![]() ![]() An extra resistor at the cathode helps maintain balance even when the loads do differ e.g. The cathodyne remains balanced provided the external loads are identical. Now load each with 1K - you get 1V output from both! ![]() So the external circuit sees an output impedance of 100 at the first, and 100K at the second. ![]() The first one has a series resistance of 100ohms, the second a series resistance of 100Kohms. Consider two perfect voltage sources, one develops 1.1V, the other develops 101V. If this is too complicated, think of it in another way. As the currents are the same, identical loads will develop identical voltages. high impedance) which happens to track the cathode current. The anode output approximates to a current source (i.e. Then the cathode output approximates to a voltage source i.e. Think about building one with a valve which has high gm (so low output impedance as a cathode follower) and also high mu (so high anode impedance, made even higher in this case by the large cathode resistor). It also has identical output levels if the load impedances are the same. The cathodyne does have markedly different output impedances at its two ports. It is very easy to confirm this by either a spice simulation or measurements in a real amplifier, I have both simulated and measured the output impedances and can confirm that it is correct, it is interesting to notre that the output impedance is so low, in my case with a 12BH7 and 15k anode resistor you get ~300ohm.Īdding a resistor from the cathode will introduce imbalance and increase distortion. The reason for the mistake in calculation is that the output impedance of the anode and cathode is calculated separately which of course is wrong as both sides are loaded, calculating the output impedances with both sides loaded give the result above. That a faulty description is circulated in many web forums and even in some books doesn't make it more less faulty. This subject has been discussed many times on this forum and the correct calculation of the ouput impedances is described in many books e.g Morgan Jones Valve amplifiers 3rd edition. ![]() The ouput impedance can be calculated as Rout = RL*ra/(RL*(µ+2)+ra) Where RL is the anode or cathode resistor, ra is the anode resistance of the tube. Sorry but you have interpreted the text wrong, that text confirms that the the output impedances ARE equal, Quote "In the case of the phase splitter circuit Zk=Zl=Z" (page 2 in the middle). ![]()
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