How to stop a Reverse Blocking Thyristor ?

I have a small board with 2 small push button tactile switches. The intention is to have them control an LED light : 1 switch for 'on' and one switch for 'off'.

The circuit is 12V DC.

Actually there are a number of LEDs, but for the problem at hand, we'll keep it simple and say there is only 1.

The 'on' switch is connected to a C106D1G Reverse Blocking Thyristor ( Gate pin ). The Cathode pin connects to ground, and the Anode pin to the LED, the LED to a resistor, and the resistor to 12V+.

My desire is to have the 'on' operation a simple push of the button ( as opposed to a toggle switch ). The setup as I have it does work for the 'on' operation.

What I don't know how to do, is to wire the bottom tactile switch to interrupt the circuit between the Ground and the Cathode pin.

Is there any way I can use basic components ( resistors / transistors / etc ) to make button 2 interrupt the Ground signal to the Gate ?

Perhaps you could use a MOSFET instead of a Thyristor. Then you could use a simple bi-stable transistor circuit to control it.

Put second switch between anode and cathode of thyristor. If you press it, current will be rerouted from thyristor, causing it to goes in "off" state.

Magician: Put second switch between anode and cathode of thyristor. If you press it, current will be rerouted from thyristor, causing it to goes in "off" state.

Magician, you deserve your title. MAGIC solution. Thank You.

Thanks also to John for the idea.

Magician:
Put second switch between anode and cathode of thyristor. If you press it, current will be rerouted from thyristor, causing it to goes in “off” state.

OK. So now I need to expand the circuit ( as I said before, there are a number of LEDs ) to 2 LED lights, but with a common ‘off’ button.

Please excuse the basic diagram attached.

If I wire the “Lights Off” button to the Cathode and Anode of both thyristors (as in the diagram), won’t it cause both thyristors to activate when either of the top 2 ‘on’ buttons is pressed ?

diagram15.jpg

It would. To prevent it, create a diodes logical “OR” as on picture.
3 & 4 goes to individual thyristors anodes, switch instead of resistor.
Diodes have to be Schottky, it’s not clear how much current your LED consume.
If it too low, voltage drop across thyristor would be small, compare to voltage across diode,
and thyristor wouldn’t go to “off”.
In this situation, individual low power mosfet across each thyristor necessary. All gates MOSFETs combine together and connect to ← switch → to +12V rail.

250px-Diode-OR2.png

Hi Magician.

The details of the LEDs are, for the first switch :
3 x 4 array uses 12 LEDs exactly
Source +12.43V
4 sets of 3 LEDs with 150 ohms resistor on each set.

  • each 150 ohm resistor dissipates 60 mW
  • together, all resistors dissipate 240 mW
  • together, the diodes dissipate 768 mW
  • total power dissipated by the array is 1008 mW
  • the array draws current of 80 mA from the source.

The details of the LEDs are, for the second switch :
3 x 1 array uses 3 LEDs exactly
Source +12.43V
1 set of 3 LEDs with 150 ohms resistor.

  • the 150 ohm resistor dissipates 60 mW
  • the diode dissipate 192 mW
  • total power dissipated by the array is 252 mW
  • the array draws current of 20 mA from the source.

I have 1N4007 diodes : are they suitable ?

Alternatively, I have some 2N3904 transistors - could those be of any use ?

From what you said in your previous post, I’ve modified what I “think” you suggested, but can’t see how this would prevent both sets of lights going ON when either button is pressed.

Addition : sorry, but thinking about what you said, is the following ( second attachment ) diagram correct ? I think the Anode to the LEDs has to go to the LEDs before the addition of the diode ?

diagram15b.jpg

diagram15c.jpg

The schematic 15c almost correct, you only "reversed" diodes. Diode 1N4007 is regular silicon p-n junction, and it difficult to say if it forward voltage low enough compare to thyristors Von. Can you measure voltage across a thyristor when it's on with DMM? Data sheet has not included this info. I'd suggest, make quick test, turn on led , than put a diode across anode-cathode of the thyristor and see if it goes off, if not, reverse a diode and try again. If it's not working, you can built "interrupt" switch as you suggest initially:

switch to interrupt the circuit between the Ground and the Cathode pin.

with one transistor and resistor, 1 - 10 k ohm from +12 to base, so it's "on" all the time, emitter to ground, collector to cathodes. Off-switch connected to base-emitter.

Hi Magician

OK. Solution completed.

I did the test as you suggested with the diode over the A-C pins of the thyristor, but no result there.

Next I added a 2N3904 transistor as suggested.

This nearly worked, but had a strange side effect that I do not understand. If I turn on ‘light 1’, the LEDs light up. If I then press the top button for ‘light 2’, the LEDs for light 1 go off, and the LEDs for light 2 come on. Would be a great feature if I wanted to turn on either LEDs “1” --OR-- LEDs “2”.

My setup is that the middle button for ‘light 1’ is to power only 3 out of a line of 15 LEDs. The top button ‘light 2’ is to turn all 15 LEDs on.
So ‘light 1’ is used for emergency lighting when running off backup power, and ‘light 2’ for normal full mains power usage.

I overcame the side effect by adding a 1N4007 diode between the 2 anode links to the LEDs.

So, the ‘light 1’ button turns on 3 LEDs, and the ‘light 2’ button turns on the same 3 as well as another 12 LEDs.

Attached is the final drawing, hoping that someone else can find use of it.

Again, my sincere thanks for your help, advice, explanations and guidance - without you experts around, I would be lost before I started.

Regards

diagram15d.jpg

Ooopppss, my mistake. I look for h21 of the transistor 300, but it maximum value. Minimum is 30. Change resistor to: 30 x 12V / 100 mA = 3.6 kOhm 10 k ohm is too much. Should works w/o diode.

Magician: Ooopppss, my mistake. I look for h21 of the transistor 300, but it maximum value. Minimum is 30. Change resistor to: 30 x 12V / 100 mA = 3.6 kOhm 10 k ohm is too much. Should works w/o diode.

Thanks Magician. Trying to understand the info you've given. Is the 100mA taken from the 'total current' that my LED lights will draw ? Is the '30' in your calculation the "DC Current Drain" from the datasheet ( in "On Characteristics" ) where Ic = 100mA ?

It does work without the diode, but then it illuminates either light1 or light 2, but not both at the same time. But that's with the 10K resistor in place. I have not tried it with the lower valued resistor.

I had already assembled the board with the 10K (and the diode) and it is working 100% as required.

Do you think that it would do any harm or long term damage to components to leave the 10K in place ?

The lights can draw either 20mA or 100mA (depending on which switch is on). Would this also affect the value of the resistor ?

Is the 100mA taken from the ‘total current’ that my LED lights will draw ?

Yes

Is the ‘30’ in your calculation the “DC Current Drain” from the datasheet ( in “On Characteristics” ) where Ic = 100mA ?

Yes

Do you think that it would do any harm or long term damage to components to leave the 10K in place ?

Not quite. But two negative effect:
1). Transistor dissipates power, as it stay in active area. Power less than it can tolerate,
P = 100 mA x 6 V < 625 mW if temperature 25 C, but care should be taken if T goes up to 75 C and above. Look at the chart on p.4
(6 V is rough estimate for maximum voltage = Vsuply - Vleds = 12 - 3 x 2 V = 6)
2). As some voltage drop across emitter-collector, current provided for leds and consequently their brightness would be less than pre-calculated, and it probably would fluctuate due temperature variation, that is better to avoid.

The lights can draw either 20mA or 100mA (depending on which switch is on). Would this also affect the value of the resistor ?

Yes, to keep transistor “saturated”:
I(base) should be more than: I(collector) / h21
I(base) = Vsupply / R; I(collector) = leds current
and R = h21 x Vsupply / I(leds)
I simplified equation, not taking in account Vb-e, but error less than 5%.

2N3904BU-Fairchild.pdf (111 KB)

Hi Magician

Thanks again for the reply. Unfortunately, a lot of the info that you gave in your last post is, at present, beyond my comprehension. I am working on it.

However, if I get the general gist, it boils down to the fact that a transistors current that it conducts ( E to C ) is dependant on the current supplied to the Base - in other words, it is Variable.

So the trick is to make sure that the transistor Base receives enough to power the 'light2' (100mA) circuit at full potential, but at the same time does not cook itself when 'light1' is used with only 20mA current draw through the E-C pins ?

If this is the case, and IF ( I am assuming here ) the C106D1G Reverse Blocking Thyristor has the feature of either ON or OFF ( as opposed to Variable ON as with the transistor ) would it not be better to replace the transistor with the Thyristor, keeping the Thyristor permanently on ( like we are doing to the transistor ) and have the 'off' switch connect the Cathode and Anode pins to de-activate the Thyristor ?

but at the same time does not cook itself when 'light1' is used with only 20mA current draw through the E-C pins ?

There is no danger when load consume less or 0 current. My advise, read more good book on electronics or i-net; http://en.wikipedia.org/wiki/Transistor

would it not be better to replace the transistor with the Thyristor, keeping the Thyristor permanently on ( like we are doing to the transistor ) and have the 'off' switch connect the Cathode and Anode pins to de-activate the Thyristor ?

It wouldn't turn off lights.

Interesting tutorial to read, that shows one more option to turn off thyristor using capasitor: http://www.sentex.net/~mec1995/tutorial/triacs/triacs.html