ROBERTLATHAM1 Posted January 30, 2006 Report Share Posted January 30, 2006 hey guys here goes another stupid question. i was looking at the marshall dsl 401 scheamatic and the signal on valve 3(last tube of pre amp before inverter) appears to be comming from the cathode of valve 3 to feed the phase inverter tube 4. the plates are still feed from b+ voltage(450vdc or there about) the first plate (pin 1) looks to directly feed the second grid of valve 3. the second plate of valve 3 goes straight toB+ with no junctions. the first plate of valve 3 has 1 junction to b+ and to the second grid of valve 3. the first grid pin2 is fed from earlier gain stages. woo! ok now my question is how does the signal feed through the cathode and not the plates for the signal gain? (signal is only tapped off the second cathode of valve 3). i have seen this on other marshall amps. the 8100 valvestate uses the same tube configuration on the boost channel. my question still stands "how and why does this benefit the signal. thanx robert latham Quote Link to comment Share on other sites More sharing options...
JoeAArthur Posted January 30, 2006 Report Share Posted January 30, 2006 hey guys here goes another stupid question. i was looking at the marshall dsl 401 scheamatic and the signal on valve 3(last tube of pre amp before inverter) appears to be comming from the cathode of valve 3 to feed the phase inverter tube 4. the plates are still feed from b+ voltage(450vdc or there about) the first plate (pin 1) looks to directly feed the second grid of valve 3. the second plate of valve 3 goes straight toB+ with no junctions. the first plate of valve 3 has 1 junction to b+ and to the second grid of valve 3. the first grid pin2 is fed from earlier gain stages. woo! ok now my question is how does the signal feed through the cathode and not the plates for the signal gain? (signal is only tapped off the second cathode of valve 3). i have seen this on other marshall amps. the 8100 valvestate uses the same tube configuration on the boost channel. my question still stands "how and why does this benefit the signal. thanx robert latham The first plate should be going through a resistor, normally 100K before connecting up with B+. This provides the signal for the second grid. This is a common configuration. The second tube is operating as a "cathode follower"... and the output signal is obtained from the cathode, via a resistor (normally another 100K) to ground. It has a high input impedance providing a light load on the first tube, which is operating as a normal common cathode. A cathode follower is a "non inverting buffer", meaning the output phase is the same as the input... and the configuration provides a gain of slightly less than unity (i.e. < 1) There are disputes as to the benefits. Some think it only increases the number of tubes required. Others think that the cathode follower allows the common cathode circuit preceeding it to operate at maximum gain since the output load that stage sees is essentially only the plate resistor. Quote Link to comment Share on other sites More sharing options...
Mike Sulzer Posted January 30, 2006 Report Share Posted January 30, 2006 And the other advantage is that the output impedance of the cathode follower is low, determined by the tube, not the cathode resistor. For example, it would be good to drive a long cable that has high capacitance to avoid losing high frequencies. In some circuits it is used to drive the tone stack. The impedance looking into the tone stack could vary with frequency and change as the settings of the controls are changed. If you drive it with a low impedance, it behaves in a more ideal manner, and you get the full boost and cut that you should. Quote Link to comment Share on other sites More sharing options...
ROBERTLATHAM1 Posted January 31, 2006 Author Report Share Posted January 31, 2006 And the other advantage is that the output impedance of the cathode follower is low, determined by the tube, not the cathode resistor. For example, it would be good to drive a long cable that has high capacitance to avoid losing high frequencies. In some circuits it is used to drive the tone stack. The impedance looking into the tone stack could vary with frequency and change as the settings of the controls are changed. If you drive it with a low impedance, it behaves in a more ideal manner, and you get the full boost and cut that you should. so the signal is merley pased and conditioned not boosted and the impedence is low so the tone stack is better able to filter the signal. this is the way the tube in the marshall valvestate 8100 is configured so im assuming here that it was used just for the tube conditioning and not for high gain(saturation)? and this is the cathode follower in a pure tube amp. Quote Link to comment Share on other sites More sharing options...
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