Hi, is the gain the same for this two amplifier circuits  ? - Page 1
Forum section: electronics / beginners

author: nyame
author_rank: Regular Contributor
timestamp: September 08, 2018, 04:20:20 pm
content: Hi, please is the gain the same for this two amplifier circuits ? did i calculate the gain correctly ?please what is role of the following component marked with arrow ?

author: JS
author_rank: Frequent Contributor
timestamp: September 09, 2018, 01:13:20 am
content: Looks like we are doing your homework for you, they don't have the same gain, the 80k resistor is to take out the DC from the cap, so you just read AC with the volt meter. The cap and the two resistors is a network to get a different emitter impedance for signal and DC, so you can control bias independent from signal gain.JS

author: nyame
author_rank: Regular Contributor
timestamp: September 09, 2018, 03:31:39 am
content: Thank u for the reply, please u are not doing my home work for me, I was given that uncompleted circuit to build with the same gain, I have build it up to this level. Components I never had better understanding of their roles are the once I requested for explanations, also please how can I adjust the gain to be the same as shown in the uncompleted circuit thanks

author: JS
author_rank: Frequent Contributor
timestamp: September 10, 2018, 08:31:11 am
content: The gain will be given by the ratio of Rc and Re, with the circuit with the cap at the middle of Re the resistor in parallel with the cap doesn't count for signal.JS

author: nyame
author_rank: Regular Contributor
timestamp: September 10, 2018, 10:32:07 am
content: Thanks for the reply, please the gain for the uncompleted circuit is 3.9 right ?

author: JS
author_rank: Frequent Contributor
timestamp: September 10, 2018, 02:31:25 pm
content: YesJS

author: nyame
author_rank: Regular Contributor
timestamp: September 10, 2018, 03:47:48 pm
content: OK please kindly go through if i correct the circuit to meet the gain of the original circuit

author: JS
author_rank: Frequent Contributor
timestamp: September 11, 2018, 05:51:17 am
content: Yes, looks like a reasonable first approach.More accurate will be considering a few things about the transistor and the 80k load, just considering the 80k resistor the gain is about 3.83, adding the transistor properties will make it a bit under that but the transistor is not even specified so you can ignore those I guess.JS

author: nyame
author_rank: Regular Contributor
timestamp: September 11, 2018, 06:44:59 am
content: The transistor is BC 108. Transistor gain is 165

author: JS
author_rank: Frequent Contributor
timestamp: September 11, 2018, 06:51:43 am
content: I'm not doing the math right now, but the consideration is usually the amplifier gain (transistor in this case) to be much higher than the loop gain (the one set by the resistors) makes the loop gain to be pretty close to the final value. For most applications 165 is much higher than 3.8 and I say for most applications, because if you are dealing with precision that's not the case, as you won't see exactly 3.8 but 3.79 something and it will already be wrong.JS

author: nyame
author_rank: Regular Contributor
timestamp: September 11, 2018, 07:37:57 am
content: 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

author: JS
author_rank: Frequent Contributor
timestamp: September 11, 2018, 07:48:51 am
content: 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

author: Zero999
author_rank: Super Contributor
timestamp: September 11, 2018, 10:14:16 am
content: Quote from: nyame on September 11, 2018, 06:44:59 amWhere 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 RCin parallel with the RLmultiplied by the transistor's transconductance, i.e. the change in ICvs VBE.At room temperature the transconductance of a BJT is equal to:gm= ICmA/25.6mVLet'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) = 2.6VVE= VB- VBE= 2.6V - 0.6V = 2VIE= VE/RE= 2/(386+235) = 3.22mAICis near enough equal to IE, as the Hfe is high.RLin parallel with RCRC|L= (1800*80*103) / (1800+80*103) = 1760Now the open loop gain:gm= 3.22/25.6 = 0.126 = 126mSAV_OL= RC|L*gm= 0.126*1760 = 222Finally the closed loop gain:AV_CL= AV_OL/(1+AV_OL*β)β = RE/RC|L= 386/1760 = 0.219AV_CL= 226/(1+222*0.219) = 4.48That's only a little less than RC/RE, which is 4.66

author: nyame
author_rank: Regular Contributor
timestamp: September 11, 2018, 03:21:26 pm
content: 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 circuitplease find the attached below

author: Zero999
author_rank: Super Contributor
timestamp: September 11, 2018, 07:18:12 pm
content: Yes, removing the load will increase the gain. This circuit has an effective output impedance equal to RC.