. Electronic apparatus for biological research. Electronic apparatus and appliances; Biology -- Research. Figure Load, minimum value 3000 ft 250 ii. Load Figure V^ = 2Fj = 510 V, the provisional design is in Figure , and the ballast resistance is 110 mA X 210 V = 2,300 ohms. At full load 40 mA is dehvered to the tubes and 70 mA to the load. On no load all 110 mA pass through the tubes which must be rated to pass this current. The practicability of the design depends on whether tubes can be found to pass up to 110 mA. If they can only be found to pass up to, say, 85 mA then the curre
. Electronic apparatus for biological research. Electronic apparatus and appliances; Biology -- Research. Figure Load, minimum value 3000 ft 250 ii. Load Figure V^ = 2Fj = 510 V, the provisional design is in Figure , and the ballast resistance is 110 mA X 210 V = 2,300 ohms. At full load 40 mA is dehvered to the tubes and 70 mA to the load. On no load all 110 mA pass through the tubes which must be rated to pass this current. The practicability of the design depends on whether tubes can be found to pass up to 110 mA. If they can only be found to pass up to, say, 85 mA then the current output is restricted to 25-70 mA. Let us see how it behaves. On first applying K„, the potential difference across the load rises to 3,000/(3,000 + 2,300) X 510 = 278 V, so that the tubes strike with 23 V to spare. Thereafter, for an incremental tube resistance of 250 ohms—a typical value—the forward stabilization ratio dVJdV^ is, exactly 250 . 3,000 250 + 3,000 2,300 + 250 . 3,000 250 + 3,000 which is not far from 250/(2,300 + 250) which is approximately equal to 1/10. Such a performance is greatly inferior to that of a voltage reference tube as regards stability of the output, but is often worth having. The backward stabilization will be poor, for the equivalent circuit of the arrangement is substantially that of Figure , and there will be a change of output voltage of 70 mA X 250 ohms = 17-5 V between 0 and 100 per cent of full load. The tube hfe is likely to be short. Tubes cannot efficiently be run in parallel to help matters. Now consider a stabilizer to supply 210 V at 0-7 mA, using the same tubes. The full-load load resistance is now 30,000 ohms. As before IRj^ — 300 V, so / must be 10 mA or more. If it is 10 mA and the load current is 7 mA, only 3 mA would flow through the tubes and it is doubtful if they would work properly at such a low current. Let us assume the lower current hmit for these hypothetical tubes is 10 mA. Then / would have to be 17 mA 123
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