With Ic=100mA, you don't need 15mA of base current to saturate the transistor. 10mA is adequate.
Rb=(3.3-0.7)/10mA=260Ω.

 

Hello Markd77,

</
If the LED is too bright, increase the value of the collector resistor.
The reason is that the collector resistor is known and doesn't change much with temperature or current.
\>

In reference to T7... just for my personal information, I can also higher rb to decrease the LED's brightness... right?

 

Hello Markd77,

</
If the LED is too bright, increase the value of the collector resistor.
The reason is that the collector resistor is known and doesn't change much with temperature or current.
\>

In reference to T7... just for my personal information, I can also higher rb to decrease the LED's brightness... right?

You can, but for practical circuits there are many reasons to increase the collector resistor and very few to raise the base resistor.
Another of the many is that running it at saturation means that the transistor generates less heat, so you can use a smaller, cheaper transistor. The total amount of power used is similar but the resistor takes most of it instead of it being more evenly shared.

 

Hello Audioguru,

</
Why do you keep picking the hFE?
\>

</
Your transistor does not have a Vce of 10V,
\>

Okay, let's please take this one step at a time here....

Yes you are right, I keep ranting about HFE... but it's only because unfortunately, I don't seem t get it. And I think I know why... this saturation mode is totally throwing me off. Let's look at a sample where the transistor is to operate in linear region mode instead. Let's use T8 sample diagram... and I have re-attached it here so we don't have to go back and forth.

Please let's make no assumptions when I say "I don't seem to get it". Bear in mind, I know that it is a constant given by the manufacturer which is to be applied as an ib gain multipliyer for ic. I am just confused as to which one to use in my calculations when confronted with the HFE table in spec sheet.

I think T8 shows why I don't understand HFE. In reference to T8, let's suppose that the base current is 1 ma as calculated in T8.png. Also, as you can see, Vce is 1.61V.

****** BUT LET'S STOP RIGHT HERE A SECOND!!!!

I know about the ic = 1ma and Vce = 1.6V **BECAUSE** I used 100 as HFE in my calculations. But if we go back when I initially started to calculated T8 diagram when I didn't know that ic = 1ma and that Vce = 1.6V, I then only knew about ib by doing:

ib = (VCC-VD1-Vgama)/100K

So, when I did the above calculation, ib resulted to approximately 1ma. Now at this very point, for the HFE, are you saying I should match my 1ma to the following HFE in the spec table:

IC = 1.0mA, VCE = 10V | 50

and solve for ic by using 50 as a HFE... like this:

ic = 50 x 1ma ?


Is this what you mean?


Thanks for your help!
r

 

The LED should be connected to its current-limiting resistor and driven from the collector of the transistor as a switch instead of as a slowly ramping emitter-follower like you show.

A transistor used as an on-off switch is supposed to be saturated or turned off but yours is not saturated, it is half-way and some transistors will not turn on enough because they might need the base current to be 1/10th the collector current as shown on almost every datasheet for small transistors.

FORGET ABOUT AN hFE OF 50, use an hFE of 10 in your calculation for the base resistor value so that ANY transistor will saturate as a switch.

Don't you understand that some transistors have a low hFE and other transistors have a high hFE even if they have the same part number? Some have a "typical" hFE.
Don't you want your circuit to ALWAYS work instead of failing when the hFE of a transistor is low??
What if the transistor manufacturer has a poor yield and EVERY transistor has a low but passing hFE?

EDIT: I added some contrast to your very faint sketch.

 

<\
A transistor used as an on-off switch is supposed to be saturated or turned off but yours is not saturated, it is half-way and some transistors will not turn on enough because they might need the base current to be 1/10th the collector current as shown on almost every datasheet for small transistors.

FORGET ABOUT AN hFE OF 50, use an hFE of 10 in your calculation for the base resistor value so that ANY transistor will saturate as a switch.
/>

Yes, but T8 was intended to be used in beta mode!!! I don't just
want to do circuits in saturation mode you know....

Isn't T8 in beta mode.... In T8, I am simply trying to supply
small currents to a led but don't necessarily require full led
current.... It's just tests that I am doing.

Thanks for adding contrasts to T8.

In T8, since "I think" it's in beta mode, then why doesn't
"Ic = HFE * ib" not hold true.....

See calculations!

Thanks

 

Have you measured the voltage across the 10K resistor? That would tell you the true base current, I think you may have calculated it by adding other voltages and for example Vbe isn't always 0.7V.

 

Hi Markd77,

Just to make sure, there are two main circuits we have been looking at here... T7 which was the saturation one and T8 which is the one that I want to operate in linear mode.

Please note, we are now talking about T8! You can view this at post #24 or #25.

One more thing, I know that my circuit is a little untypical by connecting the LED at the emitter as opposed to the collector. But for now, I started it this way and I will experiment with the collector convention later on... so let's keep it the way I have it in T8 to avoid any further confusion.

In ref to T8, you mean the voltage across the 100K right? well, okay, here are all the values based on physical measurements while diagram T8 is connected to a 3.33 VDC source:

- V across the 100K: 1.037V
- Vbe is 0.645V
- Rb = 99200 ohms
- Rc = 20 ohms
- Vrc = 0.027V

Using the above measured values I calculate ic:

ic = HFE (3.33-0.645-1.69)/99200
ic = 100(10.03 x 10^-6) ..... or 100(0.00001003A)
ic = 1.003ma

However when I measure ic, it is 1.37 ma ???

Is this difference acceptable?

Now back to the HFE (sooooo discouraged about this HFE.....)... I have been really trying .... honestly... articles... old books, videos.... and so forth...

If you look at this video:

http://www.youtube.com/watch?v=DLl7-CmVT7w&playnext=1&list=PLFC0DD356D1FD0196&feature=results_video

The tutor in this video actually mentions that the spec sheet of a certain transistor (2n3906-04) offers a gain of 60 for a current of approximately 50 ma at Ic. He then goes onto say that if he has 1 ma at the base he can control 60 ma in Ic. So knowing that he needs only about 30 ma in Ic, he supposes that when applying 0.5 ma at ib, it would therefore produce the 30 ma he needs at Ic. But to assure a strong saturation he chooses to provide 1ma for ib.

Now, I know that his example is for a saturation circuit and that T8 is for a beta (linear range) circuit. But in T8, I used a gain of 100 as HFE, and pretty much works given the above calculations. So let me continue...

The tutor in the video, requires 30 ma at Ic, and so proceeded in calculating his ib in accordance to a gain of 60 offered by the part. In T8, don't I want a current of approximately 1ma!!!!!! But I did my ib calculations using the gain that offers a whopping 150 ma???

In other words, for T8, I was wondering if I should of reasoned it as the tutor did. Like this:

"I need apprx. 1ma as my collector current and I see that the PN2222 spec sheet offers a gain of 50 for a current of 1ma at collector."

But wait a minute here, having 1/50th of a ma as ib would simply put me back in saturation mode or at least pretty much ... right? So I designed around the offered HFE at 150ma which is the famous gain of 100 according to spec.

So allowing 1/100th of a ma at ib would give me more "linear range in my collector current!
So I know that we don't pick the HFE, but we design around the HFE offered according to required Ic.... is this the idea here.... cause I am exhausted!!!! :-(

help! thanks all!
r

 

The tutor in this video actually mentions that the spec sheet of a certain transistor (2n3906 or 2N3904) offers a gain of 60 for a current of approximately 50 ma at Ic. He then goes onto say that if he has 1 ma at the base he can control 60 ma in Ic.

No.
You and the tutor did not read the datasheet. The current gain IS NOT 60. Look again, the MINIMUM current gain is 60 so it could be 60, 100, 150, 200, 250, 300 or who knows how much.

So knowing that he needs only about 30 ma in Ic, he supposes that when applying 0.5 ma at ib, it would therefore produce the 30 ma he needs at Ic. But to assure a strong saturation he chooses to provide 1ma for ib.

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).

You still don't understand that current gain, hFE or beta is not a fixed number. Each transistor has a different amount and it changes with the amount of current and with the temperature.
Negative feedback is added so that all different transistors operate the same.

Your circuit is wide open with no negative feedback.

 

</
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).
/>

Okay, this is understandable. I have done saturation, but I would have to go back and see if I did it at 1/10th the collector current... I think I did it more than that... but as soon as I get a chance I will go back to T7 and see what I did. I am much slower than you as these calculations are not as obvious to me... I really have to think about it ...

</
You still don't understand that current gain, hFE or beta is not a fixed number. Each transistor has a different amount and it changes with the amount of current and with the temperature.
/>

I don't doubt what you are saying... At this point I have to plead total ignorance.
You are telling me that transistors ex: part# PN2222, are manufactured and available on the market and all have a minimum and a maximum beta (according to spec) and that this value varies so much that inserting a resistor RE in the emitter circuit known as negative feedback is required to stabilize the HFE. Interesting.

</
Your circuit is wide open with no negative feedback.
/>

So then my circuit which has no negative feedback cannot really be trusted in linear range or beta mode... right? However,
can I simply use it for saturation mode?

When a newbie to transistors like me watches a video like the following one, it is hard for me to wrap my mind around the fact that beta is so variable to the point that a special feedback circuit must be done.... really, nowhere does he mention this.... he just uses HFE at 100 without a negative feedback resistor... so just wondering ... how sure is he that HFE is really 100?

Now, Audiogurru, I know you have been repeating it endlessly that beta varies, but I was under the impression that I could get away with it or ignore it based on how the following video explained the common emitter circuit.

Please view!


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


r

 

However, can I simply use it for saturation mode?

Yes.

\(h_{FE}\) "hybrid Forward Emitter" is the hybrid parameter for a Bipolar Junction Transistor in common emitter configuration. Its the DC current gain of the BJT in common emitter configuration i.e.. the ratio of collector current to base current. Notice in \(h_{FE}\) the "FE" are written in upper case it means DC current gain when "FE" is written in smaller case \(h_{fe}\) it is the AC current gain as BJT's also work with AC signals. The \(\beta\) symbol can also be used for current gain. In my earlier post I have already said about forward \(\beta\) and reverse \(\beta\).

You already know that different BJTs have different \(\beta\),not only that same BJTs from same manufacturer will have different \(\beta\) ,for example two identical BJTs 2N2222 from Philips will have different \(\beta\) ,and manufacturer cant do anything to help that its how semiconductors works, what they can guaranty is that the \(\beta\) of two identical BJTs will fall with in a range ,and that range is specified in datasheet as \(minimum - \beta\) and \(maximum - \beta\).

Now this \(\beta\) factor for a BJT in common emitter configuration is very important as common emitter configuration is a configuration in which you can have both current and voltage gain in a large amount in BJT's active region and which is controlled by the factor
\(\beta\),but this \(\beta\) is not a constant factor for a BJT ,it changes dynamically while the BJT is operating in active region, the \(\beta\) increases. In linear amplifiers circuit of BJT we have a operating point also known as Q point which is very important for a linear amplifier ,it should be stable ,now if the \(h_{FE}\) of a BJT is not constant then the Q point will be not stable, and as a result the amplification will be distorted or clipped. So we add up some short of additional circuit to compensate the change in \(h_{FE}\) ,and this additional circuit is nothing but negative feedback. Their are many kinds of negative feedback circuits like emitter feedback ,collector feedback and mostly used voltage divider circuit which is the best among all the feedback circuits for a BJT,in its calculation you can actually omit the \(h_{FE}\) factor.

While for saturating a transistor ,we have a different type of \(\beta\) ,known as forced \(\beta\),while a BJT is working in saturation region its beta doesn’t increase like active region, the factors which increase \(\beta\) in active region have very small effects in saturation region. Normally most American small signal BJTs will have \(\beta\) of 10 for saturation and most European small signal BJTs will have \(\beta\) of 20 for saturation ,and many Power BJTs will have \(\beta\) of 5 for saturation , but its just a typical value from manufacturer. Calculating proper value for forced \(\beta\) is bit different, but the above values always worked for me.

Hope that clears a bit.

Good Luck

 

I drawed a circuit to help you measure the Ib and Ic.
If you have any question, you can post it.

 

R1 and R4 don't have any purpose in the circuit. You can measure voltage across R3 and find Ic current. Ic = (Voltage across R3)/R3

P.S. Jony130, I have not had a chance to do your circuit yet... but just wondering, why are putting a diode (D1) at the base ... is it for voltage attenuation purposes or just so I can visualize the brightness of the LED based on various RX resistors?

LEDs are only for indicators purposes.

 

I show how adding a series emitter rersistor reduces the collector voltage change when the hFE, Vbe and temperature are changed and show another form of negative feedback.

 

Hello Debjit Roy,

That was nicely put and interesting ....

Thanks

 

Hello Audioguru,

wow!, I never would of thought that transistors operating in their linear range are so unstable... Thanks for providing some circuits that show this.

Also you have provided some circuits that resolve this issue such as the series emitter resistor, the voltage divider at the gate and the collector feedback circuit. Let me ask you, I am sure there is lots of calculations behind these negative feedback circuits. Are there any articles/videos that would show the basic calculations for each one of these circuits (without capacitors... just the transistor and resistors) which show how to stabalize the ic and Vce. Basically, lets take the first one you have attached .. "the series emitter resistor". A tutorial which would show how to calculate the the emitter resistor so to obtain proper stabilized ic and vce values?

Thanks again!
r

 

Hello Debjit Roy,

That was nicely put and interesting ....

Thanks

You are welcome.

In AAC we have a very good ebook covering the basic with very low math,you can read the Semiconductor part http://www.allaboutcircuits.com/vol_3/index.html
or if you know much about semiconductors and BJT's you can read directly about the Biasing calculations ....
http://www.allaboutcircuits.com/vol_3/chpt_4/10.html

If you want to calculate with hybrid equations and lot of maths and know most out of it,then some engineering text will be best as per me.

Good Luck

 

Hello Audioguru,

wow!, I never would of thought that transistors operating in their linear range are so unstable... Thanks for providing some circuits that show this.

You ain't seen nuthin yet!
A transistor without negative feedback at a fairly high output level produces SEVERE distortion. When an emitter resistor is added then the distortion is much less but the voltage gain is also reduced.

You know what a pure sinewave is supposed to look like? Look at this:
The series emitter resistor provides DC negative feedback so that the DC is fairly stable if the hFE, Vbe and temperature are changed. For high AC voltage gain the capacitor parallel to the emitter resistor bypasses it so that the voltage gain is about 180 and the distortion is about 40% (it sounds horrible with audio).
When the capacitor is removed from the emitter resistor then the resistor provides AC negative feedback which reduces the voltage gain to about 9 and reduces the distortion to about 3% (still pretty bad but much better than before).

A second transistor can be added with overall negative feedback around both transistors for plenty of voltage gain and low distortion. Or an opamp can be used.

 

Here is a simple current regulator I use. It's pretty good for higher powered LEDs like your IR LED. It puts about the same current through the LED even if the power supply voltage changes, or if the LED forward voltage changes (which they do when they get hot).
It uses forward active mode and emitter feedback among other things.
http://forum.allaboutcircuits.com/showthread.php?t=39303
It's explained briefly here:
http://en.wikipedia.org/wiki/Current_source

 

wow... okay... I just tried my first collector feedback circuit.....

I will post in another thread.... this one is getting to long... so look out for me in a thread called "Collector feedback bias..."

Thanks all
r