# Help in understand TRANSISTORS.

Discussion in 'Homework Help' started by KansaiRobot, May 1, 2015.

1. ### KansaiRobot Thread Starter Active Member

Jan 15, 2010
318
5
Hello everybody and thanks always for your help.

This time I would like to ask a more general question related to transistors.
Say you have the following two schematics:

1) What is the difference in functionality between the two? Assuming I want to control the Lamp, which of the two circuits is most recommended?? What are the pros and cons of each one...

I am not good at doing calculations so please tell me if the following are correct.
If V= 10V , Ic is 100mA and the β of Q1 and Q2 is 100

circuit 1:
Ib= 100mA/100 =1mA
R1=(10V-0.7V) /1mA = 9.3kΩ

circuit 2:

Ib= 100mA/100=1mA
I3= 10*Ib = 10mA
I2=Ib+I3=11mA
therefore
R3= 0.7V/10mA= 0.07kΩ=70Ω
R2=(10V-0.7V)/11mA= 0.845kΩ= 845Ω

Is R3 necessary to maintain stability or something??

2. ### ScottWang Moderator

Aug 23, 2012
4,934
777
Go to check the hFE of any of bjt when it get into the saturation status.

3. ### KansaiRobot Thread Starter Active Member

Jan 15, 2010
318
5
OMG, now I am confused again. (this is more related to my other thread but...ok..)
If I check the hFE in saturation, then the 2N2222 which I thought rather adequate for my needs, fails too cause in the datasheet it says:

Collector Emitter Saturation Voltage
IC= 150mA, IB= 15mA
IC= 500mA, IB= 50m

so hFE is not written but can be calculate as Ic/Ib = 10!!!! and I need at least 18!

Now I dont know what to do... I thought I found an adequate transistor

---------
Related with the original question of this thread, any comment??

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
First you must distinguish two thinks.
1 - BJT working as a amplifier.
2 - BTJ working as a BJT a switch (in saturation). And in saturation region Ic = hfe*Ib do not hold any more.

For your circuit you need to work in saturation region. And both of your circuit will work if you properly select the resistor values.
And R3 (pull-down resistor) was add to ensure that BJT will be turn-off if you open the switch.

Last edited: May 1, 2015
5. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
1. supply voltage is 12V
2. load is a 12V @ 0.5A lamp = 24Ω (Ignoring cold resistance for now)
3. transistor is an average 2N2222 with a β of 200.

I am going to show you what happens if the base resistor is varied logarithmically from 100Ω tpo 10KΩ. Note that all of the plots are a function of the base resistor R1. Match legend color to plot color, and units are shown on the Y axis (right or left).

Most important in understanding the transistor as a switch is by looking at the red plot V(c), the collector voltage. For all values of the base resistor R1 less than ~1.6KΩ, the transistor is turned on about as well as it can be, with a collector voltage less than ~400mV. That
means the lamp is turned on as hard as possible, with (12-0.4)V across it, making the lamp current I(R2) yellow plot a bit less than 500mA.

Note that for all values of the base resistor R1 greater than ~1.6KΩ, the red plot V(c) collector voltage begins increasing, and the transistor is not turned on hard enough. This means that the voltage across the lamp is decreasing, the current through it I(R2) is decreasing, and most importantly, look at the violet plot (complex expression in the lower plot pane) which is the power wasted (Watts) in the transistor. Note power is less than 0.2W for values of R1 between about 300Ω and 1.6KΩ.

Note that if a 2N2222 is being used as a switch to switch 1/2A as in this example, at 0.2W, it will get quite hot to the touch.

So how small should we make R1? Note that as R1 is decreased below about 1K (cursor position), the power wasted in the transistor increases (due to the excess current being injected into the base-emitter junction) for virtually no improvement in any of the other plots.

In a future posting, I will discuss choosing R1 in the context of pulling out a 2N2222 that has a lower than average β.

6. ### MrAl Distinguished Member

Jun 17, 2014
2,573
522
Hello,

As you are finding out now from the other posts, the transistor base needs more current to stay in saturation than it does in more linear operation where we have Ic=ib*Beta. For a transistor with a gain of 100 it may fall as low as 10 in saturation. This means that if 1ma base current was good enough before, now we need 10ma to keep it in saturation, so for the circuit 1 above that means a resistor value roughly 10 times less than before. If it was 9k before then it is around 900 ohms now.

For a switching circuit, the lower resistor however need not be too small. That is, it can be much larger than the upper resistor (the 9k or 900 ohm resistor). The purpose of that resistor is to keep the transistor off during times when it should be off and for temperature extremes. Because of this, the calculation is:
R=0.4/iL

where iL is the collector base leakage current of the transistor and can be found on the data sheet.
For example, for 10ua leakage current, the value of the lower resistor is:
0.4/0.000010=40k

so you see this resistor can be much higher than the driver resistor of 9k or 900 ohms.

This assumes that the switch used does not have any leakage though. If it does, that has to be taken into account too and the resistor made smaller. Usually physical switches are pretty good though especially for low voltage applications.

The second use of that lower resistor would involve the turn off speed of the transistor. A lower resistor would sweep the charge carriers out of the base region faster than a larger resistor, so a small resistor value would be better. But for a light bulb application you probably dont have to worry about this unless you intend to use a pulse width modulation in place of the switch. Then the turn off speed affects the power dissipation of the transistor so you want it to turn off fast.

So the two main reasons for using a lower resistor are:
1. To keep the base emitter voltage less than 0.4v even with max transistor leakage.
2. To turn the transistor off faster (possibly combined with a little reverse bias supply voltage for fastest possible turn off speed, and possibly with some anti saturation circuitry).

BTW, if you need more gain you can always look into combining two transistors into one in a configuration called a Darlington Compound, or just one transistor that drives the other. The gain is then much higher.

7. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Mr Al, look at my V(c) vs R1 {or I(R1) vs R1} and tell us what the base current needs to be to "keep the transistor in saturation"?

8. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
There's a cart and horse issue when it comes to the transistor beta in saturation.

It's easy to think that, oh, when this transistor is saturated the beta turns out to be 10.

The reality is that the transistor is considered to be in saturation once the beta falls to 10.

There's nothing magical about that number. It's an arbitrary but nearly universal definition of 'saturation' for small signal BJTs.

If you need to design around a beta of 20, then that is almost certainly "close enough" for most practical purposes and you can expect your Vce to be pretty close to Vcesat.

The presence of R3 ensures that the charge on the bc and be capacitances is discharged quickly resulting in a quicker turn-off time. It also makes in nearly impossible for EMI to produce anough current in the base circuit to cause the transistor to conduct and it gives a path for collector leakage current to go rather than letting it turn into base-emitter current. For your purposes none of these are likely to matter too much. The downside of R3 is that your circuit has to be able to deliver several times the base current that it normally would in order to ensure that you turn the transistor on.

9. ### MrAl Distinguished Member

Jun 17, 2014
2,573
522
Hi Mike,

I can say with absolute and complete certainty that the required base current is: "26 railroad".

Throwback from Monty Python

In all seriousness, if you read my previous post first paragraph you'll see i actually dictated the operating parameters, so there was no wiggle room. It's like a homework question that reads, "Given a transistor with gain of 10 in saturation, what base current is required with a 100ma load?" The answer is of course 10ma or more.
Of course i assumed a 100ma load because that's what the OP seemed to be using, and as WBahn pointed out it is a common approximation to think in terms of roughly a gain of 10, with a tolerance of +100 percent and -50 percent.

But if i did not understand your point correctly, feel free to clarify and i'll give it another look.

I'd also like to clarify another points about the lower resistor "R3".
This resistor only needs to be small when we want fast turn off. If we dont need fast turn off (simple on and off, no real PWM) then it can be large...so large that it is almost irrelevant in the calculation of the base current (almost, not exactly), or at least responsible for only a small portion of the current from the upper resistor (iR3=0.7/R3) as an approximation.

10. ### KansaiRobot Thread Starter Active Member

Jan 15, 2010
318
5
So if I want to use PWM, do I necessarily have to use the circuit 2 with the lower resistor???

I will use today saturday to read again my textbook on transistors to see if I understand this time

One question: If I want to use a transistor as a switch, does it has to be necessarily on saturation??

Last edited: May 1, 2015
11. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Neither circuit is set up for PWM and the answer depends on the details of how you provide the PWM signal to the transistor. If your PWM signal is active-HI and active-LO (as opposed to open-collector, for instance) then you won't need pullup/pulldown resistors at all. If your PWM is open-collector then you will have to have a pull-up resistor otherwise it won't work at all.

Your last question begs the very question I was talking about earlier -- what is "saturation" to begin with? Related to this is the question of what it means, in the context of a specific application, for a transistor to act "as a switch"? For some applications the voltage across the transistor has to be very low in order for it to be a good enough switch, while other applications a transistor can be a good enough switch if it is able to reduce the voltage across it by a few percent. Whatever it means for your application, the transistor just needs to be reliably driven far enough as to meet whatever requirement applies. From a typical design standpoint we like to drive the transistor hard into saturation just to remove as many variables as possible -- at the expense of sacrificing optimal performance in many different areas that, in most situations, don't matter that much.

12. ### KansaiRobot Thread Starter Active Member

Jan 15, 2010
318
5
I will take this opportunity to ask question while I study my textbooks

1) My textbook says

However in every datasheet I ve seen there are several gains for different conditions...so it is not constant... how do I reconcile these two positions?

2) My Textbook says
However when I see a datasheet like:
https://www.sparkfun.com/datasheets/Components/2N3904.pdf

in saturation , first the Vce is not 0! but 0.2V why?
and more surprisingly at saturation Ic can be 10 or 50 mA. According to the book this (which??) should be the maximum but then in Hce it says that Ic = 100mA also is possible! Why? how??

Last edited: May 1, 2015
13. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Simple. Your text book is starting you off with a simple model of a transistor so that you can learn the basics. Once you are proficient with them, they will introduce some of the real-world variables and let you come to terms with them, and then they will introduce some more. It's just like when you studied physics and, at first, there was no friction or air resistance.

Since I don't have your textbook open in front of me, I can't tell you want it says and doesn't say. Note in your quote that you indicated that it is the maximum current "given the stated values of the collector resistor and supply voltage". So it is external components and conditions that establish the maximum current. The data sheet is just giving you information that applies to several difference maximum currents.

KansaiRobot likes this.
14. ### MrAl Distinguished Member

Jun 17, 2014
2,573
522
Hi,

To your last question, a resounding, "NO".

But as you probably know by now, when it is in saturation it consumes the second least amount of power that it can, and first, lowest power is when it is completely 'off'.
That doesnt mean you have to have it in saturation, and for the fastest switching speed you actually have to STOP it from saturating on purpose or else your switching speed could be severely limited. To to this you use an anti saturating network which steals base current from the transistor just before it enters saturation. Most of the time you wont need this though because you can probably get fast enough speed with the normal switching times of the transistor, even though it is allowed to enter saturation.

Also just to note, not all transistors are created equal. Most of these little guys saturate around 0.2v or maybe a little lower or higher depending on current, but you can get some very quality transistors that saturate at very low voltages like 0.1 and lower, depending on device and current. One maker is (or was) Zetex, which make (or made) very very low sat bipolar transistors.

Side note:
It looks like Zetex was acquired by Diodes Incorporated in 2008.
Their transistors are some of the lowest sat transistors i've ever seen.

Last edited: May 2, 2015