value into the R2/R1 formula? |
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