Theoretical gain of common emitter amp

**caffine** · 19th April 2009 10:56 PM

Hey guys

how do i calculate the theoretical gain of my common emitter amplifier

my values are as follows
VCC = 12V Beta = 239
VE = 1.2 BV = 1.9 VBE = 0.7 VE = 1.2 RE = 1200 IE = 1mA IC = 1mA
RC = 6kOhmz R1 = 148kOmhz r2 = 28.68KOhmz Re = 1.19KOhmz
C1 10.81 uF C2 = 9.99uF c3 = 93.5uF

any help provided would be aprreciated, some of these values maybe wrong im not very good with electronics hence why i am asking on here

any help provided would be greatly appreciated

**crutschow** · 19th April 2009 10:57 PM

Submit a schematic.

**audioguru** · 19th April 2009 11:32 PM

Your transistor probably does not have any negative feedback so it will be extremely distorted at high output levels even if it is not clipping.
If its load is 60k ohms or more then its voltage gain is about 120 but the severe distortion will make it difficult to measure.

I didn't look to see if it is properly biased.

**Hero999** · 19th April 2009 11:35 PM

It's just the usual gain equation.

$Av_{CL}= \frac{Av_{OL}}{1+ \beta Av_{OL}}\\<br /> \beta = \text{Negative feedback factor} = \frac{R_C}{R_E}\\<br /> Av_{OL} = \text{Open loop gain, the hfe of the transistor gain}\\<br /> Av_{CL} = \text{Closed loop gain.}<br />$

**caffine** · 20th April 2009 12:15 AM

Diagram as requested

**audioguru** · 20th April 2009 12:55 AM

There are three capacitors and the one across the emitter resistor has a high value so I think there is no negative feedback. Just high gain and distortion.

**Hero999** · 20th April 2009 09:24 PM

Yes, it's totally open loop, the gain is equal to the hfe of the transistor.

Remove C1 and the gain will fall to about 5 and the output waveform will be less distorted.

**audioguru** · 20th April 2009 11:05 PM

Originally Posted by Hero999

Yes, it's totally open loop, the gain is equal to the hfe of the transistor.

hfe is current gain, not voltage gain.
Voltage gain is simply RC/(RE+Re) where RC is the collector load resistor in parallel with an external load.

**Hero999** · 21st April 2009 08:16 PM

Sorry, that doesn't make any sense because you haven't mentioned the curren gain of the transistor which does affect the voltage gain.

**audioguru** · 22nd April 2009 12:59 AM

There are many articles that say the hfe or hFE does not affect a transistor's voltage gain.
The first article in Google is from Amature Radio magazine:
THE TRANSISTOR AS A VOLTAGE AMPLIFIER

**Mikebits** · 22nd April 2009 01:02 AM

I am sure your correct, but that is confusing. Is not Ie dependent on Hfe?

**audioguru** · 22nd April 2009 01:12 AM

It is the angle of the slope of the base voltage to the collector voltage that shows the voltage gain of a common emitter transistor. When the hfe is high then the slope stays the same but it shifts downward. When the hfe is low then the slope stays the same but shifts upwards.

**Mikebits** · 22nd April 2009 01:41 AM

Originally Posted by audioguru

It is the angle of the slope of the base voltage to the collector voltage that shows the voltage gain of a common emitter transistor. When the hfe is high then the slope stays the same but it shifts downward. When the hfe is low then the slope stays the same but shifts upwards.

Are you saying Vgain has a function where m is same, something like this function.

Using slope of line equation Y=mx+b

**audioguru** · 22nd April 2009 02:47 AM

Yes, but the curves are curved at the low current end.
A transistor without negative feedback is very distorted when it approaches cutoff.

**The Electrician** · 22nd April 2009 11:11 AM

Originally Posted by audioguru

There are many articles that say the hfe or hFE does not affect a transistor's voltage gain.
The first article in Google is from Amature Radio magazine:
THE TRANSISTOR AS A VOLTAGE AMPLIFIER

The author of this article makes a small error just above Figure 2. He says:

"Base current equals collector current (or emitter current) divided by Hfe and hence, with near constant voltage across the base/emitter forward biased junction, input resistance (Rb) is multiplied by Hfe. Thus we get:"

Actually, base current equals collector current divided by Hfe, or emitter current divided by (Hfe+1), but not emitter current divided by Hfe.

This means that the input resistance seen at the base of a common emitter amplifier with no (or bypassed) external emitter resistance is .026/Ie * (Hfe+1), not .026/Ie * Hfe.

So his gain expression needs to be multiplied by Hfe/(Hfe+1), and the gain of a common emitter amplifier does depend on Hfe. The dependence is slight, and becomes completely negligible for reasonably high values of Hfe, but the slight dependence is really there.

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