Question about direct coupling a anode follower into a cathode follower.

[Sch3mat1c]

Dude, graphing a CF isn't the simplest thing because of the NFB element. But it's not necessary anyway, at least until you get to the nitty-gritty. Just set it at a convienient voltage (determined by the preceeding stage as mentioned) and current (determined by the voltage Vp SY was using, and Rk of the CF). It's a simple, stable circuit element that can be broken down with Ohm's law. As SY also said, no one cares about the 2V grid to cathode.

Tim

 

[G]

Dude, graphing a CF isn't the simplest thing because of the NFB element. But it's not necessary anyway, at least until you get to the nitty-gritty. Just set it at a convienient voltage (determined by the preceeding stage as mentioned) and current (determined by the voltage Vp SY was using, and Rk of the CF). It's a simple, stable circuit element that can be broken down with Ohm's law. As SY also said, no one cares about the 2V grid to cathode.

Tim

The problem I'm running into is finding the right resistor values. I may just put a 2 watt 50K carbon pot in // with a 50K resistor on the CF cathode and adjust the bias that way. The 20K resistor is only dissipating .5 watts so that should work fine. Dude.

 

[SY]

Which resistor values are you trying to calculate? The cathode resistor for the CF? If so, I've outlined how in pretty complete detail. The biasing takes care of itself!

 

[G]

Keep it simple. The CF draws essentially no current from the first stage. So for the purpose of DC analysis of the first stage, you can ignore the CF completely. Using load lines or CAD or whatever, you can determine what the plate voltage of that stage is going to be. It will be determined by the cathode resistor, the plate resistor, and the B+ rail. They're all part of the equation. But the point to take away is that you can calculate what you need about this stage in isolation. Let's call the voltage at the plate of this stage Vp. Doesn't matter what it is, we can go calculate it.

Now, let's move to the CF. Clearly, the voltage at the grid will be Vp because it's direct coupled from the previous stage. So with that as a given, we can look at this stage in isolation. The voltage at the cathode will be Vp plus a couple of volts, which are needed for biasing. But here's the trick: the required bias voltage in this circuit is very low compared to the drop across the CF's cathode resistor! So we can, to a very good approximation, say that the cathode voltage is also Vp. Sure, sure, it's actually going to be Vp plus something like two volts. But Vp is going to be something like 60-70 volts, so saying that 62 = 60 is not too bad an approximation.

Now... with this in hand, we can either calculate the value of the CF's cathode resistor from the desired current in that stage OR we can figure out the current in that stage from a given cathode resistor. Let's say, for argument, that Vp is 60 V. So the voltage at the cathode of the CF will be slightly higher than 60, but not significantly so. Then for the 20K value of CF cathode resistor, the current will be about 60/20 or 3 ma. You don't have to worry about whether that CF is going to need to sit 2 volts above Vp or 3 volts above Vp or whatever- the current will take care of itself. The variation in required biasing voltage will cause the current to be not exactly 3 ma, but who cares if it's 2.94 ma instead? Not me.

Is that a little clearer?

Hi Sy. You are saying the cathode voltage figure that I put in the schematic is wrong and that it will be different because of feedback. That means I can build it as shown but the voltages will not be what I calculated them to be, but the circuit will work as drawn. Correct? If I set the cathode resistor to drop 100v then the cathode voltage will actually be about 102v because of the feedback. That's what I'm getting from this post. Thanks for replying.

 

[SY]

If you're talking about the 6SN7 preamp schematic, yes, that's it exactly. It will work fine, the cathode will take care of itself by biasing up a couple of volts.

 

[SY]

BTW, though the 6SN7 circuit you drew will work, it's not optimal. The first stage will be running too little current to be linear and the output Z will be higher than your stated target. Frank's design will be better in these respects.