Author Topic: common emitter amp  (Read 3241 times)

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Offline NASKTopic starter

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common emitter amp
« on: August 09, 2017, 04:33:24 am »
I want to design common emitter amp  having with 100 gain. my calculation is attached but design not amplify that much. therefore could i know what wrong in my calculation ? |O 
 

Offline ElektronikLabor

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Re: common emitter amp
« Reply #1 on: August 09, 2017, 06:17:39 am »
A gain of 100 is pretty high; most npn Transistors won't do that.
Try several stages of amplification; e.g. first stage with gain 20 and second one with gain 5.

If you want only one stage; you can try out a darlington transistor, which has a much higher current gain.
« Last Edit: August 09, 2017, 06:20:11 am by ElektronikLabor »
 

Offline Zero999

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Re: common emitter amp
« Reply #2 on: August 09, 2017, 08:16:07 am »
A gain of 100 is pretty high; most npn Transistors won't do that.
Try several stages of amplification; e.g. first stage with gain 20 and second one with gain 5.

If you want only one stage; you can try out a darlington transistor, which has a much higher current gain.
I think you've confused voltage and current gain.

BJTs can be configured with a voltage gain, well in excess of their current gain. A Darlington transistor may have a higher current gain, than a single transistor, but it actually has a lower transconductance, therefore will give a lower voltage gain, when configured as a common emitter amplifier.

I want to design common emitter amp  having with 100 gain. my calculation is attached but design not amplify that much. therefore could i know what wrong in my calculation ? |O 
I can hardly read that schematic. It's so fuzzy and small.

Try setting the DC voltage gain to something more manageable, so the voltage across the collector resistor can be stabilised to half the supply voltage. Add a bypass capacitor and resistor across the emitter resistor, to increase the AC gain to the desired level.

See attached example.

« Last Edit: August 09, 2017, 08:41:07 am by Hero999 »
 

Offline NASKTopic starter

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Re: common emitter amp
« Reply #3 on: August 09, 2017, 09:52:27 am »
thank you for the answers :D :D :) !!
Hero999, your  attached example circuit Icq equals almost 0.46mA. why did you use that much of small current ,  can I use 5mA for Icq.  and your gain is  almost 200, if start from Av=Rc/Re' equation can I achieve this given solution. 
« Last Edit: August 09, 2017, 10:14:50 am by NASK »
 

Offline Zero999

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Re: common emitter amp
« Reply #4 on: August 09, 2017, 10:37:35 am »
thank you for the answers :D :D :) !!
Hero999, your  attached example circuit Icq equals almost 0.46mA. why did you use that much of small current ,  can I use 5mA for Icq.
No particular reason. You can increase the current to 5mA if you like and reduce all the resistor values to compensate.

Quote
your gain is  almost 200, if start from Av=Rc/Re' equation can I achieve this given solution.
You forgot to take the fact that the BJT doesn't have a limitless transconductance. In this case, the easiest way to account for this is imagining an internal emitter resistor, with a value of 25mV/Ic.

Re' = 25mV/Ic = 25mV/0.46mA = 54R
Re = Re' + R5 = 93R                             ;We can ignore R2, as it's much larger than R5.
Av = 10000/93 = 108

If you download LTSpice (it's free and runs under WINE if you use Linux/Mac OS) you can simulate the .asc file attached to my previous post.
« Last Edit: August 09, 2017, 10:41:01 am by Hero999 »
 

Offline phliar

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Re: common emitter amp
« Reply #5 on: August 10, 2017, 12:41:59 am »
The best video on designing a BJT common-emitter amp is the one from The Signal Path:


Returning to electronics after a 25 year break.
 

Offline orolo

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Re: common emitter amp
« Reply #6 on: August 10, 2017, 07:41:55 pm »
The computations look a bit fuzzy, but it seems you usted a grounded emitter transistor with 500 ohms collector resistor. That kind of amplifier is very unpredictable and temperature dependent (might easily saturate with a slight temperature increase) besides having low input impedance. Of course, if you choose to use emitter degeneration, for a gain of -100 you need a x100 colector resistor, which will be unpractical. For example, for a 100 ohms emitter resistor, you'd need a 10k collector one, who would steal 50 volts quiescent voltage! For these situations you have two choices: either cascade two moderate gain amplifiers, as suggested, or use an active load: instead of the collector resistor, use a pnp or jfet current source. In the last case you'll have gain to spare, so with a bit of feedback you'll achieve both correct gain and high input impedance.

In any case, more that one transistor is required.
 

Offline NASKTopic starter

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Re: common emitter amp
« Reply #7 on: August 17, 2017, 01:22:36 pm »
I designed circuit emittor bypassing capasitor value is almost 2uF accoding this calculation
Xc=1k/10 = 100R
c=1/2?fXc= 1/(2?*1khz*100R)=1.5uF 
when I use calculated value, amplification getting low , but I use 100uf it getting correct amplification. how to overcome this issue.
 

Offline Zero999

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Re: common emitter amp
« Reply #8 on: August 17, 2017, 02:17:15 pm »
I designed circuit emittor bypassing capasitor value is almost 2uF accoding this calculation
Xc=1k/10 = 100R
c=1/2?fXc= 1/(2?*1khz*100R)=1.5uF 
when I use calculated value, amplification getting low , but I use 100uf it getting correct amplification. how to overcome this issue.
What do you mean by low? What's the predicted and actual amplitude? How does it compare with what you get at 5kHz?
 

Offline NASKTopic starter

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Re: common emitter amp
« Reply #9 on: August 18, 2017, 02:10:56 am »
Using 100uF capacitors 5mv signal amplify 1.06 V @ 1KHz, When using calculated capacitor’s value 2uF, 5mv amplify 350 mV @ 1KHz.  Required value for amplification is 1V.
Using 100uF capacitors 5mv signal amplify 1.06 V at 5KHz.

attached pdf shows results and diagram
 

Offline Zero999

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Re: common emitter amp
« Reply #10 on: August 18, 2017, 08:17:57 am »
I designed circuit emittor bypassing capasitor value is almost 2uF accoding this calculation
Xc=1k/10 = 100R
c=1/2?fXc= 1/(2?*1khz*100R)=1.5uF 
when I use calculated value, amplification getting low , but I use 100uf it getting correct amplification. how to overcome this issue.
Where did you get that formula from?

Try this:
RE' = 0.025*IE
C = 1/(2pi*FC*RE')

Where FC is the 3dB point, when the voltage is 0.707 of the maximum. If you don't want any attenuation than make FC much lower, than the frequency of interest.

With an FC of 500Hz and the component values you've used, I get around 8.8µF for C.
« Last Edit: August 18, 2017, 08:20:11 am by Hero999 »
 

Offline IanMacdonald

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Re: common emitter amp
« Reply #11 on: August 18, 2017, 08:44:19 am »
There will be two problems with achieving a gain of 100 this way. BJTs are essentially current amps,  and driving the base with a voltage, with little or no series resistance, will give rise to nonlinearity due to the b-e diode law conduction. That, and the gain will vary quite a bit across samples so no use for a production run if it must be 100.

Better to use a simple discrete op-amp.  A two transistor complementary arrangement should be adequate. Only slightly greater component count and far better results. The feedback potential divider then determines the gain to within a few percent regardless of hfe tolerances. 



The gain here wil be too high, increase the 100R.
 

Offline Zero999

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Re: common emitter amp
« Reply #12 on: August 18, 2017, 09:50:22 am »
There will be two problems with achieving a gain of 100 this way. BJTs are essentially current amps,  and driving the base with a voltage, with little or no series resistance, will give rise to nonlinearity due to the b-e diode law conduction. That, and the gain will vary quite a bit across samples so no use for a production run if it must be 100.

Better to use a simple discrete op-amp.  A two transistor complementary arrangement should be adequate. Only slightly greater component count and far better results. The feedback potential divider then determines the gain to within a few percent regardless of hfe tolerances. 
I must point out that the Hfe does not determine the voltage gain of the circuit. It's proportional to the transconductance (which is temperature and current dependant) and the collector resistor.

I doubt this is intended to be a useful circuit. It's most likely a college assignment.
 

Offline BigBoss

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Re: common emitter amp
« Reply #13 on: August 18, 2017, 02:54:22 pm »
Here are the results of your Common Emitter Degenerated Amplifier.I have used MathCAD to calculate and "iterate" the equations.
Ref: "Analysis and Design of Analog Integrated Circuits. Paul R. Gray, Paul J. Hurst, Stephen H. Lewis, Robert G. Meyer.John Wiles and Sons, 4th Edition" Page 197-200
« Last Edit: August 18, 2017, 02:56:12 pm by BigBoss »
 

Offline firewalker

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Re: common emitter amp
« Reply #14 on: August 18, 2017, 10:10:49 pm »
Become a realist, stay a dreamer.

 


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