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# Square Waves Produced by Clipper Action

Author: J.B. Hoag

Suppose a comparatively large sinusoidal voltage is applied to the grid of an amplifier tube. When the grid goes negative beyond the cutoff point, plate current ceases to flow and the bottoms of the sine curve are clipped or squared off. When the grid goes positive to the saturation region beyond the upper knee of the characteristic curve, the plate current is limited and the tops of the sine curve are clipped off.

When sharp cutoff tubes are used, the bottoms of the sine waves are squared off with reasonable sharpness. On the other hand, the tops of the waves are not sharply squared off because of the grid current which flows when the grid goes positive and extracts electrons from the plate circuit, as well as develops a C-bias across the input resistors.

 Fig. 30 A. A circuit for changing 60-cycle sine waves into 60-cycle square waves by clipper action. (Martin, Electronics, July 1941)

Figure 30 A shows a circuit which gives both square tops and bottoms. A comparatively large sinusoidal potential of 166 peak volts from the 60-cycle line is applied to the grid of the first half of the twin-triode tube 6F8G. During the positive half-cycle, the grid would become very positive were it not for the resistor R. When the grid becomes slightly positive, grid current flows through R (and through the usual grid resistor R1). Then a negative voltage is developed on the grid which acts as a variable C-bias as in Fig. 30 B to limit the positive voltage on the grid. The resultant plate current then has a slightly rounded top as shown in the figure.

 Fig. 30 B. Action of the first tube in Fig. 30 A

During the negative half-cycles, electrons on the grid condenser, C of Fig. 30 A, flow to ground through resistor R1 and develop a C-bias of essentially constant amount, as indicated at C in Fig. 30 B. The constancy is due to the fact that the discharge time of C through R1 (0.1 sec.) is large compared with the period of the input voltage.

When the grid of tube 1, Fig. 30 A, goes positive, its plate current rises, the upper end of R2 becomes more negative, and this voltage change feeds through C2 to the grid of tube 2. In other words, there is a phase reversal of 180° between the plate current of 1 and the grid voltage of 2.

The time it takes to discharge C2 through the grid leak of the second tube is chosen quite long so that a comparatively fixed and large C-bias is established on 2, as indicated in Fig. 30 C.

 Fig. 30 C. Action of the second tube in Fig. 30 A.

Its value is controlled by the strength of the input voltage and is such that the positive peaks just drive the grid of 2 slightly positive.

Note, in Fig. 30 C, that only the square bottoms of the plate current of the first tube are used to give plate current in tube 2. Hence both the tops and bottoms of the current through R3 are sharply square. Note, also, that the time constant of the output circuit is very large, so as to retain this squareness.

Last Update: 2009-11-01