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Question about loadline in Morgan Jones

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Alan0354

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#1
The questions refer to page 460 to 463 in Morgan Jones 3rd edition. This is about the 13E1 conditions.

1) Why for triode, the optimal load is 2ra?

2) The ra given in p460 for 13E1 is 130ohm each tube. So 2ra=260 ohm. Why in page 462, it said from Fig.6.40 that ra=282ohm? How can you read ra from the graph?

3) In p463, it said each valve must see 625ohm. But for CT push pull primary, the anode to anode should be 4 X 625= 2.5Kohm. This is because double the winding gives 4 times the impedance. Why is the book said from anode to anode is 4 X 312=1.25Kohm? This is in p463.

I thought you draw load line from the quiescent point to the max current point on the Vg1=0V to get the load resistance. Then the anode to anode impedance of the primary of the OPT is 4 X the load resistance from the loadline.
Thanks
 
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#2
1. It's the load for maximum power, not necessarily "optimum."
2. ra is gotten from the graph by running a tangent at the proposed operating point and getting its slope. Remember, by definition, ra = dVa/dIa at constant Vg. It is often somewhat higher than the datasheet value and, of course, varies markedly with current.
3. Read the second sentence on that page.
 
#3
Alan0354 said:
The questions refer to page 460 to 463 in Morgan Jones 3rd edition. This is about the 13E1 conditions.

1) Why for triode, the optimal load is 2ra?
Click to expand...

It isn't really. For the best linearity, triodes like the lightest possible loading. That applies to small signal voltage amps as well as large signal power amps. (Attached)

3) In p463, it said each valve must see 625ohm. But for CT push pull primary, the anode to anode should be 4 X 625= 2.5Kohm. This is because double the winding gives 4 times the impedance. Why is the book said from anode to anode is 4 X 312=1.25Kohm? This is in p463.
Click to expand...

This applies to Class AB, Class B, and Class C only, since one side is driven into cutoff as the other reaches its Ip(max). If the PP output stays in Class A, then the primary of the OPT needs to be 2 X Rload/phase since neither side is ever cutoff, and the Class A final remains a balanced, two phase, system at all times during the signal cycle.

If one side cuts off, it disappears from the circuit. In that case, all that remains is the active half, which needs to be matched to the load. In that case, for both sides to see Rload/phase, the whole primary needs to match 4 X RL.
 

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#4
SY said:
1. It's the load for maximum power, not necessarily "optimum."
2. ra is gotten from the graph by running a tangent at the proposed operating point and getting its slope. Remember, by definition, ra = dVa/dIa at constant Vg. It is often somewhat higher than the datasheet value and, of course, varies markedly with current.
3. Read the second sentence on that page.
Click to expand...

1) I don't get why 2ra load get max power.
2) The graph in Fig 6.40 doesn't show the top, but is it true to draw from the 400V and 237mA point to max current specified in the data sheet of 800mA at Vg1 = 0 curve?
3) I read through the whole thing over and over, If the author claimed the load for each tube should be 625ohm, the anode to anode impedance of the primary should be 4 X 625=2.5K. Whether is 4 X 312 = 1.25K comes from?

Thanks
 
#5
Miles Prower said:
This applies to Class AB, Class B, and Class C only, since one side is driven into cutoff as the other reaches its Ip(max). If the PP output stays in Class A, then the primary of the OPT needs to be 2 X Rload/phase since neither side is ever cutoff, and the Class A final remains a balanced, two phase, system at all times during the signal cycle.

If one side cuts off, it disappears from the circuit. In that case, all that remains is the active half, which needs to be matched to the load. In that case, for both sides to see Rload/phase, the whole primary needs to match 4 X RL.
Click to expand...

Sorry I totally don't get it. You can make the amp into Class A or Class AB with the same primary impedance. You change the bias current to change the class. You want class A and you bias at 240mA, you want to hit the Vg1=0 and 480mA. Then you ensure both tube conduct 360 deg and become class A. If you want max power, then draw from the operating point to the max current allow ( 800mA from 12E1) at Vg1=0 to get the load line for Class AB.
 
#10
Anyone has link to data sheet of 13E1 other than AEI. I search and only found AEI, it did not has graph for triode strapped mode, so I can't even start to work on the load line to verify anything. All I have is the Morgan Jones that still does not make sense to me. The graph in Morgan Jones does not reach to high enough current to draw the load line to Vg1=0.
 
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