The BC547 and the MUN5211DW1 are completely different so that the schematic only look the same. The insertion of a pre biased dual transistor's abbreviated symbol into BC547 discussion on saturation is just confusing what could have been obvious became difficult by changing horses in midstream your.socks may get saturated.

https://www.rohm.com/electronics-basics/transistors/principles-digital-transistors

 

The BC547 and the MUN5211DW1 are completely different so that the schematic only look the same. The insertion of a pre biased dual transistor's abbreviated symbol into BC547 discussion on saturation is just confusing what could have been obvious became difficult by changing horses in midstream your.socks may get saturated.

https://www.rohm.com/electronics-basics/transistors/principles-digital-transistors

Quite Good Information....
Thanks a lot ...

 

Hello there,

One of the things you should do at least once is model the transistor as a current controlled current source with two diodes. One diode is the base emitter diode and the other diode is the base collector diode. This simple model provides an insight as to what happens when the transistor is in active mode, enters saturation mode, and continues in saturation mode. The different modes are determined by noting the bias of both diodes.

The possibilities are:
1. BE diode reverse biased, BC diode reverse biased.
2. BE diode reverse biased, BC diode forward biased.
3. BE diode forward biased, BC diode reverse biased.
4. BE diode forward biased, BC diode forward biased.

So you can see that there are four different possibilities because there are two diodes and two different ways of biasing each one.

Of these four #2 is rarely encountered except under unusual circuit conditions so the other three are more important to talk about at first.

For #1, the transistor is considered to be off.
For #3 the transistor is in the linear active mode.
For #4 the transistor is in saturation.

So you can see that the different modes are not that hard to determine in theory. What we need though are the models for both diodes and the transistor beta. The beta determines the current gain in mode #3, while the current gain plus the knowledge of both diode models determines the current gain in mode #4. Modes #3 and #4 are the most important with #1 considered a bit trivial.

To calculate the output of any circuit with a transistor in it biased for modes #3 or #4 you just analyze it like any other circuit with diodes and a current controlled current source. You can then determine just about anything you want when dealing with a DC circuit.

One of the operating points usually of most concern is when the transistor enters saturation. This happens as soon as the BC diode becomes forward biased. Because this only happens when the collector emitter voltage falls below a certain point, the effect is that the BC diode starts to 'steal' current from the BE diode, and so the overall apparent gain of the transistor seems to drop. That makes the Beta look lower than it really is.
This is really simple. For example, say we have to start with 1ma circuit input current and a Beta of 100, so we have 100ma collector current in mode #3. Now say we increase the input current to 2ma, we expect the collector current to rise to 200ma but instead the voltage across the collector emitter falls below the point where the BC diode starts to forward bias. What happens now is some of the input current that was going through the BE diode get rerouted through the BC diode (and through the collector to emitter), and that means the BE diode gets less current. So say the BC diode current goes to 0.5ma. That means the BE diode only gets 1.5ma of that total input current of 2ma now, and that means the collector current drops to just 150ma. So instead of seeing 200ma collector current with 2ma input current to the circuit, we only see 150ma, and we see also the collector emitter voltage drop. The transistor is now in saturation and the degree of saturation depends on the two diode models. If the models were such that the BC diode conducted 0.6ma, then the BE diode would only get 1.4ma and so the collector current would then only be 140ma not 150ma.
With 2ma in and 150ma out, the apparent Beta dropped to 75 down from 100.
With 2ma in and 140ma out, the apparent Beta dropped to 70.
So the two diode models play an important role in this, but more importantly we can see more intuitively what is happening when the transistor goes from linear active mode into saturation. The simple answer is that the BC diode now being forward biased steals some of the base current and so the overall circuit Beta looks lower than the transistor beta. Another way to look at it is that now all of the circuit input current does not go through the base emitter diode so the collector current does not reflect the value of that current and the original Beta.

When solving the equations with one current controlled current source and two diodes and maybe a couple resistors we usually have to solve them numerically because of the exponential diode models. This isnt too much of a problem usually. For most circuits there probably wont be a symbolic solution. I'll try to get back with a calculation example later but there is already one here in this forum that i did last time the basic question of saturation came up here. It is really just basic circuit analysis with exponential component models. The exponential models complicate things a little but numerically it turns out to be not super difficult.

 

Would suggest you the following read for calculation purposes:

http://electricalacademia.com/electronics/transistor-switch-theory/

 

waiting someone to respond ???

Thanks

The transistor is on when the square wave is almost ground.
It does not return to 3.3 volts because of the LED.

 

The transistor is on when the square wave is almost ground.
It does not return to 3.3 volts because of the LED.

Sorry, didn't see the replies after the plea for help.

 

The transistor is on when the square wave is almost ground.
It does not return to 3.3 volts because of the LED.

Thank you all for your valuable reply and time.
Its indeed a great learning !!!

Thanks again.!!!