What are common issues when using 2N2222 in a BJT NPN amplifier?

In summary: R2 should be 1.9... kohmIn summary, when working on a BJT NPN amplifier with a 2N2222 model, the user encountered three problems: no AC voltage at output Vce, difficulty determining hoe resistance, and unsure if schematics for small signal analysis were incorrect. After clarifying some details, it was determined that the design was flawed and not worth pursuing. The user was then given a new task by their professor to compile a common emitter amplifier with specific parameters and chosen a BC547 transistor. The user was in the process of determining the necessary values for resistances R1 and R2.
  • #1
Ivan Antunovic
111
4
Hello people,

I am working on a BJT NPN amplifier the model that I've choosen is 2N2222 : http://www.e-voron.dp.ua/files/pdf/tranzistor/2N2222.pdf

I am having 3 problems here:

1) am not having any AC voltage at output Vce seems like I have set wrong operating point?

2) I am trying to figure out how to determine hoe resistance of this transistor from the graph hybrid output resistance hoe=dIc/dUce at operating point Q

3)Is my schematics for the small signal analysis wrong?

I hope pictures are not blur.
 

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  • #2
Okay they are blur
BJT_NPN_Shema.jpg

DSC_0792.jpg

http://postimage.org/
 
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  • #3
Ivan Antunovic said:
Okay they are blur
BJT_NPN_Shema.jpg
and Xce should be in parallel with Re my bad
DSC_0792.jpg

http://postimage.org/
 
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  • #4
Ivan Antunovic said:
Hello people,

I am working on a BJT NPN amplifier the model that I've choosen is 2N2222 : http://www.e-voron.dp.ua/files/pdf/tranzistor/2N2222.pdf

I am having 3 problems here:

1) am not having any AC voltage at output Vce seems like I have set wrong operating point?

What's your frequency and amplitude? Your RC and RT are both very low.
2) I am trying to figure out how to determine hoe resistance of this transistor from the graph hybrid output resistance hoe=dIc/dUce at operating point Q
forget about hoe. It's negligible.
3)Is my schematics for the small signal analysis wrong?
No idea. Depends on your software. Your schematic in post 2 does not show a signal source ...
 
  • #5
rude man said:
What's your frequency and amplitude? Your RC and RT are both very low. forget about hoe. It's negligible.
No idea. Depends on your software. Your schematic in post 2 does not show a signal source ...

Okay forgotten to say about voltage source it's sinus with amplitude 100 mV and frequency 50Hz
 
  • #6
Ivan Antunovic said:
Okay forgotten to say about voltage source it's sinus with amplitude 100 mV and frequency 50Hz

I know that hoe is so small but I am trying to do things atleast as correct as I can I do this for my soul.There is Ic-Uce curve in the link that I've given but it's in logaritmic paper how to I convert it from dB to Ampers and Volts?
 
  • #7
Due to C1 being so large you should be seeing a horribly distorted sine wave at the collector of about 200 mV.
At the output there is very little ac since the time constant formed by RC, RT and CV2 is very low (cutoff freq. about 2 KHz).

This is a really bad design so worrying about hoe is a total waste of time.
What does your simulator say?

.
 
  • #8
rude man said:
Due to C1 being so large you should be seeing a horribly distorted sine wave at the collector of about 200 mV.
At the output there is very little ac since the time constant formed by RC, RT and CV2 is very low (cutoff freq. about 2 KHz).

This is a really bad design so worrying about hoe is a total waste of time.
What does your simulator say?

.
I have taken schematics from the internet,voltmeter shows no voltage at output and oscilloscope shows no signal at output
 
  • #9
Ivan Antunovic said:
Okay forgotten to say about voltage source it's sinus with amplitude 100 mV and frequency 50Hz

For a signal frequency of 50Hz your coupling capacitors are way too small; they are attenuating the signal more than the 2222 is amplifying it. What is the reactance of a 1uF capacitor at 50Hz?
 
  • #10
NascentOxygen said:
For a signal frequency of 50Hz your coupling capacitors are way too small; they are attenuating the signal more than the 2222 is amplifying it. What is the reactance of a 1uF capacitor at 50Hz?
it's 3813 ohms.Okay then make it 10kHz so the reactance goes down by 200
 
  • #11
Forget about this task it has very bad resistances.

I got a new task from my professor:

Compile common emmiter amplifier(NPN) with the following default parameters:

VCC = 12V, power transistor 50 mW max. output signal between 3-5V, VRE = 1V.

Determine Computing and measuring AV, AI, Rin, Rout, the upper and lower limit frequency, THD.

Then, instead of RL connect the speaker and try something heard on the speaker.

I have choosen transistor BC547.

I assumed that the voltage at RL is 5V and from PRL=VRL^2 / RT I get that RL=250 ohms.I measured that Vbe=0.68V.

I am trying to figure out what should be resitances R1 and R2
transistor.jpg
image upload no limit
 
  • #12
Ivan Antunovic said:
Forget about this task it has very bad resistances.

I got a new task from my professor:
Compile common emitter amplifier(NPN) with the following default parameters:
VCC = 12V, power transistor 50 mW max. output signal between 3-5V, VRE = 1V.
Determine Computing and measuring AV, AI, Rin, Rout, the upper and lower limit frequency, THD.
Then, instead of RL connect the speaker and try something heard on the speaker.
I have choosen transistor BC547.
I assumed that the voltage at RL is 5V and from PRL=VRL^2 / RT I get that RL=250 ohms.I measured that Vbe=0.68V.
I am trying to figure out what should be resitances R1 and R2
R1 and R2 are chosen to give you the required 1V at the emitter. Consider the base dc current also.
The professor meant to limit your dissipation to 50 mW, not that the transistor cannot sustain more than 50 mW, probably.
EDIT:
I didn't see the 1V VRE requirement. So this will not be a real common-emitter circuit.
What are your upper and lower frequencis?
You can probably connect the speaker in series with the collector resistor.
 
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  • #13
Ivan Antunovic said:

I got a new task from my professor:

Compile common emmiter amplifier(NPN) with the following default parameters:

VCC = 12V, power transistor 50 mW max. output signal between 3-5V, VRE = 1V.
The specs say Vcc = 12V, yet your calculations use Vcc of 15V. Why? :oldconfused:
 
  • #14
NascentOxygen said:
The specs say Vcc = 12V, yet your calculations use Vcc of 15V. Why? :oldconfused:
My bad I was really tired at evening when doing this that I didn't even notice it was 15 V.
 
  • #15
How do you mean it won't be real common emitter circuit when it has a resistor RC connected directly to the collector electrode?R1 and R2 are voltage dividers I know that they are used to set Quiescent Point.If R1 and R2 are voltage dividers of Vcc and if I want 13.35 V at R1 and 1.65 V at R2,R1 should be 8.09 times larger than R2.

Edit(after drinking a coffee :D):
Btw I am using BC547 for this circuit https://www.fairchildsemi.com/datasheets/BC/BC547.pdf maybe I should take some data from the datasheet and try calculating resistances.For example there is a Ic-Vce curve and I tried to make load line and then it's a piece of cake for finding the resistances.

Load line equation is:
Ic=-Uce/Rc+Re + Ucc/Rc+Re (Asumming that Ic=Ie)the maximum Vce voltage is Vcc which is 12 volts ,maximum current Ic will flow when Ucc/Rc+Re and therefore it's y-intercept of the curve.But still I don't know Rc and Re so therefore I can't find y-intercept...maybe I should consider a worst case scenario and take values for Vce ,Ic and Ib from the datasheet?
 
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  • #16
If you're aiming for 50mW dissipation, and you bias for VCE of around 6-7V, then what will be the DC value of IC?
 
  • #17
NascentOxygen said:
If you're aiming for 50mW dissipation, and you bias for VCE of around 6-7V, then what will be the DC value of IC?

Well I don't know since I am trying to figure out how to find R1 and R2 since they determine the bias IB and then IC.If I assume that Vce is around 6 V then VRC=VCC-VCE-VRE=12-1-6= 5 V but still I don't know the value of the resistance RC?
 
  • #18
power dissipation = voltage x current

- which voltage?
- which current?

Remember that power into the base is a negligible fraction of total power, for these small signal amplifiers.
 
  • #19
NascentOxygen said:
power dissipation = voltage x current

- which voltage?
- which current?

Remember that power into the base is a negligible fraction of total power, for these small signal amplifiers.
I think he meant power dissipated at the output load resistance
 
  • #20
Yes power into the base is negligible but I didnt understand first part of your post?
 

What is a Transistor NPN BJT 2N2222?

A Transistor NPN BJT 2N2222 is a type of bipolar junction transistor (BJT) that uses two layers of n-type material and one layer of p-type material to amplify and switch electronic signals.

What is the function of a Transistor NPN BJT 2N2222?

The main function of a Transistor NPN BJT 2N2222 is to amplify or switch electronic signals. It acts as a current-controlled switch, allowing a small current to control a larger current. It can also be used as an amplifier to increase the strength of a weak signal.

What are the characteristics of a Transistor NPN BJT 2N2222?

The Transistor NPN BJT 2N2222 has three terminals - the emitter, the base, and the collector. It has a maximum voltage rating of 40V and a maximum current rating of 800mA. It has a low collector-emitter saturation voltage and a high current gain, making it suitable for a wide range of applications.

What are the common uses of a Transistor NPN BJT 2N2222?

The Transistor NPN BJT 2N2222 is commonly used in electronic devices such as amplifiers, oscillators, and switching circuits. It is also used in digital logic circuits and as a driver for other transistors or components.

How do I identify a Transistor NPN BJT 2N2222?

The Transistor NPN BJT 2N2222 can be identified by its part number, which is usually printed on the device. It has a distinctive TO-92 package, with the emitter lead shorter than the collector and base leads. It also has a flat side on the top of the package, which can be used to orient the transistor correctly.

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