The electronic analogue of the push-button switcher allows you to selectively turn on one or another load from several possible ones, while simultaneously disabling the previously turned on load of another control channel. Unlike a mechanical switcher, it is possible to turn off the previously selected load by pressing the control button again.

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The electronic device described in the article can replace a multi-button switcher. The electronic analogue of the push–button switcher uses bistable elements (Figure 1), which are turned on by applying a high-level voltage to the input of the element and turned off by applying a low level.

Figure 1 Electrical circuit of a bistable element controlled by the discharge of the capacitor C1 to the resistance R1.

To turn on and off the bistable element, the discharge of the capacitor C1 to the resistance R1 is used. As shown in Figure 1, during the initial moment of the discharge process, a short press on the control button S1, turns on the bistable element and the load resistance. A longer press on the control button S1, turns off the load.

The S1 button press duration that is sufficient to turn off the load is defined as toff[sec] ~ 2R1[Ohm]C1[Farad] and, at the nominal values indicated in the diagram, is 0.15 seconds.

In order to be able to control the switching of one of several loads using several control buttons, a dependent load switching scheme is used. Figure 2 shows a diagram of a three-button analog of a mechanical load switcher.

Figure 2 Electrical diagram of a three-button electronic analog of a mechanical load switcher.

This electronic analogue of the push–button switcher allows a short press on one of the control buttons S1–S3 to turn on the selected load Rload1-Rload 3, simultaneously disabling other loads. Unlike mechanical switches, pressing the control button for a longer time will ensure that the previously selected load is turned off.

The circuit in Figure 2 uses 6 diodes D1–D6. In general, a circuit switching with n loads will require n bistable elements (See Figure 1) and n(n–1) diodes. So, for n=1, diodes are not needed; for n=2, it is necessary to use 2 diodes; for n=3, 6 diodes are needed; for n=4, 12 diodes are required, etc.

If diodes D1–D6 are removed from the device in Figure 2, the electronic analogue of the mechanical switcher will be able to turn on and off each of the loads Rload1–Rload3 independently of each other by pressing the corresponding button S1–S3.

What’s missing from Figure #1? What is the element in the Box labeled “1”?

Apparently the mystery box contains a CD4050 non-inverting buffer.

But what interests me is the transistor Michael has apparently found that has saturated beta (~1000?) sufficient to sink 10s of mA sufficient to drive an LED and a relay, with only 10s of uA of base current input!

Stephen, that’s right. At a supply voltage of 10 V, the base current is 10 V / 100 kOhm = 0.1 mA. Then, at beta = 100, the collector current is 10 mA.

But your article mentions no specific E+.

At an (equally plausible) supply voltage of 5V, the base current would be (5 – Vbe)/100k = ~43uA, and the operating current of a typical (small) 5V relay would be ~50mA. Add something (10mA?) to light the LED and the necessary beta becomes 60mA / 43uA = 1395

Stephen, of course, at a supply voltage of 5 V, the base current will be (5 – Vbe) / 100k = (5 – 0.7) / 100k ~ 43 uA, and the load current at beta = 100 will be ~ 5 mA. It is quite obvious that if it is necessary to increase the maximum current in the load, it is necessary to reduce the resistance of the resistor R2.

Why not connect the transistor’s collector to E+, put Rload between emitter and ground, and eliminate R2 altogether? As a bonus you could also then do without the diode, thus eliminating two components.

Nevermind. I see that wouldn’t work in fig. 2.

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