Tube Amps / Music Electronics
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|previous: Joe Fuzz This is a circuit reduction of the ... -- 8/11/2000 3:53 AM||View Thread|
|8/11/2000 7:09 PM|
|Randall Aiken||Re: Analysis Requested: Vox AC30|
Okay, Joe, here's your quick course :
The circuit is half of a long-tail pair phase splitter, the other half being used as the second channel input (both inputs are used, instead of the "normal" Fender/Marshall type that uses global negative feedback).
Returning the grid resistor to the midpoint between the bias resistor and the "tail" resistor is called "bootstrapping", and increases the input impedance to the stage. In the case of the long-tail pair, the signal at the cathode is ideally half the input signal voltage, because the impedance seen looking into the cathode of each side is equal to (Ra + Rl)/(mu + 1) = (62.5K + 100K)/(100 + 1) = 1.6K, and since each side has the same output impedance as the input impedance of the cathode of the other side, the signal is divided in half, provided the tube halves have equal mu's.
Since half the input voltage appears at the cathode, the voltage at the junction of the 1.2K and 47K resistors will divide by the voltage divider rule, giving a voltage at the junction (assuming a 1V input) of 0.5V * 47K/(1.2K+47K) = 0.488V. Since 1V was applied to the top of the 1M resistor, and there is 0.488V at the bottom of the 1M resistor, there is an AC signal current flowing in the 1M resistor equal to (1V - 0.488V)/1M = 512uA. This means the effective input impedance is equal to 1V/512uA = 1.95 Megohms. This larger input impedance will allow you to use a smaller coupling cap than you would normally think is necessary, because the input impedance is approximately twice the value of the grid resistor.
In addition to providing a high input inpedance, this configuration is necessary to properly bias the tube. If the grid resistor were returned to ground, both the grid and cathode would be at ground potential, and no DC bias would be developed for the tube to properly operate.
The way it is connected in the AC30, the bias voltage is developed across the 1.2K, which properly biases the grid with respect to the cathode. The shared 1.2K is equivalent to separate 2.4K cathode resistors on each half, and gives around 1.4mA of total shared current between the two halves (700uA for each side). Since there is no DC current flowing in the 1M grid resistors, all this 1.4mA must also flow in the "tail" resistor. This then creates an offset DC voltage of 1.4mA*47K = 65V. This gives the stage some operating headroom so it can properly amplify the signals. In addition, the larger the value of the "tail" resistor, the better the balance between the two stages, because the large value resistor acts like a constant current source.
Since there is negative feedback in the circuit, the gain will decrease and the equivalent internal plate resistance will increase. The total resistance at the cathode as seen by the first tube is the 1.2K bias resistor in parallel with the 1.6K resistance seen looking into the cathode of the other section. This gives an equivalent parallel impedance of 686 ohms. Using this, the new internal plate resistance value with feedback, Ra', is equal to Ra + (mu + 1) * Rk' = 62.5K + (101)*686 = 132K.
The equivalent input voltage is not the input applied to the grid, rather it is the differential voltage between the grid and cathode, because the cathode is not bypassed, so the equivalent input voltage is only half the original input (the 1V input minus the 0.5V at the cathode in the previous example). This gives a voltage gain from input to output of 0.5*mu*Rl/(Rl + Ra') = 0.5*100*100K/(100K + 132K) = 22.
For more on this type of circuit, go to the advanced section of the tech info pages at http://www.aikenamps.com and read the papers entitled "The long-tail pair" and "Designing long-tail pairs - the load line approach".
|Randall Aiken I screwed up. I said:|