Short DC Power-Line Pulses Afford Remote Control

Texas Instruments CD74HC14

If you face the challenge of adding a second, independently controlled light source to an existing ceiling lamp controlled by a wall switch, you may find that stringing a second power line is impossible. First, you can replace the wall switch by the circuit in Figure 1. Pushing the on switch S1 or S2 for approximately 1 sec inserts the 12 V zener diodes D1 or D2 in series with the hot wire of the power line. During the push, the polarity-dependent conduction of the zener diodes creates a small positive (negative for D2) dc component across the line and only slightly reduces the line's 120 V-ac component. A control circuit at the lamps' site reacts selectively to the polarity of this dc pulse and controls the power to the two lamps. The required power rating of the two zener diodes depends on the load current. The short duration and low duty cycle of the activation are helpful. The 1N2976 diodes in Figure 1 are rated for continuous dissipation of 10 W.

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This circuit creates dc pulses for use with control circuitry located at the load.
Figure 1. This circuit creates dc pulses for use with control circuitry
located at the load.

Figure 2 shows the first part of the control circuit located at the lamps' site, including the two leads of the power line, W1 and W2. Current through capacitor C1 and resistor R1 creates a 60-Hz square wave across the 6 V zener diode, D3. Diode D1 and filter capacitor C2 generate a dc supply voltage of VDD = 5 V for the control circuit's active elements. Two two-stage RC filters connected to W2 create V1 and V2, with reference to W1. The filters attenuate the 120 V-ac voltage between W1 and W2 to a subvolt level in V1 and V2. An extra zener diode, D4, creates a positive 5 V dc bias in V2. The filter outputs V1 and V2 drive inverting Schmitt triggers IC1 and IC2. Inserting zener diode D1 by pushing S1 in Figure 1 changes V1 from 0 V to 5 V and V2 from 5 V to 10 V. Inserting D2 in Figure 1 by pushing S2 changes V1 from 0 V to –5 V and V2 from 5 V to 0 V. Note that the input-protection diodes of IC1A and IC1B limit the voltage swings of V1 and V2. The output V3 of IC1C responds to pushing S1 by a positive transition and has no response to pushing S2. The output V4 of IC1B responds to pushing S2 by a positive transition and has no response to pushing S1.

This control circuit uses dc pulses from the circuit in Figure 1 to drive triacs in the circuit in Figure 3.
Figure 2. This control circuit uses dc pulses from the circuit in Figure 1 to drive triacs in the circuit in Figure 3.

Figure 3 shows the second part of the control circuit located at the lamps' site. Signals V3 and V4 in Figure 2 drive the clock input of toggle flip-flops IC1A and IC1B, respectively. For clarity, Figure 3 doesn't show the connections of the flip-flops of /Q to D and the Set terminal to VDD. When you push switch S1 in Figure 1, the positive transition in V3 toggles flip-flop IC2A. Similarly, when you push S2, the positive transition in V4 toggles flip-flop IC2B. Thus, you can independently control the states of flip-flops IC2A and IC2B by pushing S1 and S2, respectively. To drive the two lamps, the Q outputs of the flip-flops drive the gates of triacs TR1 and TR2 via coupling resistors R2 and R3. The MT2 terminal of each triac drives the lamps, L1 and L2, respectively. Pushing S1 changes the state of lamp L1; pushing S2 changes the state of lamp L2. Thus, you have independent control of both lamps on a single power line. In this application, you want to keep each lamp's terminals safely connected to the hot and neutral wires. Therefore, you make W1 the hot wire and W2 the neutral wire.

Triacs independently control two loads based on signals from a pair of wall switches.
Figure 3. Triacs independently control two loads based on signals from a pair of wall switches.

With the control circuit, the state of the flip-flops becomes uncertain if, after an interruption, the ac power returns. This situation is unacceptable because the lamps could turn on and stay on for an uncontrollable length of time. Therefore, you add a power-up reset circuit (Figure 3). To guarantee a safe reset also for short interruptions, the reset circuit must quickly pull down the flip-flops' Reset terminal, which is independent of VDD's slowly dropping level. Diode D2 (driven by the 60-Hz square wave across zener diode D4), capacitor C3, and resistor R4 act as an auxiliary rectifier supplying voltage V5. When power experiences an interruption, V5 drops to 0 V much faster than VDD. V5 drives the cascade of inverting Schmitt triggers IC1D and IC1E, which then quickly pull down the flip-flops' Reset terminals via diode D3. When power returns, the Reset terminals pull up slowly via resistor R5. The R5C4 time constant guarantees that the flip-flops' Reset does not release before VDD reaches its full value.

Note that you can use this Design Idea in other applications. For example, if you omit the circuit in Figure 3, the transitions in V3 and V4 can drive an up/down counter that can perform an auxiliary control function for a device that the ac line powers. If you insert an additional conventional switch in series with the circuit in Figure 1, it can turn on and off the power to the device. If the application requires control signals near ground level, wire W1 should be the neutral lead of the power line, and W2 should be the hot lead. However, make sure that the powered device is not an inductive load because it can short out the controlling dc pulses.

Materials on the topic

  1. Datasheet Naina Semiconductor 1N2976B
  2. Datasheet Digitron 2N6072B
  3. Datasheet Texas Instruments CD74HC14

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