Loadlines, Operating Points, B+, etc.

[dbishopbliss]

I've had something of a revelation recently, and I want to confirm my understanding.  There is a fairly popular site that introduces many concepts about designing and building amplifiers that was the foundation for my understanding of Loadlines, Operating Points, B+, etc.  However, after one of PJ's replies to an earlier question I posted, and reading a number of other texts, I believe the site's explanation is incorrect.  Therefore the information below may seem elementary, but I'm coming from a position of having to unlearn everything I had learned before.  Will someone please confirm or correct my new understanding.  

The attached image contains the plate curves for a 300B tube.  The yellow line represents the maximum plate dissipation of 40W.  The blue line represents the "typical operating conditions" taken from a 300b datasheet - 4K Load with an operating point of 350V at 60mA.

Here is where I was confused.  According to the other site, the B+ should be 350V.  However, my new understanding is the B+ is determined where the load line intersects the Voltage axis where Ia = 0.  In other words, B+ = ~600V.  350V is the plate voltage which is the value after the load resistor, while ~600V is the B+ taken before the load resistor (correct?).

Assuming, I am correct I notice a couple of things.  First, the 4K load is below the maximum plate dissipation.  To get more power out of the tube I could use a 2.5K load (the red line) and an operating point of 350V at 100mA.

However, the problem with the above load and operating points is that the B+ is fairly high.  I don't know enough yet to determine if this is really a problem, but I suspect it is because most electrolytic capacitors are rated around 450V and even many film caps are rated for 600V.  Not only that, but it is not that common to find power transformers that can deliver 600V (they are certainly there, but just not as common as lower voltage transformers).

So, the alternative is to lower the B+ and move the load lines and operating points to the left.  However, doing so will again decrease the output of the tube and move the operating conditions to a less linear operating area of the tube.

I'm sure there are books that can be (and have been) written about this, but I want to be sure what I now think is correct, is correct.  Thanks for your feedback.

[Paul Joppa]

First, you must distinguish between the plate voltage and the power supply voltage.

Incidentally, the curves you posted are usually called plate curves; there are different curves called grid curves. Since all the curves show plate voltage, plate current, and grid voltage, these names are kind of arbitrary.

Now then. Those curves have plate-to-cathode voltage as the horizontal axis. In a conventional voltage amplifier with a cathode resistor to provide bias, and a plate resistor for a load, the power supply must also provide those voltages. What you have drawn shows the power supply assuming the cathode is grounded, in other words there is a separate negative voltage supplied to bias the grid. As shown, the load resistor dissipates 14.4 watts of heat, plus any AC signal power if the tube is amplifying an AC signal.

Now comes the tricky part - AC impedance and DC resistance. In a conventional series feed power amplifier, the plate gets its voltage through the output transformer primary winding. That winding has a certain DC resistance, let's say around 150 ohms. At 60mA that would drop 9 volts, which is the DC voltage difference between the plate and the power supply, still assuming the cathode is grounded and the grid has its own negative voltage supply. The output transformer is dissipating only 0.54 watts of heat.

At audio frequencies, the inductance of the primary winding produces a very high impedance, which is in parallel with the loudspeaker impedance as adjusted by the transformer turns ratio. Let's assume you have a 4K:8 ohm transformer and have put an 8 ohm resistor across the secondary. Then the transformer primary appears, at audio frequencies, to be a 4K resistor. Almost all of the AC signal power is dissipated in the 8 ohm resistor (or the loudspeaker, if it replaces the 8 ohm resistor).

Now you see why transformers are used. Almost no DC power is wasted as heat, and almost all of the signal power goes into the loudspeaker instead of a resistor.

It's the energy stored in the transformer's magnetic field that is able to drive the plate to 600 volts at the peak of the AC waveform, even though the power supply voltage is only 359 volts.

[dbishopbliss]

Just when I think I understand...  I've included a simple circuit diagram to help clarify my questions, because your comment about the winding having a DC resistance has me confused.  Assume a single-ended grounded-cathode triode circuit as shown in the diagram.  Also assume that the values I'm presenting are for illustration purposes only.  

• Did I draw my load line correctly for a 4K load, 350V/60mA operating point?
• Given the operating conditions above, will voltage at point B = 350V?
• Given the operating conditions above, will voltage at point A = ~600? (I'm thinking no now)

P.S.  I updated my original post to say, "plate curves".  

[Paul Joppa]

    * Did I draw my load line correctly for a 4K load, 350V/60mA operating point?
Yes.

    * Given the operating conditions above, will voltage at point B = 350V?
No. It will be 424 volts at B, and 74 volts at the cathode.

    * Given the operating conditions above, will voltage at point A = ~600? (I'm thinking no now)
No. If the DC resistance of the output transformer is 150 ohms, then A will be at 433 volts.

[dbishopbliss]

So, the first thing I've learned is that transformers are different than resistors.  I was mistakenly thinking that a 4K transformer load was the same as a 4K resistor load.  How do I determine the DC resistance of the winding, simply using my DMM?

Assuming 424V at point B, I can see how you arrive 433V at point A if you assume that the DC resistance of the transformer is 150 Ohms. 

9V = 150R * 60mA
433V = 424V + 9V

Assuming 74V at the cathode, I can see how you arrived at 424V at B (424=350+74).  However, where did 74V come from?  I was thinking it came from where the operating point intersected the curves, but that appears to fall half-way between the -60 and -75V curves so I would have used a value around 67V.

[Paul Joppa]

...How do I determine the DC resistance of the winding, simply using my DMM?

... I can see how you arrived at 424V at B (424=350+74).  However, where did 74V come from?  ...

1) You got it!

2) The same place the 350v and 60mA and 4K came from - the January 1950 Western Electric specification sheet for the 300B. The curves you showed look like they may have been an actual measurement of a specific 300B sample. There is a certain amount of variation from one tube to another.

[dbishopbliss]

Looking at the plate curves from the Western Electric datasheet I can see where the 74V comes from.  I have redrawn my original diagram so others can see as well.

Thanks for your help.  I hope my questions are useful to others.