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#1 |
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diyAudio Member
Join Date: Nov 2009
Location: Ontario
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Lately, I've been itching to make an amp out of a transmitter tube (eimac 4-125a).
The graphs in the datasheet are "constant current characteristics" as opposed to the traditional "plate characteristics" curves. Is there any way to work with/around these graphs? The graphs may be seen here: http://tubedata.tubes.se/sheets/140/4/4-125A.pdf |
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#2 |
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diyAudio Member
Join Date: Jan 2008
Location: Bridgeville, CA
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I plot load lines directly on the constant current current charts as in the 4CX250 example below. Plot points on the constant current lines at plate voltages corresponding to the dynamic load resistance, in this case 2K ohms since it's 8K plate-plate and class B.
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#3 |
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diyAudio Member
Join Date: Nov 2009
Location: Ontario
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Thanks for the reply. What formula are you using to derive the voltage on each current line? The standard V=IZ formula doesn't seem to do it...
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#4 |
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diyAudio Member
Join Date: Dec 2001
Location: Hickory, NC
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something more like: V = B+ - I*Rload
__________________
Ohms Law V = I R |
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#5 |
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diyAudio Member
Join Date: Jan 2008
Location: Bridgeville, CA
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It's ohm's law applied to each current step, plus the quiescent current. Just like a load line on the more usual plate curves, but transposed onto the constant current lines.
I start by placing the first point on the quiescent op point at 50mA, 1500 volts. Since this is a class B amp, I only plot anode voltage for increasing current, in delta steps up from the idle point. The second line is at 100 mA which is a delta of 50ma from the previous point. The delta voltage for 50 mA * 2000 ohms is 100V. The second point therefore goes on the 100 mA line at quiescent voltage minus 100 volts (1400 in this case). The next constant current line is 200 mA, which is a delta step of 100 mA from the previous. The delta voltage is 100 mA * 2000 ohms or 200V. Thus the third point goes on the 200 mA line at 1400 -200 or 1200 volts. And so on... The same process is used for class A amps, plotting points outward in both directions from the idle point. Cheers, Michael |
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