B Eliminator Output Resistances.

Radio Record, Volume II, Issue 13, 12 October 1928, Page 27

B Eliminator Output Resistances.

How They are Calculated. rps following method of calculating resistances placed across’ the positive and negative terminals of a B eliminator will be helpful to constructors. This refers to the "potential divider" type, in which several fixed taps are provided to give the standard voltage for each stage of the receiver. Diagram 1 shows the resistances of an eliminator, and we will suppose that tap No. 1 is to be 45 volts to operate the detector. The loss current through R38 is usually about 3 milliamps. If tap No. 1 is to give 45 volts, then the drop across R8 must be 40. The resistance of R38 will be equal to the voltage across it divided by the current through it, which is in this ease 45 divided by 0.008 (3 mils.), which gives 15,000 ohms as the value of R38. The voltage at tap No. 2 is to be 90. and as the voltage drop across R3 is 45, the drop across R2 will also be 45.

in order to make the total voltage between negative B and tap No. 2 equal to 90. The current fiowing through the resistance R2, will be equal to the loss current of 3 milliamps, plus the ‘Yeurrent drawn by the detector, 1 milliamp. Therefore the value of R2 . will be equal to the voltage (45) across it, divided by the current through it, which is 0.003 plus 0.001, or a total of 0.004 amperes. This gives a value of 11,250 ohms for R2. QUPPOSE that the total drain from the 90-volt tap is 10 milliamps, the total flowing through R1 will be equal to 10 plus 1 plus 3, or 14 milliamps. If the maximum voltage available from the power unit is 180, and the voltage at terminal No. 2 is to be 90, then the drop across R1 must be 90, and 90 volts divided by 0.014 amps. gives 6400 ohms as the value of Resistances must be capable of carrying the load without overheating, but the wastage is very small in a B eliminator. The load in watts being handled by a resistance is determined by multiplying the resistance in ohms by the square of the current in amperes. This gives for the above illustration, watts through R38, 0.1385; through R2, 0.18 watts, and through Ri, 1.25 watts. If it is desired to add further resistance externally to the detector supply, in order to reduce voltage toa more suitable value than 45 volts, this can_ be effected by means of a variable hi#h resistance with 2 mfds. across positive and negative.