Hi.
In this circuit,pressing on sw1 will trigger the scr and the led(10mA)will turn-on.
In order to turn off the scr(and the led),i use a bjt transistor,which bypass the scr.
I understand that when the transistor will be saturated(0.2Vce)all the current will bypass the scr and the scr will turn-off.My question is:What is the max Vce value that i can plan for the transistor which will still be able to turn-off the scr?(/making the current flow through it and not through the scr?).
Thanks.

 

Why not skip the transistor and have Switch 2 connect between the SCR and ground (just like the CE is doing in the transistor)?

 

Just put the switch across the A K thyristor terminals.

 

Hi btebo.
You right,it is possible,but i am interesting more in the calculation's theory aspect of the circuit.I mean the aspect of the maximum Vce that the transistor can be,which more than that,the scr won't turn off.

 

I'm just a beginner but I would suspect the max Vce need would be about 7.8V... When the transistor is on, it will connect the LED through the transistor. Assuming a Vf of the LED is about 1.2V, 9V - 1.2V = 7.8V.

I also would like to be corrected if I'm wrong by some of the experts on this site!

 

What is the max Vce value that i can plan for the transistor which will still be able to turn-off the scr?

About a diode drop.

 

It is not possible to calculate exactly what you are asking for based on the part data sheets. The required information is not in there.
However, knowledge is power. Learn about how an SCR works, it is made of 2 PN junctions in series and they are not saturated when on. Therefore Vtm will always be much higher than Vce(sat) so you can have high confidence that your circuit will work. An SCR will unlatch when the current through it goes below the minimum required to hold it latched. Look at the spec, what is that minimum?

SCR's have a minimum latching current requirement. You have mentioned that the LED current is 10mA. Check the spec, is that enough to ensure the SCR will latch on when triggered?

Ifixit

 

About a diode drop.

Do you mean that if the Vce of the transistor will be more than 0.7V,the current will keep flowing through the scr,but if the Vce will be less than 0.7V,for example:0.5V,the scr will turn-off?

...Therefore Vtm will always be much higher than Vce(sat) so you can have high confidence that your circuit will work...

Yes,in a saturate state,but my question is to figure out,how much
can i maximum increase the Vce(not saturated state)of the transistor,until the point that the current will start/prefer to flow through the scr?

...An SCR will unlatch when the current through it goes below the minimum required to hold it latched. Look at the spec, what is that minimum?

SCR's have a minimum latching current requirement. You have mentioned that the LED current is 10mA. Check the spec, is that enough to ensure the SCR will latch on when triggered?

The minimum holding current is about 0.3mA.
btw,i had already read the datasheet of the C106 SCR and i know that i must
reduce the current to the scr below 0.3mA and it will happen,probably,when the transistor will be saturated,but main point is: how much
can i maximum increase the Vce(not saturated state)of the transistor,until the point that the current will start/prefer to flow through the scr or in other
words,until the point that the current through the scr will be over 0.3mA?

 

Do you mean that if the Vce of the transistor will be more than 0.7V,the current will keep flowing through the scr,but if the Vce will be less than 0.7V,for example:0.5V,the scr will turn-off?

No. It depends on the current in the SCR.

Using this image I pulled randomly from the web:

The voltage between anode and cathode needs to be low enough to turn off TR1.

Unless you're using a sensitive gate device that can be turned off with a pulse of the appropriate polarity on the gate.

 

As I said before, it is not possible to calculate exactly what you are asking for based on the part data sheets. The data sheet says the Vtm will be 2.2V max at 4 Amps and at 25C. That voltage will be less than that at 10mA but we don't know what that value is. The transistor Vce will be this value as well as it begins to conduct and take the 10ma current away from the SCR.

I think you are asking the wrong question, this is about currents, not about Vce or Vtm voltages. When Ice is > 9.7mA (10mA minus Ih typ) then the SCR will unlatch.

Ifixit

 

View attachment 137199
Hi.
In this circuit,pressing on sw1 will trigger the scr and the led(10mA)will turn-on.
In order to turn off the scr(and the led),i use a bjt transistor,which bypass the scr.
I understand that when the transistor will be saturated(0.2Vce)all the current will bypass the scr and the scr will turn-off.My question is:What is the max Vce value that i can plan for the transistor which will still be able to turn-off the scr?(/making the current flow through it and not through the scr?).
Thanks.

You can order special ultra low VCEsat transistors, but I've never seen transistors used this way.

The old industry standard way is to add a commutating thyristor. a second SCR with a resistor load is cross coupled by the commutating capacitor between the 2 anodes.
When your load SCR is on, that end of the capacitor is pulled down. the commutating SCR is off, so its end of the cap is pulled up to Vcc.

When you trigger the commutating SCR, it pulls the positive charged end of the commutating capacitor down - as long as the charge lasts; the other end of the commutating capacitor tries to go below GND.

For all intents and purposes - the circuit is symmetrical. firing the load SCR resets the commutating SCR just as it does to the load SCR.

 

The transistor needs to be rated for a VCE of 9V or more because that's the voltage across the collector and emitter when the SCR is not on.

 

The transistor needs to be rated for a VCE of 9V or more because that's the voltage across the collector and emitter when the SCR is not on.

Every once in a while I encounter esoteric RF power transistors rated as low as 12V..................

Normally the lowest I find is power MOSFETs on PC motherboards - occasionally as low as 20V, but 30V seems to be the norm.

Those MOSFETs don't have VCEsat, the on RDSon is usually a few tens of milli Ohms. The older ones were cropped tab SMD versions of TO220 with Id rating as high as 95A. The developing trend is for more energy efficiency and smaller power MOSFETs - the size down from the TO220 package can take less footprint than a TO92. The last ones I salvaged look similar to SOIC8, I got the data sheet and remember they're around 30 - 50A, but didn't memorise the package designation.

 

You right,it is possible,but i am interesting more in the calculation's theory aspect of the circuit.I mean the aspect of the maximum Vce that the transistor can be,which more than that,the scr won't turn off.

Hi
It seems to be your interested more in designing a circuit to do this kind of process, more than just a solution of how to make it work.

So here's an example of how this was designed to your specs.
You could look over this and see the processes involved in designing a unique circuit.

Step 1) I breadboard the SCR part of the circuit. Using your value of 9 volts for VCC. Assuming a 2 volt drop across the LED, and 700mV across the SCR, I calculated a value of 620 ohms for R2.

Then I fired on the SCR, and took a voltage drop reading across R2 to verify around 10mA current.

step 2) Then I took the 700mV across SCR and divided 10mA into it to give me a apparent resistance of around 70 ohms for the SCR.
Then I place across the SCR a 75 ohm current sink rersistor to see if it will shut off the SCR, it didn't, so I decreased it to 62 ohms, and the SCR was turned off.

Now this next step is very important, very easy to overlook it.
Since you are desinging this trasnsitor circuit to NOT be a switch but a current sink at a threshold voltage, very minimal amount of current sink, then you need to follow these next steps.

I took a voltage reading across the current sink resistor, (650 mV) then I measured the resistor value (61.7 ohms) now calculate the current through the resistor (650mV / 61.7 ohms ) = 10.53mA

That means I need a transistor to sink 10.53mA and have a volt drop of 650mV from collector to ground, which puts it in bypass of the SCR.

Now your design specs are that the transistor shall not saturate.
Ok so I'll choose to put 600mV across the transistor (Vce) and the remaining voltage drop (50mV) across an emitter resistor, so as to establish a bias current needed to be the sinking current through the load.

So that emitter resistance is calculated as (50mV / 10.53mA) = 4.7 ohms.
Now since some of that current will go through the base circuit, I need to supply more current from the emitter, so instead of 4.7 ohms I'll choose 4.3 ohms resistor for the emitter bias resistor.

Now I want to bias the transuistor on when it is switched on by a base voltage divider.
So I take that (10.53mA x 4.3 ohms) = 45.2mV. which is the emitter voltage.
Then I add the Vbe of 700mV to that (45.2mV + 700mV) to give a base voltage of 745.2mV.
next I choose a base to ground resistor that is 100 times the emitter resistor which is 430 ohms. (assuming a beta dc of around 100)

Now calculate for the top pullup resistor, fromn base to supply by, first solving for the divider current. The base voltage of (745.2mV / 430 ohms ) = 1.73mA for divider current.

Now take the 9V supply voltage and subtract the base voltage (9V - 745.2mV) = 8.25V dropped across this pullup resistor.
Now take this (8.25V / 1.73mA) = 4.7k ohms.

Now I built it using those values, and the SCR was not turned off, the reason is because of the tolerances in resistor values. So this is where you have a ballpark figure of resistor values and its just a matter of tweaking the base resistors until desired results.

So thats what I did, I held my volt meter across the collector emitter terminals of the transistor, than began tweaking the supply pullup resistor down in value because the Vce was greater than 600mV target value, so I ended up with a trimmer value of around 4530 ohms to put around 600mV across the transistor when it was switched on.

So performance is as follows the SCR is fired on and remains on, the switch to the base of the transistor current sink is pulsed on, the LED remains on as long as this switch is closed, when the switch is opened the LED is extinguished, and the whole system is ready for another input.

So this is the process that can be taken to design a circuit you were asking to be done, how to design for a non saturated transitor switch, to turn off an SCR through its load.

 

See. The Necessary additional resistor. Without resistor of no shutdown.

 

...I think you are asking the wrong question, this is about currents, not about Vce or Vtm voltages. When Ice is > 9.7mA (10mA minus Ih typ) then the SCR will unlatch.
Ifixit

I agree with you,that the current is the main player here,as if the current through the scr will be less than 0.3mA the scr will unlatch.
But i understand that the Vce of the transistor has an influence about that.
If The Vce will be at the wrong value the current will not flow through it with
sufficient amount to unlatch the scr.So,when you said:

...The data sheet says the Vtm will be 2.2V max at 4 Amps and at 25C. That voltage will be less than that at 10mA but we don't know what that value is.The transistor Vce will be this value as well as it begins to conduct and take the 10ma current away from the SCR.

Did you mean that:if the Vf of the scr will be 0.8V,for example,with the 10 mA and the Vce of the transistor will be 0.9V the current will not flow through the transistor at all.if the Vce of the transistor will be 0.8V as the Vf of the scr,some current will start to flow through the transistor,but not enough current to unlatch the scr.And if the Vce of the transistor will be 0.75V,for example,all the 10mA current will flow through the transistor and the scr will turn-off.Is that what you meant?

 

You're getting close. The voltage across the SCR (Vf) is always equal to the voltage across the transistor (Vce), after all, they are tied together. What changes is the current, as the transistor turns on the collector current increases and the SCR current decreases until the SCR unlatches. Vf and Vce will be whatever they will be, I would guess them to be around 1.4 Volts with all 10mA flowing in the SCR and approximately 1 volt with 10mA flowing in the transistor.

Review transistor theory, base current controls collector current (Ice), not collector voltage (Vce).

 

I agree with you,that the current is the main player here,as if the current through the scr will be less than 0.3mA the scr will unlatch.

Where is 0.3mA coming from?



How do you know if you have a typical device versus one that will hold at a lower current? Or what temperature the circuit will be operating at? At Tj=110C, holding current could be less than 0.07mA...

 

How do you know if you have a typical device versus one that will hold at a lower current?

When designing a circuit to turn off a latched SCR you should assume that it is required to reduce the current to zero to guarantee unlatching.

xchcui: Don't forget to put a 1K resistor from gate to cathode to ensure the SCR does not re-latch when the transistor is turned off. Refer to post #15.

 

When designing a circuit to turn off a latched SCR you should assume that it is required to reduce the current to zero to guarantee unlatching.

xchcui: Don't forget to put a 1K resistor from gate to cathode to ensure the SCR does not re-latch when the transistor is turned off. Refer to post #15.

No one seems to have spotted that the SCR can be unlatched by breaking the supply current - push to break buttons aren't all that hard to find, and it saves a transistor.