[SOLVED] Newbie : data sheet reading for 2n2222 transistor

i am trying to understand the basics ... please help me out.

according to the data sheet there are two values given to the Vce and Vbe with Ic = (150ma and 500ma) and Ib = 15 and 50ma).
which value i can choose ? for example :
what value i should choose for Vce and Vbe for the below circuit ? if i want Ic = 20ma.
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how will the values change if i keep a white LED of 3.5 voltage drop near Rc ???????

thanks in advance

chose the values according to the highest Ic or Ib current that can flow in the circuit
and have a look at note 1
what does it say??

Collector current is dependent on the value of Rc. with a 5 volt supply and Rc=250 ohms, collector current cannot exceed 20ma. That is a more practical way of controlling the current than attempting to fine-tune the base-emitter voltage.

---------- Post added at 14:30 ---------- Previous post was at 14:04 ----------

Limiting collector current with a resistor also means that the transistor will behave in a more predictable fashion. Note that with a higher collector current, the base voltage needed to put the transistor into saturation rises.

---------- Post added at 14:39 ---------- Previous post was at 14:30 ----------

...and when the base voltage is zero, collector current is zero. But when base voltage reaches the saturation point, collector current goes up way beyond 500ma, and the transistor goes up in a puff of smoke.

First of all, you should understand that the voltages given on that part of the datasheet you reproduced are saturation voltages Vce(sat) and Vbe(sat), not just Vce and Vbe.

When we use a mechanical switch to control a voltage source to a load, we generally think of the switch as perfect. That is, when the switch is closed (on), 100% of the voltage goes to the load. This is very nearly true of a switch in good condition. But a real-world switch is never quite perfect. It always has a small resistance and the load current, in flowing through the switch contact point, causes a small voltage drop in the switch so that not quite 100% of the supply voltage reaches the load. This voltage drop is usually negligible.

However, when a transistor, especially a bipolar junction transistor (BJT) like the 2N2222 is used as a switch, it's not as close to perfect as a mechanical switch. The voltage drop is significant and, depending on how heavy the load is and other circuit conditions, it can be a substantial portion of the total supply voltage. That drop is the saturation voltage. It is an inherent characteristics of the particular transistor and cannot be reduced.

Vce(sat) is roughly proportional to collector current. So, for your circuit, deduce Vce(sat) from the value closer to your 20mA load, that is, Vce(sat) for Ic = 150mA. Scale that down to 20mA and you get about 50mV max for a 2N2222 and 40mV for a 2N2222A.

Vbe and Vbe(sat) have a more complex relation to Ic and Ib. For practical designs (as opposed to textbook derivations which are often not of much use), base-emitter voltage is best estimated by experience gained from looking at various datasheets that include graphical presentations. Fortunately, it does not vary too much from unit to unit. Besides, base current is usually supplied from a much higher voltage than Vbe so that variations in the actual value are swamped by the overall voltage. For switching applications, taking it as 0.7V for Ic of several mAs and 0.8-1.0V for some hundreds of mAs is usually close enough in practice.

The above explanation is a condensed and somewhat simplified one. You will gain a better understanding as you get more and more involved in electronics.

Regarding your last question, I don't want to give you some offhand - and possibly misleading - answer. But it's way past midnight here now and I've been missing a lot of sleep lately. So I'll continue tomorrow unless someone else had already done the job by then.

how will the values change if i keep a white LED of 3.5 voltage drop near Rc ??

Click to expand...

Your problem is simple since you look to let the transistor be full off or full on.
(1) full off means Ic=0 hence Ib=0
(2) in the case of full on (Vce close to zero), we usually start from the output:
We have at the output an LED with a forward volatge of 3.5V (V_led) to be driven by 20mA current (I_led).
Therefore
Ic = I_led
Vcc = V_led + Ic*R1 + Vce
R1 = (Vcc - Vled - Vce) / Ic
Where:
Vcc = V1 = 5V
V_led = 3.5V
Vce = 0.2 (estimated since it is related to Ib, usually Ib is chosen to let Vce be less than 0.4)
Ic = 20 mA

R1 = 65R
The nearest standard value is 68R

Now to let Vce low (transistor in saturation) we can assume that Ib = Ic/20 (20 could be 50 down to 10, this depends on the transistor type and the current gain at Ic in the linear state).

Ib = 20/20 = 1mA
R2 = (V2 - Vbe) / Ib = (5 - 0.7) / 1 = 4.3 K
R2 = 3.9 K as a standard value.

I hope this helps you a little

Kerim

Pjdd

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thank you for the reply you made easy to understand but i am looking forward for your next explanation

---------- Post added at 11:11 ---------- Previous post was at 11:10 ----------

wa1kij - thank for the reply ....

---------- Post added at 11:12 ---------- Previous post was at 11:11 ----------

fareen -- according to the data sheet -- ic=500ma,ib=50ma,,,so Vce must be 1.6v or 1v ??? y do we have to select the highest values ?

---------- Post added at 11:12 ---------- Previous post was at 11:12 ----------

Kerim - thank you for the reply... but how did you estimated Vce as 0.2 ?

---------- Post added at 11:55 ---------- Previous post was at 11:12 ----------

Kerim - thank you for the reply... but how did you estimated Vce as 0.2 ?
kerim : Now to let Vce low (transistor in saturation) we can assume that Ib = Ic/20 (20 could be 50 down to 10, this depends on the transistor type and the current gain at Ic in the linear state).

can you please tell me why we should take 20 ???????????? i mean if we get vce(sat) = 50mv....y we should multiply by 20 to get 0.1 ???

.. ok i made a simple circuit to myself understand more... please correct me if i made any mistakes.

I want Ic = 20ma, Beta = 75(gain) , Vbe = 0.7(standard)
Vce = 400mv for Ic=150ma ==> Vce = 50mv for Ic = 20ma,

Calculations :

1) Rc=(Vcc-Vce)/Ic = (5-0.05)/2ma = 247 ohms.

2) Ib = Ic/75 = 0.2ma

3)Rb = (Vbb-Vbe)/Ib = (5-0.7)/0.2ma = 21Kohms.

this is what i got the results...
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if i keep a white LED(3.5v drop) at the collector side ==> Rc = (Vcc-Vce-Vled) / Ic = (5-0.05-3.5) / 20ma = 72 ohms......
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am i right ??

OK, let's see what we can do about that last question.

For the moment, let's continue with our analogy of a mechanical switch and a transistor used as a switch. Imagine a switch which does not have a snap action, but instead has considerable resistance if you push it lightly and make better contact (lower resistance) when you push harder. At some point, the resistance will reach its minimum and no longer decrease even if you keep pushing harder. That minimum resistance condition is analogous to transistor saturation and the push force is the base current.

To simplify things for the next part of the explanation, omit the LED and have only a resistor (R_L) as the collector load. If we start with a very small base current, it will cause a certain amount of collector current to flow (Ic = Ib*beta). That Ic will cause a voltage drop in R_L. The voltage drop is Ic*R_L and the remaining voltage is across the transistor's collector-emitter terminals.

For example, if we start with Ib of 20uA and beta is 100, Ic will be 100 times 20uA which is 2mA. If R_L = 100 ohms, the voltage drop across it = 2mA*100 ohms = 2V. If Vcc =5V, after dropping 2V across R_L, we'll have 3V remaining at the transistor collector. Since the emitter is at ground or 0V, Vce is 3V under that condition.

As we increase the base current, Ic will also increase until almost all the available voltage is dropped across R_L. The only voltage remaining across the transistor will be the saturation voltage Vce(sat) which is beyond our control.

For the moment, let's go back to our hypothetical non-snap mechanical switch above. It will be very difficult to apply just enough pressure on the switch to reach minimum contact resistance. So what we'd normally do is to apply enough extra pressure to make sure that the contact is in minimum resistance condition.

In the same way, it's almost impossible to calculate what would be just enough base current to cause the whole supply voltage to drop across R_L. So we apply enough extra base current to ensure that practically all the supply voltage is dropped by the load resistor R_L and the only voltage remaining across the transistor c-e is Vce(sat).

So how do we know how much base current to apply? A BJT amplifies the base current by a certain factor. But that current amplification factor (beta) varies from unit to unit even for transistors of the same type number. It also changes with temperature, with Vce and with current levels. A particular transistor may have beta of 100 at Vce of 5V and Ic of 10mA. But beta for the same transistor will change at different voltage and current levels. It drops quite a lot as we lower Vce down to saturation levels.

The usual practice is to use Ic/10 as Ib for medium beta transistors like the 2N2222, and Ic/20 or Ic/30 for high-beta transistors like a BC547B. These ratios are rule-of-thumb figures, not precisely calculated ones, just as we would apply an estimated amount of pressure on the non-snap switch.

Now let's start calculating practical values for your circuit:
We've already estimated that, at Ic = 20mA, Vce(sat) for a 2N2222 is about 50mV or 0.05V. V_LED = 3.5V, making a sub-total of 3.55V. That leaves (5 - 3.55)V or 1.45V to be dropped by the resistor.
R = V/I = 1.45V/20mA = 72.5 ohms. The nearest standard value is 68 ohms so that the actual current will be 1.45/68 = 0.021A = 21mA.

Using our rule-of-thumb formula, Ib should be about 21mA/10 or 2.1mA.

Again estimating Vbe as about 0.7V, the base resistor has to drop 4.3V at a current of 2.1mA.
4.3V/2.1mA = 2.05k. The nearest common value is 1.8k or 2.2k.

Whew! I wanted to make the reasoning behind the calculations understandable and knew that it would take a fairly long post. That's why I didn't tackle it at that late hour last night. If something's still not clear enough, just ask.

kdg007 said:

View attachment 67030

i am trying to understand the basics ... please help me out.

according to the data sheet there are two values given to the Vce and Vbe with Ic = (150ma and 500ma) and Ib = 15 and 50ma).

View attachment 67033

Click to expand...

They are giving data for two different variations of the model, one is the 2N2222 and the other is the 2N2222A. You need to select the data that applies to which ever one you have.
It should be noted that these are maximum values under specific test conditions. If you try this out you are likely to get slightly better results and they may vary depending other factors in the circuit.

DartPlayer170 - thank you

---------- Post added at 22:26 ---------- Previous post was at 22:22 ----------

Pjdd -- thank you .... choosing a proper Vbe and Vce value has been a hard part for me... you explanation helped me a lot

i am trying to understand the next step of the circuit what effect does R3 make in this transistor ???
i know it looks like a voltage divider... i want to know if we keep the same values Rc = 68ohms, Rb = 2.2kohms.... why R3 has been introduced ?
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thanks in advance

The most common reason, but not the only one, for placing a resistor at that position is not to act a voltage divider with Rb, but to bleed off any collector-to-base leakage current and prevent it from accidentally turning the transistor on. It can often be omitted with low-power silicon transistors working at normal temperatures as the leakage is usually very small. But it may be desireable or even essential under certain conditions.

In the absence of a b-e resistor, the c-b leakage current I_CBO goes into the b-e junction and acts just like an external base current input. For example, the maximum I_CBO of a 2N2222 is given as 10nA at 25°C. If this current is amplified by the transistor's own current gain, it's still less than 1uA at the collector. (The resultant collector current depends on the current gain beta which is much lower at such current levels than at some mAs). A collector-to-emitter leakage current of less than 1uA is negligible for most (though not all) applications.

However, at 125°C, the I_CBO of a 2N2222 may be as high as 10uA, not nA. Multiply that with a beta of, say, 100 and you have 1mA of collector current even when it's supposed to be off, and some transistors may have even higher leakage currents.

If we put a resistor between b and e, I_CBO now has an alternate path. It causes a voltage drop of R*I_CBO across the b-e junction. If we choose a proper resistor value, that drop will be too low to turn the transistor on. For example, I_CBO of 10uA and a 10k resistor will place 100mV across b-e. This is much too low to cause any significant collector current.

The value of R3 should be chosen such that the worst-case I_CBO will not cause enough voltage drop to turn the transistor on. At the same time, the resistance should not be so low that it will divert too much current from a normal base drive. In our example, we supply about 2mA of base drive in the on condition. Vbe(on) is about 0.7V so that 0.7V/10k or 70uA is diverted by the 10k resistor. The remaining 1.93mA goes into the base and is still high enough to ensure full switch-on.

Thus, we choose a proper value for R3 while considering transistor characteristics and other circuit conditions. High-power transistors have higher leakage curents than low-power ones. Such transistors used in high-power stages may need to have b-e resistors of 100 ohms or less.

Thank you for the explanation.so, if i keep a 10k resistor in BE ( in the image) it will be ok ? but when i calculate , the base is getting only 129ua... not 2ma !!!! the R3 is diverting 73uA but still the base is only getting 129uA..... y is that ?

That's because you're using 21k as Rb instead of ~2.1k. The drive current is now 10 times smaller than in our previous calculation, and after subtracting the current diverted by the 10k resistor, that leaves only that small current going into the base.

And how did you get 73uA through R3 instead of the 70uA we calculated before? Since we don't know the exact value of Vbe, we used the estimated value of 0.7V. 0.7V/10k = 70uA.

i have re-calculated from the scratch .. got the r3 value as 10k and the r1 as 2.2k ... not i got the desired current values.. for r3=70uA and the base is getting 1.95ma..
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i would try the same idea with BC107,BC107B.
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thank you very much.. you helped me a lot
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i keep learning more about these circuits..