Transistor working region - MMBTA06

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
Hi all,

I have a transistor which need to act as a switch.

This is my transistor : https://www.onsemi.com/pub/Collateral/MMBTA05LT1-D.PDF - MMBTA06

I need to make sure that this works in either saturation or cut off region only. It must not work the active region.

In order for the transistor to work in the active region, the Base emitter region should be forward biased and the base collector junction should be reverse biased.
So, if I donot maintain the above condition, the transistor would not enter in the active region, right?

Is there any graph there in the datasheet which is indicative of the maximum limits(Ic vs Vbe sat graph, of Vce vs gain graph etc..) to make sure that I don't enter the active region of the transistor? Please help me understand.
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
hi EM,
Its helpful if you post an image of the section of the circuit showing where the 'switch' is going to be connected and how it is controlled.
E
Hi,

This is my circuit. I need to control the transistor connected directly to the LED string. Input Voltage is 23V. Remitter is 33E.
upload_2019-1-9_15-36-19.png
 

DickCappels

Joined Aug 21, 2008
10,152
From the fact that you have a current regulator (the lower transistor) you want the transistor with the LEDs in the collector circuit to be in its linear region all the time so the current can be regulated rather turned on or off. If the transistor is knep in saturation then the LEDs will work safely over a very small range of power supply voltages. Not a good way to run LEDs except in special cases.

upload_2019-1-9_17-13-20.png
You can see that as the base current increases the collector voltage decreases. Use the collector current you are most interested (in the circuit you posted the current will be approximately Vbe/Rsense = 0.7V/33 = 32 ma (very hard to read the resistor value because of the resolution of the schematic image so please correct if I got that wrong.). If indeed it is 32 ma you should use the second curve from the left for guidance, but wait! There is good news -since the bottom transistor regulates the emitter current (and thus the collector current) at 21 ma and you can see that tvoltage changes significantly with current at 21 ma, so the circuit should work as is.

(Edit:) The chart shows the collector voltage entering saturation as the collector voltage vs current becomes close to being asymptotic to the horizontal.
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
hi,
Please mark in the component values on this clip, the original is impossible to read.
A quick look suggests that the transistor is wired as a current limiter, not a switch.??
E
View attachment 167430
I am sorry. I am trying to understand the function of the circuit. I think it is allowing the current through the LED when ON and not allowing when it is switched off. So, it thought the transistor should act as a switch(so it must be either in the cut off or saturation region). And since, in the transistor active region will make the transistor work as an amplifier, and since we dont require any amplification of current in this section, I thought it must work as a switch only. I didn't understand how you are saying it as a current limiter. Kindly excuse. Please let me know how the transistor is working as a current limiter in the circuit. I have attached the image for better resolution
 

Attachments

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
From the fact that you have a current regulator (the lower transistor) you want the transistor with the LEDs in the collector circuit to be in its linear region all the time so the current can be regulated rather turned on or off. If the transistor is knep in saturation then the LEDs will work safely over a very small range of power supply voltages. Not a good way to run LEDs except in special cases.

View attachment 167429
You can see that as the base current increases the collector voltage decreases. Use the collector current you are most interested (in the circuit you posted the current will be approximately Vbe/Rsense = 0.7V/33 = 32 ma (very hard to read the resistor value because of the resolution of the schematic image so please correct if I got that wrong.). If indeed it is 32 ma you should use the second curve from the left for guidance, but wait! There is good news -since the bottom transistor regulates the emitter current (and thus the collector current) at 21 ma and you can see that tvoltage changes significantly with current at 21 ma, so the circuit should work as is.

(Edit:) The chart shows the collector voltage entering saturation as the collector voltage vs current becomes close to being asymptotic to the horizontal.

Thank you very much for the clear explanation. But i was not able to understand that how the transistor works as a current regulator. Can you please explain me. And in the Vbe/Rsense, which transistor Vbe are you taking?
 

DickCappels

Joined Aug 21, 2008
10,152
(Edit:) Additional words in first sentence for clarification)
As the voltage across the emitter resistor of the transistor with the LEDs in its collector circuit approaches approximately 0.65 volts the bottom transistor starts drawing current away from the upper transistor, thereby regulating the emitter current to 0.7V/33 ohms. Here the emitter voltage is the one with the LEDs in the collector circuit.

If you really want the transistor to perform the function of an on/off switch you can use this:

upload_2019-1-9_17-44-22.png
This does not offer much regulation. The value of the resistors is

= ((total diode drop + Vce) - Power supply voltage) / emitter voltage.

Select a base resistor to get a base voltage/10 to assure saturation.
 
Last edited:

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
hi,
Where did you get the circuit, its configured as a constant current source with current control level drive.
It is not used as a switch.
E
So, it will not work in the cutt off/saturation region?

It will always work in the active region? But i thought the transistor will work in the active region only for amplifying purpose. but here we are not amplifying anything, right?
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
(Edit:) Additional words in first sentence for clarification)
As the voltage across the emitter resistor of the transistor with the LEDs in its collector circuit approaches approximately 0.65 volts the bottom transistor starts drawing current away from the upper transistor, thereby regulating the emitter current to 0.7V/33 ohms. Here the emitter voltage is the one with the LEDs in the collector circuit.

If you really want the transistor to perform the function of an on/off switch you can use this:

View attachment 167432
This does not offer much regulation. The value of the resistors is

= ((total diode drop + Vce) - Power supply voltage) / emitter voltage.

Select a base resistor to get a base voltage/10 to assure saturation.

Thanks. but why do we need to regulate the current in this circuit by using the transistor in the linear region?
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
hi,
It is working in the Active region because it needs to control the current flowing thru the LEDs, to a set current level.
E
but the resistor only regulates or sets the maximum current through the LED string right?

Please excuse for sub standard/silly questions. I am unable to understand clearly. please come a level down and explain a little more details for my benefit. thank you
 

DickCappels

Joined Aug 21, 2008
10,152
but the resistor only regulates or sets the maximum current through the LED string right?

(Some text removed for clarity)
Yes, that is correct. If you reduce the voltage to the circuit at some voltage the LEDs will dim because there is not enough voltage to drive them at full current. Above the voltage at which the LEDs draw the intended current the transistor to which the LEDs are connected will begin to absorb "excess voltage" to limit the collector current to the value set by the resistor on the emitter of the transistor to which its collector is connected.

The transistor current source section of this article (URL below) goes into some detail.
http://hobby.dapj.com/2010/12/drive-led-with-constant-current.html
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
Yes, that is correct. If you reduce the voltage to the circuit at some voltage the LEDs will dim because there is not enough voltage to drive them at full current. Above the voltage at which the LEDs draw the intended current the transistor to which the LEDs are connected will begin to absorb "excess voltage" to limit the collector current to the value set by the resistor on the emitter of the transistor to which its collector is connected.

The transistor current source section of this article (URL below) goes into some detail.
http://hobby.dapj.com/2010/12/drive-led-with-constant-current.html
thank you for the patient explanation. really helped me understand better. thank you
From the fact that you have a current regulator (the lower transistor) you want the transistor with the LEDs in the collector circuit to be in its linear region all the time so the current can be regulated rather turned on or off. If the transistor is knep in saturation then the LEDs will work safely over a very small range of power supply voltages. Not a good way to run LEDs except in special cases.

View attachment 167429
You can see that as the base current increases the collector voltage decreases. Use the collector current you are most interested (in the circuit you posted the current will be approximately Vbe/Rsense = 0.7V/33 = 32 ma (very hard to read the resistor value because of the resolution of the schematic image so please correct if I got that wrong.). If indeed it is 32 ma you should use the second curve from the left for guidance, but wait! There is good news -since the bottom transistor regulates the emitter current (and thus the collector current) at 21 ma and you can see that tvoltage changes significantly with current at 21 ma, so the circuit should work as is.

(Edit:) The chart shows the collector voltage entering saturation as the collector voltage vs current becomes close to being asymptotic to the horizontal.
Hi. MMBTA06 is the bottom transistor. I so badly mixed the part numbers. Forgive me. Upper transistor is BCX56

How much current will flow through the base of the bottom transistor so as to measure the Vce from the mentioned graph in the above thread. From what I calculated 0.7/33=21.2mA is the collector current in the upper transistor. So, when the voltage across the emitter resistor (33E) reaches above 0.7V, the bottom transistor will start to conduct right? So, the entire collector current of the upper transistor will now be the entire base current for the bottom transistor , right?
from my circuit, i calculated the collector current of the bottom transistor as, (5-0.25)/4700=1.01mA ((Voltage from micro - Maximum Vce of transistor in the datasheet)/(base resistance)).

So, for this base current and collector current for the bottom transistor, Ic=1.01mA and Ib=21.2mA, there is no plot in the graph as mentioned in the fig 9 of the datasheet of MMBTA06.

I am really sorry for these questions. If you find me really confusing, can someone tell me how to calculate the transistor parameters (both upper and bottom) like Vce, Vbe, Ic and Ib and which graphs we must check for these values in the datasheet. I am so confused. This will really help me
 
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