AC amplifier, transistor with bias network - Page 1
Forum section: electronics / beginners

author: fady232
author_rank: Contributor
timestamp: May 13, 2025, 09:30:48 am
content: Hello all, im trying to simulate and understand the circuit made with 4 resistors as bias networkNow i explained each step in the image u can check below, but i will repeat hereSo to start what i know is that i wanna make a voltage amplifier ( so in terms of current i dont care yet even if i dont know what i need to consider in order to make the amplifier actually work , so if u got any suggestion about what currents should flow in the circuit, tell me )What i know is that VCC is 20V and VIN is 5VAC what i wanna obtain is a gain of AV=2, so in output i wanna have an ac voltage with 10v peak to peakwhat i did is:1° calculating R1 and R3 in order to set VB DC voltage, in this case i saw that the rule says it needs to be half of VCC, so R1=R3=10k2° secondly i calculated the resistor on emitter, which i did by assuming IE=1mA ( this value i choose it? or i find it on datasheets? )3° to choose RC i read there is another rule to have maximum signal swing, the midway beetwen 0 and 20V of VCC is 10V , so we wanna have VC = 10VRC = (VCC - VC) / IC , for small currents IE=IC, so 1mAIn this case RC = 20 - 10 / 1*10^(-3) = 10k4° and last step is double checking the gain AV=RC/RE = 10k/9.3 = 1,07 and we can see we dont obtain AV = 2 even if we followed the ''rules"So i ask u, someone can help me clarifying my doubts? what i did wrong? what steps i need to follow? how i need to think whne building ac voltage amplifier ( current doesnt matter for me now )Thanks and please use ''simple terms''Attachm

author: xvr
author_rank: Frequent Contributor
timestamp: May 13, 2025, 12:31:18 pm
content: First, the output of the amplifier must be taken from the collector of the transistor, not from the emitter.Second, Ie is the main factor for the circuit to work properly, you can't take it arbitrarily.As you said, the gain of the circuit is determined by RC/RE - that's the first thing.Second, it's the DC voltage across RC - it should be VCC/2 to ensure maximum output AC voltage range.The steady state point is determined by the base voltage (10V), so you have the emitter voltage - 10-0.7 = 9.3V (this is the RE voltage). So the voltage across RC will be 9.3*2 = 18.6V. The total voltage (assume zero on the transistor) is 18.6+9.3=27.9V. This is greater than 20V VCC. So you can't get a gain of 2 in this configuration.You need to lower the DC offset for the base or raise the VCC

author: fady232
author_rank: Contributor
timestamp: May 13, 2025, 01:16:52 pm
content: Quote from: xvr on May 13, 2025, 12:31:18 pmso i dont set ie? which is equal to ic? the current needed for my load to work? and after that i do my calculations?

author: PGPG
author_rank: Frequent Contributor
timestamp: May 13, 2025, 01:33:58 pm
content: First.Where from is the specification that you have 5VAC VIN and what 5V in this specification means?If 5V is amplitude than you have 10V peak to peak at input and you need gain of 1.If 5V is effective voltage than you have at input 2 * 5V * SQRT(2) = 14.1V peak to peak and to get 10V peak to peak at output you need gain of 0.71.Second.To get a room for maximum AC voltage at output you don't need to have VC=half of supply voltage but in your case you have to have Vce=half of supply voltage so the sum of voltage drop at R2 and R4 should be 10V what will left also 10V for Vce.

author: fady232
author_rank: Contributor
timestamp: May 13, 2025, 01:46:19 pm
content: Quote from: PGPG on May 13, 2025, 01:33:58 pmSo 10V peak to peak, i need to obtain 20V peak to peak in output, i didnt get why vce= half of supply voltage can u explain?

author: xvr
author_rank: Frequent Contributor
timestamp: May 13, 2025, 02:08:33 pm
content: QuoteYes, Ie is equal to Ic (base current could be disregarded). But you have to ensure that Ie will be in valid range for all possible input voltages. Both minimum and maximum. And do not drop to zero and do not violate SOA of transistor.But Ie is not a base point of evaluation, I was wrong.QuoteYour load current should be much less than Ie, otherwise it will affect gain and [may be] linearity of amplifier.BTW, maximum output voltage swing in this circuit is limited. Maximum valuege is VCC (transistor is closed), minimum - 1/3VCC (transistor in saturation, RC and RE forms voltage divider). So peak-to-peak output voltage is 2/3VCC, and swing is 1/3VCC. So if you want to get 10V output swing you should provide at least 30V of VCC

author: fady232
author_rank: Contributor
timestamp: May 13, 2025, 02:34:34 pm
content: Quote from: xvr on May 13, 2025, 02:08:33 pmI dont get it, the 1/3VCC is the voltage beetwen RC and RE when the transistor is closed? but its like that if re is equal to rc.Secondly i need to set a collector current so ie that needs to be higher than the one needed by the load?

author: PGPG
author_rank: Frequent Contributor
timestamp: May 13, 2025, 03:06:04 pm
content: Quote from: fady232 on May 13, 2025, 01:46:19 pmIf Vce = half of supply voltage then you can get the same peek positive signal (when transistor current is 0 (working point current - working point current) you have the highest possible voltage at collector) as negative peak (when transistor conducts working point current + working point current its Vce=0). When you can have the same + and - peaks you can get the highest possible AC signal without distortions.I have said that it is in your case (you don't have capacitor at emitter).If you will use capacitor between emitter and GND than Vce should be half of supply voltage minus Re drop voltage (provided capacitor is enough big you can consider it being short for AC).If you divide emitter resistor into two and short only one with capacitor than only that one voltage drop you should subtract from supply when searching the working point to get the highest possible output AC signal.

author: xvr
author_rank: Frequent Contributor
timestamp: May 13, 2025, 03:10:54 pm
content: QuoteIt is voltage on RC and RE common pin (connected via closed transistor) and on output.QuoteNo, RC is twice of RE (to provide gain of 2). So voltage divider will be RE/(RE+RC) = RE/(RE+2*RE) = 1/3QuoteYes, much higher. Any load current will add constant voltage drop on RC to output.You can consider load as additional resistor, connected to output of amplifier. It will form additional voltage divider with RC and will distort all output voltages.

author: fady232
author_rank: Contributor
timestamp: May 14, 2025, 05:05:50 pm
content: Quote from: xvr on May 13, 2025, 03:10:54 pmHello allI built a common emitter amplifier with these requirements: VCC=20V, Ic=10mA Av=2What i did is calculating RC first of all with this formula VC=(VCC-RC•IC) where i chose the vc half of vcc, to have full dynamic so VC= 10V. and i found that RC=1k, now i chose RE in order to have 2 as gain, so RE=500 OhmAfter that i calculated VE as VE=RE•iC , since ic=ie and found that VE=5V.At this point know that VB= VE+VBEso now i calculated VB=R2/R1+R2•VCC so 5.7=R2/(R1+R2)•20 , now if we suppose that hfe is 100 and the IR2 current needa to be 10•IB we get that R2= 6.3kohm and R1 by doing the inverse formula we obtain R1=16kBy simulating it looks working fine, but i had a question now RE the way i calculated it isnt the best for stability of the circuit cus transistor changes its parameters also with its temperature, then how can i consider that too? thanks all.

author: PGPG
author_rank: Frequent Contributor
timestamp: May 14, 2025, 10:28:13 pm
content: Quote from: fady232 on May 14, 2025, 05:05:50 pmYou don't have full dynamic.Without input signal you have Vc=10V and Ve=5V.When you fully switch off the transistor you will get Vc=20V so you can get your output voltage to jump up 10V from starting point.When you fully switch on the transistor it will short 1k and 500R resistors. So output voltage will be 20V*500/(1000+500)=6.67V. So you can get your output voltage to jump down from starting point by 10-6.67=3.33V.You don't have symmetry as you set Vc to be half of VCC and not Vce to be half of VCC.

author: Terry Bites
author_rank: Super Contributor
timestamp: May 15, 2025, 03:27:24 pm
content: A better design sequence is:1 Choose Ic2 Set Ve~1V or so to achieve thermal stability.3 Calculate Re4 Choose Rc to centre the supply at the collector5 calculate the bias resistors last.

author: PGPG
author_rank: Frequent Contributor
timestamp: May 15, 2025, 05:16:00 pm
content: Quote from: Terry Bites on May 15, 2025, 03:27:24 pmIt will not give Rc/Re=2 what OP wants:Quote from: fady232 on May 13, 2025, 09:30:48 amTo get AV=2 and as high output amplitude as possible he needs (as I have already written) to set Vce=VCC/2 so:1. Choose Ic2. Chose R=VCC/(2*Ic)3. Take Rc=R*2/3 and Re=R*1/3.

author: Zero999
author_rank: Super Contributor
timestamp: May 15, 2025, 07:50:43 pm
content: Quote from: PGPG on May 15, 2025, 05:16:00 pmThe problem with that approach is the RCand REform a potential divider as the transistor saturates, which limits the voltage swing.A way round this is to make the gain of the CE amplifier much higher, then add an external negative feedback network, as is done with an inverting op-amp configuration.