Transistor in Active and saturation region

crutschow

Joined Mar 14, 2008
27,438
What is force beta in lecture.
How do we calculate force beta value.
"Forced beta" is simply the beta value you use to force the transistor into saturation, which is typically well below the active region value.
For standard BJT's a forced beta of 10-20 is usually used.
You can see the value used in the data sheet where they specify the transistor saturation voltage.
Beta_Force = ic/ib
and then where to get ic value :
The Ic value is the maximum current the transistor will switch when saturated.
It's determined by the load resistance and the supply voltage.
Thus, if the supply voltage was 10V and the load resistor was 100Ω, then the maximum Ic would be 100mA.
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
"Forced beta" is simply the beta value you use to force the transistor into saturation, which is typically well below the active region value.
For standard BJT's a forced beta of 10-20 is usually used.
You can see the value used in the data sheet where they specify the transistor saturation voltage.
The Ic value is the maximum current the transistor will switch when saturated.
It's determined by the load resistance and the supply voltage.
Thus, if the supply voltage was 10V and the load resistor was 100Ω, then the maximum Ic would be 100mA.
Ok thanks.

So what did i understand is if Ic_max = 100mA and lets say Ib= 1mA, Beta =20 so we are drawing only 20mA of Ic_max.
Is my understating is correct.

Earlier my understanding was Ic_max = Beta x Ib . so we can only draw max current from load is Beta x Ib and now we have to adjust Ib as per load requirement.

But you quoted quite different.

Thanks again.
 

crutschow

Joined Mar 14, 2008
27,438
Yes, you adjust Ib per the load.
In summary:
To use a transistor as a switch you use a forced beta of 10-20.
So if the maximum collector load current in the circuit (not the maximum the transistor can carry) is 100mA, then to fully turn the transistor in saturation, Ib should be at least 5mA.
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
Yes, you adjust Ib per the load.
In summary:
To use a transistor as a switch you use a forced beta of 10-20.
So if the maximum collector load current in the circuit (not the maximum the transistor can carry) is 100mA, then to fully turn the transistor in saturation, Ib should be at least 5mA.
Thank you so much Mr. Crutschow.

Now i cleared my doubt how to bias transistor in saturation region and i am able to design it. This thread help me a lot to learn about transistor.

But still I am unable to bias a transistor in active region.
How to calculation for active region.
Can we only amplify ac/PWM/SAW signals in active region. or else we can amplify dc signal as well in active region.
I firmly believe in initial calculation of circuit so i would be better if you explain by considering an examples circuit and do calculation for it.
I would be easier to understand me in one one go.

Thanks once again. !!!

Regards,
 

crutschow

Joined Mar 14, 2008
27,438
Can we only amplify ac/PWM/SAW signals in active region. or else we can amplify dc signal as well in active region.
A transistor can amplify either AC or DC signals.
An op amp is a high-gain differential DC amp, for example.

To bias in the active region you select the desired collector bias point (typically somewhere about half the supply voltage.
You then design a bias circuit to provide the desired bias.
Here's a good discussion of that.
 

sparky 1

Joined Nov 3, 2018
565
glad you followed, another example uses a 2N2369A saturation forced stuff
https://www.physicsforums.com/threads/transistor-current-gain-in-saturation-mode.865392/

The mechanical process for bias has a "do loop" because it is not an ideal transistor,
Again refer to (follow) it sort of like finding the roots of a quadratic equation only in this case the resistor values for a given voltage. Follow is following directions or you will stay in the do loop...
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
glad you followed, another example uses a 2N2369A saturation forced stuff
https://www.physicsforums.com/threads/transistor-current-gain-in-saturation-mode.865392/

The mechanical process for bias has a "do loop" because it is not an ideal transistor,
Again refer to (follow) it sort of like finding the roots of a quadratic equation only in this case the resistor values for a given voltage. Follow is following directions or you will stay in the do loop...
Thanks for your reply.
I understood about transistor saturation region.

Now i was looking to understand about transistor in active region.
It would be helpful if you make me understand about transistor in active region.

Thaanks
 

dl324

Joined Mar 30, 2015
12,928
Now i was looking to understand about transistor in active region.
It would be helpful if you make me understand about transistor in active region.
Is there something about post #17 that you don't understand?

The voltage divider on the base of the transistor biases the base to 1.67V. Assuming Vbe of 0.67V (I chose that to make the numbers from the book work), the emitter is at 1.0V. That gives an emitter current of 1mA. Since Ic=Ie (for high beta), the drop on the collector resistor is 10V, giving a collector voltage of 10V.

For convenience, the schematic is copied below:
 

sparky 1

Joined Nov 3, 2018
565
If you are willing to follow directions and do the work then
you will find that these guys on here are all on the same page.
We have to be organized and make notation on paper in order to get through this lab.
Or you can change plug in a sack of various resistors getting close but never really
understanding what these guys are trying to say. Like solving for the roots of a quadratic equation that is finding the correct resistors is more methodical as a mechanical process. Here is a demonstration of good lab method. please follow just like he shows. Find your beta and use the third circuit called voltage divider then your voltage measurements will have good bias point so you can explore the active region.

 
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Ford Prefect

Joined Jun 14, 2010
202
I have been reading this post with interest.
I quite often had a mental block when it came to reading transistor datasheets with all the numbers, abbreviations and calculating the base resistor of a transistor to saturate and switch the transistor hard on.
Then I found this website which helped me...
Transistor Base Resistor Calculator
....and basically 'ignored' the hFE values on the datasheets.

I wanted to calculate ohm value of the base resistor (Rb in the picture below) to saturate a 2N2222 transistor so that while using an OpAmp on a 12vDC supply which the OpAmp gave an output of about 10v (Vi ) to switch on a 12vDC (Vcc), 400Ω (RL), 0.03A/30mA (iL) relay.

I now use a Forced Beta (hFE) of usually 10,
(Increasing the Forced Beta will increase the base resistor value)

With the 12v relay being RL = 400Ω, Vcc = 12v, and Vi (output of OpAmp) = 10v
The base resistor on this site is calculated as 3333.33Ω
...and also assuming that the saturation voltage of the transistor (Vbe) is 0.7v.

iL = Vcc / RL
= 12v / 400Ω
= 0.03A or 30mA
So Rb = Vi x Forced Beta / iL
= 10v x 10 / 0.03A
= 100 / 0.03A = 3333.33 Ω (3.3kΩ)

HFE1.jpg

Hfe2.jpg
 
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Thread Starter

mishra87

Joined Jan 17, 2016
906
I have been reading this post with interest.
I quite often had a mental block when it came to reading transistor datasheets with all the numbers, abbreviations and calculating the base resistor of a transistor to saturate and switch the transistor hard on.
Then I found this website which helped me...
Transistor Base Resistor Calculator
....and basically 'ignored' the hFE values on the datasheets.

I wanted to calculate ohm value of the base resistor (Rb in the picture below) to saturate a 2N2222 transistor so that while using an OpAmp on a 12vDC supply which the OpAmp gave an output of about 10v (Vi ) to switch on a 12vDC (Vcc), 400Ω (RL), 0.03A/300mA (iL) relay.

I now use a Forced Beta (hFE) of usually 10,
(Increasing the Forced Beta will increase the base resistor value)

With the 12v relay being RL = 400Ω, Vcc = 12v, and Vi (output of OpAmp) = 10v
The base resistor on this site is calculated as 3333.33Ω
...and also assuming that the saturation voltage of the transistor (Vbe) is 0.7v.

iL = Vcc / RL
= 12v / 400Ω
= 0.03A or 30mA
So Rb = Vi x Forced Beta / iL
= 10v x 10 / 0.03A
= 100 / 0.03A = 3333.33 Ω (3.3kΩ)

View attachment 176387

View attachment 176391
Thank you so much Mr. Ford Prefect for your time and valuable information.
Its sound good that you follow this thread interestingly.

So many people contributed with their useful information. it took me time to understand me completely about transistor saturation region but finally i understood well.

Your information was quite straight forward which enables reader to understand in one go. Novice people like want the information straight forward.

This thread completely enable reader to design a transistor circuit in saturation region.

This thread had two parts. One part has been discussed completely but one part still left to be discussed more.

I tired to understand the biasing of transistor in active region.
Could you explain a bit more of transistor amplifier design in active mode in the same fashion you described for saturation region.
which parameters changes in active mode.

Thanks !!!
 

Ford Prefect

Joined Jun 14, 2010
202
Thank you Jony130, I wasn't entirely sure how I was going to answer as trying to explain the active region of a transistor is slightly more awkward than explaining when it is saturated. There are several videos on YouTube which explains the different states of a transistor.

Also the post by dl324 in post #23:

Insufficient base current prevents the transistor from saturating.
Is correct. The base current with the 10k resistors is VERY low.
Try changing the transistor to something like a BC547 and change the resistors from 10k to 1k as below. This will increase the base current.
The voltage at node B will be between 0v and no greater than 1.65v (3.3v / 2 resistors) but only 0.7v is enough to turn on the transistor.

Test1.jpg
 

Ford Prefect

Joined Jun 14, 2010
202
I disagree. Ths TS real-world measurement via a scope clearly shows that the BJT is in a saturation region.
https://forum.allaboutcircuits.com/attachments/20190425_114214-jpg.175897/
Yes true. I must admit that I did not look at this photo and it does look as if the transistor is in the saturation region.
But the base current is very low using 10k resistors and it would not hurt using 1k (or slightly higher ohm) resistors to ensure that base current is higher and the transistor is well into the saturation region.
 

Jony130

Joined Feb 17, 2009
5,241
But the base current is very low using 10k resistors and it would not hurt using 1k (or slightly higher ohm) resistors to ensure that base current is higher and the transistor is well into the saturation region.
You're right. But there is one problem here. TS decided to use a BJT with built up resistors.

Also, a side note because you have proposed a Bc547 as a replacement (European BJT).
For this transition the forced beta is 20.
https://www.haw-hamburg.de/fileadmi...hnik/Datenblaetter___Datasheet/BC546_kurz.pdf
So there is nothing special about "10" number.
 

Ford Prefect

Joined Jun 14, 2010
202
You're right. But there is one problem here. TS decided to use a BJT with built up resistors.
Also, a side note because you have proposed a Bc547 as a replacement (European BJT).
For this transition the forced beta is 20.
So there is nothing special about "10" number.
The only reason I used and mentioned a BC547 in the comment above is because mishra87 mentioned it in post #1 and presumed he has access to these transistors. I did not know the BC547 is a European transistor, I thought it is a cheap general purpose transistor available worldwide. I'd also never heard of a MUN5211DW1 transistor until I saw it on the schematic photo and looked it up. :D
You're right. there is nothing special about the "10" number (I presume you are talking about my comment in post #50 which I said 'I now use a Forced Beta (hFE) of usually 10'.
But if number 10 is used as forced beta with a BC547 as opposed to a beta of 20...would the transistor saturate and work?
 

WBahn

Joined Mar 31, 2012
26,398
The only reason I used and mentioned a BC547 in the comment above is because mishra87 mentioned it in post #1 and presumed he has access to these transistors. I did not know the BC547 is a European transistor, I thought it is a cheap general purpose transistor available worldwide. I'd also never heard of a MUN5211DW1 transistor until I saw it on the schematic photo and looked it up. :D
You're right. there is nothing special about the "10" number (I presume you are talking about my comment in post #50 which I said 'I now use a Forced Beta (hFE) of usually 10'.
But if number 10 is used as forced beta with a BC547 as opposed to a beta of 20...would the transistor saturate and work?
Yes, it would work.

There is nothing magical about the value of beta given in the data sheet. The transition from active region to saturation region is not abrupt. Most manufacturers use 10 as the value of beta at which they characterize the saturation behavior of their transistors simply because that is the convention that evolved decades ago and because adhering to it makes comparing transistors a lot easier. But you can use whatever value of beta you want. Choose it too high and you are too far into the active region to be able to force the beta in a well controlled fashion. Choose it too low and you waste excess power in the base-emitter junction in exchange for basically no improvement in Vcesat. Remember, a big part of the reason for operating in saturation is to minimize power dumped in the transistor. But since the Vbe is now a few times the Vce, you want to keep Ib as low as you can consistent with other goals. Running deep saturation also increases the charge that has to be removed to bring it out of saturation which slows down the switching speed.
 

Ford Prefect

Joined Jun 14, 2010
202
Yes, it would work.
There is nothing magical about the value of beta given in the data sheet.........
Yes I know it would work. Sorry but I was being a bit facetious in my comment. :D
There was no disrespect or criticism intended.
With my comment 'I now use a Forced Beta (hFE) of usually 10' meant that I start at 10 and work up if needed or required.
I had a look at the MUN5211DW1 datasheet and noticed that it has quite a low gain. :eek:
It maybe a good idea for mishra87 to use a general purpose silicon transistor (2N2222 or similar with a higher gain) instead of the MUN5211DW1.
 
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Thread Starter

mishra87

Joined Jan 17, 2016
906
Thank You all for you reply & time.

The Idea behind using MUN5211DW1 series transistor is having low cost with built up base resistor to save the space in PCB.
My personal observation 'using this transistor was it do go in saturation region but having very less collector current around 5mA.
because i have built in base resistor so i can not alter the part.

So now i have chosen a part with same series which are having base resistor R1=2.2k & R2 =47K. I will test and measure collector current then finalize part.

But one thing i still did not understand why has it has gone in saturation when base resistor was 10K ???

Thanks
 
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