COLD CATHODE DIODE 



Suppose we wish to design a stabilizer to supply 210 V at 0-70 mA. Then 

 the load resistance which corresponds to maximum current is 210 V/70mA = 

 3,000 ohms. Suppose a string of stabilizers have been chosen which will 

 strike at 255 V. Then 2 K, - F, = 510 - 210 = 300 V, and IR^ must be 

 greater than this. Since Rj^ = 3,000 ohms, / must be 100 mA or greater. 

 Let it be 10 per cent over the critical value, i.e. 110 mA, for safety. Taking 



110 mA 2-3 ki2 



510 volts 

 Unstabilized 



Figure 7.7 



Load, 

 minimum 



value 

 3000 ft 



250 ii 



Load 



Figure 7.8 



V^ = 2Fj = 510 V, the provisional design is in Figure 7.7, and the ballast 

 resistance is 1 10 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 1 10 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 7.8, and there will be a change of output voltage 

 of 70 mA X 250 ohms = 17-5 V between 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 



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