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I was reading about Silicon Controlled Switches SCS's from this book "Practical Electronics For Inventors", it mentioned that an SCS is equivalent to two BJT transistors illustrated in the following figure (section C on the right):

enter image description here

So I decided to make that equivalent circuit to see how it works, this is the layout I used:

schematic

simulate this circuit – Schematic created using CircuitLab

However when I tested the circuit the led turned on from the moment I connected the battery without even pushing any buttons:

enter image description here ![enter image description here

The led was supposed to remain off until I push the SW1 button and that after I push SW1 the led should remain on until I push SW2 which doesn't. When I push SW2 the led switches off and then on again after I release SW2.

Why is it not working?

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    \$\begingroup\$ I can't see easily if your circuit is wired correctly, but try connecting 100K from base to emitter of Q1 and see if the problem persists. \$\endgroup\$
    – Spehro Pefhany
    Oct 24, 2016 at 3:00
  • \$\begingroup\$ @SpehroPefhany Amazing.. I did as you suggested and put an 80k resistor (I don't have a 100k) between Q1's emitter and base and it worked! What did it do? I added another image to show the connections hope it'd show the connections better. \$\endgroup\$
    – razzak
    Oct 24, 2016 at 3:19
  • \$\begingroup\$ @SpehroPefhany It also remembers last switch state when I connect and remove the battery, like if the led was on before removing the battery it wud be on when i reconnect it and vice versa. How does it do that?! \$\endgroup\$
    – razzak
    Oct 24, 2016 at 3:23

3 Answers 3

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There is an uncontrolled effect or two in your circuit as shown. Many transistors have very low gain at low Ic, and low leakage, so the leakage of Q1, multiplied by the current gain of Q2, added to the leakage of Q2, multiplied by the current gain of Q1 ad nauseam does not 'run away' and cause the compound device to turn on. In your case, the gain at nA currents and leakage combination is enough to cause the device to turn on. This is normally well controlled in a SCS device so that it won't turn on even at maximum temperature when leakages are high. This may occur in thyristor devices if the device is heated out of the normal operating range.

Another effect that can cause undesired turn-on is the dv/dt as power is applied. This effect will usually be quantified on the datasheet of a real device. Any capacitance (and your breadboard has many pF) will cause a bit of current to flow into the base nodes when power is applied, the more rapidly it is applied the higher the current. If dv/dt is too high then the device can turn on immediately. You also see this effect in SCRs and triacs.

Both these effects can be controlled by shunting away a bit of base current to control the turn-on current. Real thyristor devices often have trigger currents in the hundred of uA, mA or even higher. The base voltages have to be some hundreds of mV before the device will turn on. If you put 1K resistors in each BE position you'll have a very robust device that takes a few hundred uA to turn on.

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  • \$\begingroup\$ Thanks for your answer, I never thought that leakage current can has such an effect, sorry if this question sounds nooby but how does the leakage current turn Q1 on when there's no Vbe of minimum 0.6V at its base? also the 1K resistors between bases and emitters worked perfectly, I'm wondering though how one could choose the resistance values to combat these effects? \$\endgroup\$
    – razzak
    Oct 26, 2016 at 11:44
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    \$\begingroup\$ Imagine a high value resistor representing the leakage connected between emitter and collector of Q2. The added real resistor makes a voltage divider so the base voltage does not get too high. A few hundred mV is enough for the transistor to start to turn on. The resistor value determines the current required to do that, so both the trigger and the holding current. \$\endgroup\$
    – Spehro Pefhany
    Oct 26, 2016 at 11:51
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I believe that the load (your LED / resistor) should be where you have SW2, upstream of Q1. Q1 and Q2 will act as a latching switch, driving the base of each other. Q1 needs to be able to pull the base of Q2 low to latch the switch. This is why the load must be moved.

The control will be done where you have SW1. You need to be able to pull that node high (the way you show it now), and low (not implemented in your circuit).

Dave Jones actually has a pretty nice discussion on this type of implementation at https://www.youtube.com/watch?v=Foc9R0dC2iI.

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All you must do is shunt leakage current Ice with Rbe across each Q so that it is low voltage such as 1M at low Vdc and 100k to 10k depending on sensitivity and leakage current for high Vdc with bigger R1

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