One last thing Audioguru....

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No.
Again the tutor did not read the datasheet. It and nearly EVERY datasheet for little transistors show that for strong saturation then the base current must be 1/10th the collector current (not 1/30th).
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In the following video (don't need to start from the begining... you can go to location 5:00 of 7:58):

http://www.youtube.com/watch?v=nqefeNBcowI

I wanted to make sure that I clearly understand this 10 times the collector current idiom.

In the video, when vs is 6V, the transistor is in saturation mode. So for saturation mode, we at least need 5 ma as ic which is the source circuit max current, so are you saying that the max ib required to saturate the transistor is 1/10th of 5 ma which is approximately 500ua?

ib required = beta ((Vs-Vg)/rb)
ib required = 10 ((6-0.7)/100,000)
ib required = 530ua

thanks for your help... really appreciated!

r

 

One last thing Audioguru....

</
No.
Again the tutor did not read the datasheet. It and nearly EVERY datasheet for little transistors show that for strong saturation then the base current must be 1/10th the collector current (not 1/30th).
/>

In the following video (don't need to start from the begining... you can go to location 5:00 of 7:58):

http://www.youtube.com/watch?v=nqefeNBcowI

I wanted to make sure that I clearly understand this 10 times the collector current idiom.

In the video, when vs is 6V, the transistor is in saturation mode. So for saturation mode, we at least need 5 ma as ic which is the source circuit max current, so are you saying that the max ib required to saturate the transistor is 1/10th of 5 ma which is approximately 500ua?

ib required = beta ((Vs-Vg)/rb)
ib required = 10 ((6-0.7)/100,000)
ib required = 530ua

thanks for your help... really appreciated!

r

The required Ib has nothing to do with Rb in the circuit. You already stated the relationship (above):

the max ib required to saturate the transistor is 1/10th of 5 ma which is approximately 500ua

That's all you need. You then calculate Rb:
Rb=(Vs-0.7)/Ib.

 

Another voice saying the same thing in a different way:

The gain of a transistor decreases as the collector voltage decreases.
When you have a collector resistor, it will use up some of the voltage.
When there is still 10 volts left on the collector, the transistor can still respond with the advertised gain numbers.
As the current increases, the collector resistor uses up more of the available voltage.
As the available collector voltage decreases, the gain becomes less.
If you allow enough current into the base to equal one tenth of the collector current (that the collector resistor will allow), the transistor will arrive at the collector voltage that is so low that its gain is only 10.
When you design a saturated transistor stage, you are not designing to use the gain that is advertised when the collector still has 10 volts available,
You are designing an amplifier stage that will arrive at whatever the collector voltage is that is so low that the transistor only has a gain of 10, and that is a rather small voltage.

 

Another voice saying the same thing in a different way:

The gain of a transistor decreases as the collector voltage decreases.
When you have a collector resistor, it will use up some of the voltage.
When there is still 10 volts left on the collector, the transistor can still respond with the advertised gain numbers.
As the current increases, the collector resistor uses up more of the available voltage.
As the available collector voltage decreases, the gain becomes less.
If you allow enough current into the base to equal one tenth of the collector current (that the collector resistor will allow), the transistor will arrive at the collector voltage that is so low that its gain is only 10.
When you design a saturated transistor stage, you are not designing to use the gain that is advertised when the collector still has 10 volts available,
You are designing an amplifier stage that will arrive at whatever the collector voltage is that is so low that the transistor only has a gain of 10, and that is a rather small voltage.

I'll complicate this slightly by saying that the transistor's gain when saturated is at least 10 (assuming Vce(sat) is specified with Ic/Ib=10). The gain may be more, but it is guaranteed (specified) not to be less. It will almost certainly be more than 10, but we can't choose base drive for each individual transistor we use. Besides, beta also varies with temperature, current, etc. We use the specified value (Ib=Ic/10) because we know it will work every time.
If the gain is higher (it will be), then excess base current, beyond that required for saturation, causes charge to accumulate in the base region. This must be removed before the transistor will turn off. The time required to remove this charge is called storage time.
I know - too much information. File it away for future reference.

 

That's all you need. You then calculate Rb:
Rb=(Vs-0.7)/Ib.

Thanks

</
I'll complicate this slightly by saying that the transistor's gain when saturated is at least 10 (assuming Vce(sat) is specified with Ic/Ib=10). The gain may be more, but it is guaranteed (specified) not to be less. It will almost certainly be more than 10, but we can't choose base drive for each individual transistor we use. Besides, beta also varies with temperature, current, etc. We use the specified value (Ib=Ic/10) because we know it will work every time.
If the gain is higher (it will be), then excess base current, beyond that required for saturation, causes charge to accumulate in the base region. This must be removed before the transistor will turn off. The time required to remove this charge is called storage time.
I know - too much information. File it away for future reference.
/>
I was about to ask if a little extra would render the saturation
more effective ex: 12 or 15 times the Ic ..... However you have
Made it clear!

Thank you all so much guys!
R

 

Testing quotes

Please disregard....Just testing the quote feature!
This is a test

 

In the following video (don't need to start from the begining... you can go to location 5:00 of 7:58):

http://www.youtube.com/watch?v=nqefeNBcowI

I wanted to make sure that I clearly understand this 10 times the collector current idiom.

In the video, when vs is 6V, the transistor is in saturation mode.

No it is not.
The author ASSUMED is is saturated but it is not because its hFE is much less than 100 when saturated. If you tested a few thousand transistors then a few of them might get saturated. Maybe the author got lucky and had one of the few out of a thousand.

Don't you dsesign a circuit so that ALL transistors that pass their spec's work properly instead of just a few out of a thousand?
ALL my designs work properly because I design them all to work properly, not just a few out of a thousand.

ALL datasheets for little American transistors show that they saturate properly when their base current is 1/10th the collector current.
Little European datasheets show 1/20th.
Nobody shows 1/100th.

So for saturation mode, we at least need 5 ma as ic which is the source circuit max current, so are you saying that the max ib required to saturate the transistor is 1/10th of 5 ma which is approximately 500ua?

ib required = beta ((Vs-Vg)/rb)
ib required = 10 ((6-0.7)/100,000)
ib required = 530ua

Yes, then ANY little (American) transistor that passes its spec's will saturate, not just a few out of a thousand. Little European transistors saturate properly when their base current is 1/20th the collector current.

 

So for saturation mode, we at least need 5 ma as ic which is the source circuit max current, so are you saying that the max ib required to saturate the transistor is 1/10th of 5 ma which is approximately 500ua?

ib required = beta ((Vs-Vg)/rb)
ib required = 10 ((6-0.7)/100,000)
ib required = 530ua

Yes, then ANY little (American) transistor that passes its spec's will saturate, not just a few out of a thousand. Little European transistors saturate properly when their base current is 1/20th the collector current.

The OP's answer is correct, but the method is wrong. As I pointed out, the required base current is unrelated to the 100k base resistor value shown in the video.

Here is the correct method:
Ib=Ic(sat)/beta
Ic(sat)=Vcc/Rc
Vcc=5v, Rc=1kΩ, beta=10
Ib=(5v/1kΩ)/beta
Ib=500uA

 

The author ASSUMED is is saturated but it is not because its hFE is much less than 100 when saturated. If you tested a few thousand transistors then a few of them might get saturated. Maybe the author got lucky and had one of the few out of a thousand.

Ok... I get that *10* is the magic number for sat. Yes you are correct, he assumed and then proved it by
showing "Ic > vcc/rc" FOR THIS TRANSISTOR !

The fact that he illustrated saturation mode by:
ic = beta(ib)

was it because he wanted to illustrate the effect of tripling vs?

In any case I was just using that video, with that circuit, with that particular transistor to get the basic math down for
sat mode.

Also, as you can tell I did not use '100' anywhere in my
calculation. I did respect the 1/10th the Ic .... No?

Here is the correct method:
Ib=Ic(sat)/beta
Ic(sat)=Vcc/Rc
Vcc=5v, Rc=1kΩ, beta=10
Ib=(5v/1kΩ)/beta
Ib=500uA

In all due respect towards the author of the video, for sat mod calcs, I find the above much easier than the video way.

R

 

The author showed the circuit with an input Vs of 2V. He calculated that with an hFE of 100 then the collector current ic is 1.3mA and the collector voltage Vce is 3.7V.
The transistor is not cutoff and it is not saturated, instead it is active.

The author assumed that the saturated hFE of the transistor is 100 and calculated that its saturated collector current will be 5.3mA. WRONG and WRONG.

The saturated hFE should be 10, not 100.
How in hell can the collector current be 5.3mA when the 1k load draws only 5.0mA when the transistor is a dead short?
The author selected a second input voltage of 6V to try and prove (wrongly) that it will saturate.

 

In sat, ib = 1/10th Ic, where Ic = max short circuit current!

Understood!

Thanks Audioguru