Designing EL84 Pentode AmplifierR4 = 5.6KOhm and is chosen in the usual way, in conjunction with load line and the B+ voltage of 250V. The operating point is at 130V.
When we come to calculate the gain, we find that the anode characteristic begins to curve as we reach its intersection with the load line. It is perfectly valid to treat the anode curve as a straight line, and to project this line onto our load line in order to find the gain. Projection on the axis Va of intersection of the load line and the 4V curve gives us 85V. Projection on the axis Va of intersection of the load line and the 8V curve gives us 170V. The gain is:
R5 is chosen either by a detailed perusal of the datasheet, or by observing that, in general, the anode current is a fixed ratio of G2 current. For the EL84 this ratio is approximately 9:1 (48mA divided by 5.5mA). Therefore, if the anode voltage and the G2 voltage are to be the same, the G2 resistor R5 has to be nine times larger than anode resistor R4. We select R5 at 56KOhm. Now let's calculate value of C4 in order to hold second grid at AC ground potential. Using the curves for the EL84, at Va=130V, Vg=6V, Ra=5.2KOhm, we calculate resistance of the second grid R2 at approximately 20KOhm. This resistance of 20KOhm is in parallel with R5 of 56KOhm, giving a final resistance of 15KOhm. Placing the value of 15KOhm to the formula for low cut off filter capacitor calculation gives us value of C4 equal to 10uF. F in the formula below represents a cutoff frequency and is selected to be 1 Hz.
Evaluating Gm from the datasheet, by holding anode voltage constant, and measuring the change in anode current for grid voltage, produces value of 11.3 mA/V. For the pentode, the cathode resistance Rk=1/Gm and gives us R3=100 Ohm. In turn, the value of C3 will be equal to 150 uF for a 1 Hz cutoff frequency.

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