Hello,

Okay... I am trying out a simple common emiter circuit with an 2n2222 transistor and the calculated ic does not match the actual measured ic.

My calculations gives an ic of 962ma but I only measure 391ma???

I don't believe the HFE is considered as 10 for this example because the available short circuit current would be 3.3A and since ic is less than 3.3 A then we should be in active region... confused!

Can someone please take a look at my circuit attached and see as to why I am getting this discrepantie?

Thanks all for help!

r

 

Picking a 2N2222 datasheet at random, the maximum rated power dissipation is 500mW. 962mA at 2.3V would be 2.21W. You may have simply toasted your transistor. Do you have the datasheet for your particular transistor?

 

When getting a beta of 10 in modern silicon transistors in active region, most common thing could be ,that its not the forward beta its reverse beta, it happens when you connect your transistor in reverse i.e.. you connected the collector in place of emitter and emitter in place of collector.

Or else its damaged...

In the above circuit you are working, taking hfe (forward beta) as constant (100) is wrong, hfe (forward beta) of 100 in the datasheet is just a typical value it differs a lot in real. The circuit you are using ,is best for switching small signal or small load and the transistor needs to operate in saturation region and in the calculation you use the "forced beta" to saturate a BJT.

Good Luck

 

Hello betm and thank you for replying to my question

>You may have simply toasted your transistor.

OOPS!

I am using a PN2222A and I haven't looked up the spec sheet in a long time ... I just looked it up again and here it is:

http://www.fairchildsemi.com/ds/PN/PN2222A.pdf

A while back I thought I read 1A max....

I am using a PN2222A and not a 2N2222. When I look these two part numbers up, they resolve to the same data sheet. So why do some transistors have PN2222 mark while others have the 2N2222... what's the diff?

So lets start over with another simple circuit if I may! First, in this new circuit
I have a PN2222 transistor in a common emitter configuration, however it has an TSAL6100 IR-LED connected to its emitter. Here is the spec sheet for this led:

http://www.vishay.com/docs/81009/tsal6100.pdf

According to spec, to fully benefit from its maximum infra red emissions, this LED requires a forwards voltage of at least 1.35 VDC and should draw approximately 100 ma.

Please view attachment of the new circuit. Again the ic calculations don't reflect the actual readings. Calculated ic is 103 ma, where as measured ic is 78 ma ???

I am obviously doing some error in my calculation. Do you see where I am making a mistake?

All help is appreciated.
Thanx
r

 

Hello Debjit625,

I am using a HFE of 104 cause my multi-meter has the HFE option on it ... I just realized it this morning .... LOL ))

Also, I don't think I am connecting the transistor wrong.... I think I really made sure of this.

Thanks for your input... appreciated

regards
r

 

Did you measure 1.35V across the LED? That number, like beta, will also vary widely from spec ideals, based on temperature and current. Measure Vbe and Vled and let us know what you find.

Beta of 104 is also just an approximation. Yes, you measured it, but only at one particular operating current. It actually varies by Ic and temperature.

 

Hi bretm,

I think that it was my mistake !!! I now read the 1.35V across the led.

Man! do these things vary!!!!

Okay, well, we will keep 104 HFE since at least I measured it ...
Also, the emitter PN junction is really 0.82 and not as the typical value of 0.7! So we will keep 0.82 since it too comes from a measurement.

Also, I did a small test circuit to measure the Vdrop across the led. I connected one side of the led to ground and the other side to a 1k ohm resistor. The other side of the 1 k ohm resistor is connected to +3.33 VDC. I measured 1.31 VDC across the led. As per spec of the led, vf is 1.35v. So this is pretty close too. So I used 1.31 in my calculations.

In my room right now, I read a normal room temperature of 27 deg C. and nonetheless I have to say there is still differences between theory and practical measurements.

The ic, vce and the rc v-drop are all different when measured.
Take a look at the circuit I have re-done with greater precision!

All I can say is that the resistors, v-gama(0.82) are precise values!

What do you think... are the theoretical vs practical differences in values normal in this case?

P.S. Excuse the pics for being so dark... I am taking these with my smart phone and cannot get them any brighter..


regards
r

 

Hi,
At the beginning I recommend you to build this circuit

Where are LED diodes not a IR diode.

Next change Rx resistance from 1K; 10K 100K; 220K; 470K ; 680K.
And measure voltage across R1 resistor and across Rx resistor.
And then
Ib = V_Rx/Rx and Ic = VR1/R1 ----> Hfe = Ic/Ib

Also for low Rx values the BJT will be in saturation region.
And as you shoudl know Ic = hfe *Ib don't hold in saturation. Your BJT will come out from saturation region for Rx > 100KΩ. When BJT start to come-out from saturation the collector voltage start to rise. And if Vce reaches 1V the BJT is now working in active region.

 

Hi bretm,

I now read the 1.35V across the led.

Okay, well, we will keep 104 HFE since at least I measured it ...
Also, the emitter PN junction is really 0.82 and not as the typical value of 0.7! So we will keep 0.82 since it too comes from a measurement.

r

Your BJT is in saturation this is why Vbe < 0.7V and Uce = 1.51V - 1.35V = 0.16V And in saturation Ic = β*Ib don't work any more.

PS Also don't forget that your ammeter also has a internal resistance.

 

Yeah, that's definitely saturation. In addition to the 0.26V Vce, you can see that your collector voltage (1.57V) is lower than your base voltage (2.13V), so now the CB junction is forward-biased.

 

Don't forget that the base emitter junction is just like a diode, you can put current through it with nothing connected to the collector, or if the transistor is in saturation you can increase the base current without making any difference to the collector current. That's why the beta calculation doesn't work in saturation.

 

How much collector current is used to measure hFE with your multimeter? Less than 1mA? The transistor is linear with a fairly high emitter-collector voltage so it is not saturated.
That is a lot different than the 100mA you want the transistor to conduct when it is saturated.

Then look at the typical hFE in the datasheet with a collector current of 150mA, hFE= 100 to 300 when the transistor is not saturated or hFE= 10 when the transistor is saturated.

 

Hi rougie:
If you want to measure the hfe, Vce voltage is very important, it's according to the datasheet as following:

Electrical Characteristics Ta = 25°C unless otherwise noted

On Characteristics
hFE DC Current Gain
IC = 0.1mA, VCE = 10V, hFE = 35
IC = 1.0mA, VCE = 10V, hFE = 50
IC = 10mA, VCE = 10V, hFE = 75
IC = 10mA, VCE = 10V, Ta = -55°C, hFE = 35
IC = 150mA, VCE = 10V *, hFE = 100,300max
IC = 150mA, VCE = 1V *, hFE = 50
IC = 500mA, VCE = 10V *, hFE = 40

PN2222A datasheet page 2.
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf

 

Hello guys,

It is overwhelming on how much help you have all provided... thanks!

Okay, I think I got the idea. The idea here, is the more ib we provide, the higher ic gets. Having said this, the higher ic gets, the lower vce becomes. If ic gets high enough, vce gets very close to 0V and it is said that the transistor is in saturation mode.

Therefore, it seems that "Ic = hfe*Ib" don't hold true when the transistor is in saturation mode. And saturation mode pretty much occurs when the voltage across the collector/emitter junction starts being less than 0.7V to 1V.

-QUESTIONS-
[1]- Not in reference to any attachments I have provided, when a transistor is NOT in saturation mode (meaning we are in linear region mode), we measure a VBE drop of approximately 0.7V. But at this instance doesn't the cb junction have some very small vdrop too? In other words, when we measure VCE drop, is it a combination of the BE and CB junctions?

[2]- In reference to a tutorial I watched, it mentioned that as a rule of thumb, when the ic current is greater than the max short circuit current (Max I), the transistor tends to be in saturation mode:

So I would like to know if my calculations for determining this are correct:

First, in reference to my T7 attachment, I figure out Max I:

Max I = (VCC-VD1)/RC1
Max I = (3.33-1.31) /20
Max I = 101 ma

Then using the following formula, I figure what the ic current would be:

ic = HFE x ib
ic = HFE x (Eb/Rb)
ic = 100 x (3.33 - 0.7 - 1.31)/1260
ic = 104.7 ma


and since the ic current calculation comes out greater than the MAX I current calculation, then I would assume that the transistor is in saturation mode... Is this assumption correct?

[3] - Is it okay to use a transistor in saturation mode? I mean I am using the circuit in T7 (post#7) and no components heat up and it really works very well, actually too well, since there is too much infra red coming out of the LED ... . So I would need to increase RB1 and do some tests. But I mean, there nothing wrong with using a transistor in saturation mode right?


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?

Thanks all for your help.
rob

 

Not in reference to any attachments I have provided, when a transistor is NOT in saturation mode (meaning we are in linear region mode), we measure a VBE drop of approximately 0.7V. But at this instance doesn't the cb junction have some very small vdrop too? In other words, when we measure VCE drop, is it a combination of the BE and CB junctions?

Well, yeah, in the sense that Vce always equals Vcb + Vbe, but that's not really useful to know. Vcb isn't really an interesting number in most cases, other than to tell you that you're in saturation when Vcb < 0, or edging out of forward active mode if Vcb is near 0. In an amplifier, for example, that would be a flag that there's an issue with biasing and would mean distortion.

In reference to a tutorial I watched, it mentioned that as a rule of thumb, when the ic current is greater than the max short circuit current (Max I), the transistor tends to be in saturation mode:

Pretty much. If the external circuit can provide at least the current indicated by Ib x beta, that means you're probably not going to saturate.

Is it okay to use a transistor in saturation mode?

Saturation does not harm the transistor. As long as you intended to use the transistor as a switch, it's perfectly normal.

 

Hello ScottWang,

Okay, well, I have done a new circuit, this time in linear range mode! Based on the attachement (T8), if I go according to the 2N2222 spec sheet I would have to use a HFE of 50 as per table:

IC = 1.0mA, VCE = 10V, hFE = 50

However, using 50 as HFE does not reflect ic at all?? And note that the transistor is NOT in saturation mode!

Using 100 HFE is much much closer.

And what do they really mean with the "VCE = 10V"??? my VCE is 1.61...?



P.S. oooops... here you go ... T8 is now attached!

regards
r

 

For your IR led circuit it is best to have the transistor in saturation. 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. If you are relying on the transistor in active mode then the current through the LED is much more likely to change.

 

Rougie,
Why do you keep picking the hFE?
1) The datasheet shows that the minimum hFE is 50 at 1mA but the typical and maximum hFE are not shown so the hFE can be anywhere from 50 to maybe 150 when the Vce is 10V. Your transistor does not have a Vce of 10V, instead it is saturated.
The datasheet for almost any little transistor shows that it saturates well when the base current is 1/10th the collector current because when saturated the minimum hFE is very low, near only 10.
2) You cannot select the base current by picking an hFE of 100 because you don't know the hFE. You must calculate the base current according to the minimum hFE listed on the datasheet or many of your circuits will not work.

The transistor has a Vce of 10V when it has a 20V supply, a 10k collector resistor and enough base current to make the collector current 1mA.

 

Hi rougie:

I can't see the circuit of T8, only can see the circuit of T7.

The IR LED using in the logic area very often, it's working as Hi and Lo signal in the circuits, it means that IR LED doesn't working in the liner area, as amplifier working with sinewave.

The following is explaining how the transistor working, and no matter the hFE is Hi or Lo.

If you want to using in logic area, the hFE value you can use about 50, it's close to Vce = 1.0V, Ic = 150mA.

The TSAL6100 IR LED Vf=2.6~3V, If=100mA, Ifm=200mA(used as square wave signal), when we using the components, just don't use in the max ratings, I always used about 80% of max rating for LED relative components, it can keep it's function and parts life.

We can't change the vf value, but we can reduce the If to 80%.
static circuit : If = 100mA x 80% = 80mA
square wave circuit : Ifm = 200mA x 80% = 160mA.

Vce(Sat):
Rb = (3.33V - 0.7V)/15mA = 180Ω
Although the datasheet described the vce(sat)=0.3V, but when you measure it probably less then 0.2v, so I used 0.2V to calculate.

All the other details also shows on the circuit.

VCE(sat) Collector-Emitter Saturation Voltage , IC = 150mA, IB = 15mA, Vce=0.3V
PN2222A datasheet page 2.
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf

TSAL5100,6100 Series IR LED
http://www.datasheetarchive.com/TSAL6100*%C2-datasheet.html

TSAL6100 Real Aplication schematic,PIC122F629.
http://mondo-technology.com/irdet.htm

 

I'm sorry, because I saw the wrong place, so the data was wrong.

You should measure the IR LED from datasheet, and set If = 100mA with square wave.
If you want to measure the static input voltage as DC, probably the R1 should be double and reduce the current to 1/2, but this method is not shows in the datasheet, so I can't make sure.

Forward voltage IF = 100 mA, tp = 20 ms VF 1.35V(typ) 1.6V(max)
TSAL6100 datasheet,page 2.
http://www.vishay.com/docs/81009/tsal6100.pdf