CE Transistor Amplifier

[regale]

Hi,
I am trying to amplify the ac input signal but any input value results in 0V output, Can anyone what's wrong with my circuit? I don't want to use a voltage divider bias configuration.

 

[Les Jones]

The AC input signal does not even get to the base of the transistor. It is shorted out to the +6.7 volt supply. This is assumed to be zero ohms impedance so the voltage cannot deviate from +6.7 volts. Even if the AC signal got to the base the voltage on the collector cannot change from +4 volts. You need to re-design your circuit.

Les.

 

[danadak]

Take a look at this -

Common-Emitter Amplifier

Regards, Dana.

 

[rjenkinsgb]

I am trying to amplify the ac input signal but any input value results in 0V output, Can anyone what's wrong with my circuit? I don't want to use a voltage divider bias configuration.

The right hand side of C2 should connect directly to the transistor base, not power.

However you will still not get any output as the collector of the transistor is also connected directly to a power source.

It needs a load resistor in the collector circuit, so the varying current through the transistor results in a varying voltage across that resistor.

That will give a an output signal from the collector.

The resistor value should be calculated so it has a voltage drop of roughly somewhere between a third and half the collector supply voltage, at the collector bias current you set.

eg. 150 Ohms, possibly?

 

[danadak]

Just a caution, direct connection of C2 to base with no emitter degeneration
will result in a lot of loading to input signal. As well pretty serious distortion
possibly, especially since circuit does not have proper bias.

Common-Emitter Input Impedance – 2N3904Blog

Computation of Rpi for use in above link.

How to Calculate RÏ€ (Rpi) of a Transistor

How to Calculate RÏ€ (Rpi) of a Transistor

www.learningaboutelectronics.com

Regards, Dana.

 

[atferrari]

I don't want to use a voltage divider bias configuration.

May I know why?

 

[Papabravo]

I'm a bit curious about where you got the notion that your circuit in post #1 could ever possibly work.

 

[regale]

Guys Thanks for all your replies. Really appreciated.
atferrari why I don't want to use voltage divider bias is because I want to test fixed bias configuration first.
Second I identified the mistake myself later and corrected the circuit and it finally worked with some different values.
Is there any book available that precisely focuses on circuit explanation? like what each component does etc
not just the basic theory.

 

[Nigel Goodwin]

Perhaps you would like to post your new circuit? - as the original one isn't a circuit at all, so we've no idea what you've altered.

 

[atferrari]

Guys Thanks for all your replies. Really appreciated.
atferrari why I don't want to use voltage divider bias is because I want to test fixed bias configuration first.
Second I identified the mistake myself later and corrected the circuit and it finally worked with some different values.
Is there any book available that precisely focuses on circuit explanation? like what each component does etc
not just the basic theory.

I suspect that if you actually knew how the one you don't know how it works, you would have not started this thread.

All in all, a basic CE amplifier takes minutes to calculate and less than 20 more to assemble in a breadboard.

 

[regale]

Nigel Goodwin
This is the modification and the input is 10mV.

 

[Papabravo]

You have something that "appears" to work. That is a naive viewpoint. Let me ask you if the output waveform looks at all like the input waveform. If not, it would be more correct to call it a distortion circuit rather than an amplifier. One purpose of an amplifier is to replicate the input waveform while increasing the voltage, current, or power. Get your head out of the dark place it is in and start learning about actual circuits rather than the fever dreams you have put forth.

 

[danadak]

Transistor bias calculators :

BJT Transistor Biasing Calculator

Transistor Bias Voltage Calculator - Transistor Bias Voltage Calculator - Bias Calculation,Online computing, online calculator, calculator online calculation

gislite.com

BJT Transistor Bias Voltage Calculator - Calculates for Series Resistor and Voltage Divider

calculators, engineering calculators....

calculatoredge.com

Regards, Dana.

 

[regale]

Papabravo please give reference to any book for actual circuit learning, not just numerical.

 

[audioguru]

The first thing I was taught about biasing a transistor is to never simply use one base resistor to a supply voltage.
Because each transistor has a different current gain and has a different base-emitter voltage even if they have the same part number.

Also, the current gain increases as the transistor heats up.
Therefore a voltage divider biases the base plus an emitter resistor is added to 0V.
Sometimes a biasing resistor is connecting between the collector and base for negative feedback.

 

[regale]

Therefore a voltage divider biases the base plus an emitter resistor is added to 0V.

emitter resistor to 0V ??

 

[danadak]

Step by step design :

https://www.phy.olemiss.edu/~cremaldi/PHYS321/CommonEmitterAmp%20Design-HW.pdf

https://www.csun.edu/~jou/classes/ece340/2022s/lectures/lec_9_22s.pdf

https://www.hunter.cuny.edu/physics/courses/physics222/repository/files/pdf/ElectronicsLab15.pdf

Regards, Dana.

 

[Papabravo]

Yes, a resistor from the emitter terminal to GROUND. It may or may not be fully bypassed with a capacitor to GROUND as well.

 

[audioguru]

Do it like this.
Sorry, the 220uF capacitance of the input capacitor is way too much. Try 10uF.

 

[LvW]

The first thing I was taught about biasing a transistor is to never simply use one base resistor to a supply voltage.
Because each transistor has a different current gain and has a different base-emitter voltage even if they have the same part number.

Also, the current gain increases as the transistor heats up.
Therefore a voltage divider biases the base plus an emitter resistor is added to 0V.
Sometimes a biasing resistor is connecting between the collector and base for negative feedback.

Comment 1: "Different base-emitter voltage"? A transistor does not "have" such a voltage. The base-emitter voltage Vbe is applied externally - and it has a certain relationship to the emitter resp. collector current as defined by the well known exponential formula (which, however, is remarkably temperature sensitive).

Comment 2: From the last sentence one could get the impression that such a resistor between C and B should/could be used in addition to an emitter resistor. No, this is not recommended - this would be counter productive.
It is the emitter resistor which provides negative feedback and, thus, stabilizes the device against the temperature influence as mentioned in my comment 1 above.