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.

I agree, there are lots of simpler ways to turn on and off an LED, a toggle switch comes to mind. However, I'm guessing the TS is trying to learn all about circuits so he has posted a question here about a circuit using SCR's and transistors.

 

I agree, there are lots of simpler ways to turn on and off an LED, a toggle switch comes to mind. However, I'm guessing the TS is trying to learn all about circuits so he has posted a question here about a circuit using SCR's and transistors.

Hi: The holding current for the C106 is in the range .3 to 3 ma. You need to get the forward current below this number for the Thyristor to commutate. I can't say for certain but it seems that the Vcesat you quoted (I.E. .2) volts should reduce the current in the Thyristor below this level. Ultimately, set the circuit up and measure the MT1/MT2 current when the transistor is switched on. Also measure Vcesat (The collector emitter voltage at the same time. This will ensure that the transistor is in fact in saturation.

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.

 

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.

Thanks.Very well explained how to find the right value.
From your explanation,may i understand that,while in the scr datasheet,the max Vf is 2.2V when the current is 1A,in small currents(mA)the Vf is about 700mV?
And i have noticed that when the transistor Vce was greater from 600mV,the Scr didn't
turn-off.Doesn't it mean that:since the minimum Vf that the scr can be is 700mV,the Vce of the transistor must be less than that,in order to prevent the scr turn-on,even less than 600mV inyour case?
For example:if instead of the transistor,i will connect a darlington pair as a switch and as it known,the front transistor can't be saturated as it can be minimum 700mV.Does it mean that i can't use darlington pair to switch off the scr,because that reason?

Where is 0.3mA coming from?
View attachment 137277
View attachment 137278

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...

This is what the datasheet showing:

I agree, there are lots of simpler ways to turn on and off an LED, a toggle switch comes to mind. However, I'm guessing the TS is trying to learn all about circuits so he has posted a question here about a circuit using SCR's and transistors.

Yes,i know about the toggle switch,but as you said,i am trying to understand how to do it with the transistor.

 

Hi: The holding current for the C106 is in the range .3 to 3 ma. You need to get the forward current below this number for the Thyristor to commutate. I can't say for certain but it seems that the Vcesat you quoted (I.E. .2) volts should reduce the current in the Thyristor below this level. Ultimately, set the circuit up and measure the MT1/MT2 current when the transistor is switched on. Also measure Vcesat (The collector emitter voltage at the same time. This will ensure that the transistor is in fact in saturation.

If an actual SCR matches the discrete complementary pair equivalent - the voltage the SCR pulls down to should be the VCEsat of the NPN structure in the 4-layer device plus the Vbe of the PNP structure. IWHT: even a mediocre switching transistor should get the job done.

A MOSFET has RDSon instead of VCE sat, so RDSon and RL form a potential divider - all you need is to make sure the voltage divides down low enough so the SCR is cut off.

If the TS is leaving the easy way (that I already suggested) to learn electronics - they might as well learn the industry standard commutating second SCR (which I also suggested).

 

Thanks.Very well explained how to find the right value.
From your explanation,may i understand that,while in the scr datasheet,the max Vf is 2.2V when the current is 1A,in small currents(mA)the Vf is about 700mV?
And i have noticed that when the transistor Vce was greater from 600mV,the Scr didn't
turn-off.Doesn't it mean that:since the minimum Vf that the scr can be is 700mV,the Vce of the transistor must be less than that,in order to prevent the scr turn-on,even less than 600mV inyour case?
For example:if instead of the transistor,i will connect a darlington pair as a switch and as it known,the front transistor can't be saturated as it can be minimum 700mV.Does it mean that i can't use darlington pair to switch off the scr,because that reason?

This is what the datasheet showing:
View attachment 137331

Yes,i know about the toggle switch,but as you said,i am trying to understand how to do it with the transistor.

HI: You are looking at the wrong parameter. The Holding Current is really what matters. I believe this is a thyristor so the current in the MT1/MT2 leg must be below the Iho of the particular device before it will commutate I.E. shut off. I thought from the schematic, the device was a BJT. It really doesn't matter as long as the voltage either Vds or Vce is low enough to drop the holding current below the value for the particular device you are using. Remember that once triggered on from a DC source, the SCR or Thyristor will remain in conduction until the holding current falls below Iho.

 

Thanks.Very well explained how to find the right value.
From your explanation,may i understand that,while in the scr datasheet,the max Vf is 2.2V when the current is 1A,in small currents(mA)the Vf is about 700mV?
And i have noticed that when the transistor Vce was greater from 600mV,the Scr didn't
turn-off.Doesn't it mean that:since the minimum Vf that the scr can be is 700mV,the Vce of the transistor must be less than that,in order to prevent the scr turn-on,even less than 600mV inyour case?
For example:if instead of the transistor,i will connect a darlington pair as a switch and as it known,the front transistor can't be saturated as it can be minimum 700mV.Does it mean that i can't use darlington pair to switch off the scr,because that reason?

This is what the datasheet showing:
View attachment 137331

Yes,i know about the toggle switch,but as you said,i am trying to understand how to do it with the transistor.

The effective VCEsat of a Darlington is the VCEsat of the front transistor plus a Vbe of around 0.7V - about the same as the VCEsat of the inherent NPN structure in the 4-layer device and the 0.7V Vbe of the PNP structure. The Darlington VCEsat is a bit of a grey area and I wouldn't risk it anyway.

Measure the on state voltage of the SCR and see what voltage you have to cut off - Typical VCEsat is around 0.4V but you can obtain special types with lower. The big thing about MOSFETS is; they don't have VCEsat.

 

Thanks alot to all the participants in thread.
I think that i understand,more all less,the idea.
Thanks.