| Electronics > Beginners |
| Hi, is the gain the same for this two amplifier circuits ? |
| << < (3/3) |
| nyame:
Ok thanks, so how do I work things out to match the gain of the oranginal circuit ? I thought what I worked out can still be acceptable |
| JS:
It is acceptable for a general purpose circuit, tolerances in components will usually be higher than the error for this circuit, it clearly isnt a precission application. If you want to get the exact theoretical value, the best way is using the model for the transistor, using the H parameters. Look for H parameter model of a BJT transistor. JS |
| Zero999:
--- Quote from: nyame on September 11, 2018, 06:44:59 am ---The transistor is BC 108. Transistor gain is 165 --- End quote --- Where did you get that information from? Is it the Hfe? If so, that's irrelevant to the voltage gain of the circuit, because the Hfe refers to the current gain, which is totally different to the voltage gain. The open loop gain of this circuit is equal to RC in parallel with the RL multiplied by the transistor's transconductance, i.e. the change in IC vs VBE. At room temperature the transconductance of a BJT is equal to: gm = ICmA/25.6mV Let's look at the top circuit. Calculate the approximate bias voltages and therefore the collector current: VBE = 0.6V the usual diode drop. VB = 12*2.3/(8.2+2.3) = 2.6V VE = VB - VBE = 2.6V - 0.6V = 2V IE = VE/RE = 2/(386+235) = 3.22mA IC is near enough equal to IE, as the Hfe is high. RL in parallel with RC RC|L = (1800*80*103) / (1800+80*103) = 1760 Now the open loop gain: gm = 3.22/25.6 = 0.126 = 126mS AV_OL = RC|L*gm = 0.126*1760 = 222 Finally the closed loop gain: AV_CL = AV_OL/(1+AV_OL*β) β = RE/RC|L = 386/1760 = 0.219 AV_CL = 226/(1+222*0.219) = 4.48 That's only a little less than RC/RE, which is 4.66 |
| nyame:
Hi, thanks for the reply. when i dropped the load resistor i noticed the gain also reduce from my reading i think the circuit gain is equal now with the original circuit please find the attached below |
| Zero999:
Yes, removing the load will increase the gain. This circuit has an effective output impedance equal to RC. |
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