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Tube DIY Asylum Do It Yourself (DIY) paradise for tube and SET project builders. |
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In Reply to: Re: Resistor as parafeed plate load? posted by Paul Joppa on October 30, 2001 at 19:06:30:
RL=(Eb/Ib)-2.4rp. Example: a 2A3 running 300v (max rated) at 50mA (to make 15
watts plate dissipation), the load would be (300/.050)-2.4*800 or 4080 ohms.
could you give a bit more info on what that 4080 is????i guess my question is, if we look at the 4080 load as the reflected load and assume a 1K signal (to keep the inductance out of the picture) i see the tube putting out the max undistoretd signal as the curves go, but when we consider the turns ratio, we will have more of a stepdown so the voltage output at the load will be less than a turns ratio that presents a 2000 ohm load... i understand the 2K load will cause the loadline to rotate... possibly making the AC swing available less so we may actually end up with less overall power even given the smaller turns ratio, but i was wondering if anyone knows if this is the case???
so by reducing the turns ratio by 30% we will get 30% more voltage, but will the reflected load presented to the tube change the tubes power out by more or less than 30%???
again its just another one of these things that needs lots of asteriks to qualify the speciffic situation.
dave
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Follow Ups
- Re: Resistor as parafeed plate load? - dave slagle 08:27:27 10/31/01 (1)
- Re: Resistor as parafeed plate load? - Paul Joppa 10:17:41 11/02/01 (0)
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