RC coupled multi stage amplifier - Page 1
Forum section: electronics / beginners

author: SHW1
author_rank: Newbie
timestamp: February 12, 2025, 04:11:58 am
content: I am working on a two-stage RC coupled amplifier using2N3904transistors and analyzing its gain characteristics. Below, I have provided:The circuit diagramOscilloscope output after applying a 500mV peak-to-peak sine waveTheoretical gain calculations for both stagesI need to calculate the gain at each stage and compare it with the actual outputThere is a huge difference between calculated and actual outputVR4 = 198kΩVR3 = 77.3kΩEmitter Current IE of the first transistor Q1 is 0.63mAEmitter Current IE of the second transistor Q2 is 1.85mA

author: ledtester
author_rank: Super Contributor
timestamp: February 12, 2025, 05:06:58 am
content: Shouldn't 77.3K (the value of VR3) appear in your calculation of the input impedance of the second stage?

author: SHW1
author_rank: Newbie
timestamp: February 12, 2025, 05:50:19 am
content: yess , even if i change that it doesnt make much difference for the parallel resistancei want to calculate the gain

author: ledtester
author_rank: Super Contributor
timestamp: February 12, 2025, 06:59:30 am
content: Well - your calculation agrees with LTspice...Looking at the output with a 100mVpp input shows a gain of around 51 (assuming a beta of 100).

author: SHW1
author_rank: Newbie
timestamp: February 12, 2025, 10:02:57 am
content: If the overall gain is 51, then when we apply a 500mVpp signal, shouldn't we expect an output of (250mV * 51=12.75v)?However, in the practical scenario, the output was nearly 6Vpp. output waveform attached earlier,what could be the possible reasons for this?

author: ledtester
author_rank: Super Contributor
timestamp: February 12, 2025, 02:33:50 pm
content: One is a simulation which makes assumptions and the other uses real components.It would be valuable to test each stage by itself to see if it agrees with the pen and paper approximation. That would be a good way to find out where the simulation and the real world circuit diverge.

author: MrAl
author_rank: Super Contributor
timestamp: February 12, 2025, 03:12:52 pm
content: Quote from: SHW1 on February 12, 2025, 04:11:58 amHello,You can expect differences between calculated values and real life values because these are two different aspects of circuits in general. The biggest differences come from the models of the transistors being used for the calculations.You would start by choosing a model for the transistors, then go from there with the calculations. If you don't get the real life results then you have to investigate why they don't match. In the end you find out what is different about the assumptions of the models vs real life transistors.The simplest model to start with is the current controlled current source model. This is a simplified model that allows for simpler equations because the transistors only use a few components: a CCCS, a diode, and a resistor or two. With this simple model you can gain a large amount of insight as to how these amplifiers work. You can also calculate just about everything about the circuit.You do have to remember that any model will produce results that vary from real life, and to see matching results you have to adjust the model to the real life results, but only after you have verified that the real life transistors are working properly as well as everything else in the real life circuit.The CCCS transistor model is very simple. One equation is:iC=iB*Betawhere Beta is the assumed current gain of the transistor at low frequency.You can also add a resistor or two, such as a base resistor, and for the last stage you might also add an emitter resistor of some low value because the 10uf cap creates what we call a 'stiff' impedance situation which requires at least some resistance. You can start without that though just to get a feel for how this works.Of course, a more elaborate spice model for the transistor will show the more complex behavior, but a lot of that is not needed for the simpler studies like this one. For one thing, you do not have to model the saturation characteristic of the transistors because they will always be operating on the near-linear mode.If you want to go deeper, you can look into the exponential model, but that would be a big overkill for this circuit.I also suggest starting with a one transistor amplifier circuit using the CCCS model and go on from there to the two transistor amplifier. That will get you accustomed to how to calculate the various results with the model.

author: TimFox
author_rank: Super Contributor
timestamp: February 12, 2025, 05:10:52 pm
content: For a model that depends on transistor beta, always note the huge range of beta specified by the manufacturer over the range of collector current used in the design.I was taught to start with the transconductance modeldIc/dVbe=Ic/(26 mV)and then check for the effect ofIbcalculated using theminimumvalue of beta.

author: NASK
author_rank: Contributor
timestamp: February 13, 2025, 06:51:17 am
content: Your C3 47uF capacitor in actual circuit might be faulty one, double check it. Try adding extra capacitor in parallel like 10uF.

author: MrAl
author_rank: Super Contributor
timestamp: February 13, 2025, 01:14:04 pm
content: Quote from: TimFox on February 12, 2025, 05:10:52 pmHi there,Perhaps you can show an example with a very simple circuit like an NPN with base resistor and collector resistor.When we talk about Beta there will always be a variation that has to be considered. If the data sheet says 50 to 200 then we have to check the behavior of the circuit with a Beta from 50 to 200, or possibly even outside of this range. That helps to ensure that the circuit will actually work in real life.I know a lot of people who like to use the 'gm' model, so perhaps you can show a quick example calculation so everyone can see the way this is done too.

author: TimFox
author_rank: Super Contributor
timestamp: February 13, 2025, 03:12:17 pm
content: Quote from: MrAl on February 13, 2025, 01:14:04 pmI'll generate a drawing later. The important thing is that my equation forgmis reasonably accurate over a wide range ofIcand silicon small-signal transistors, with the constant parameter 26 mV at room temperature.

author: TimFox
author_rank: Super Contributor
timestamp: February 13, 2025, 06:03:14 pm
content: Here is a simplified calculation usinggmfor a single NPN transistor amplifier, with approximations suitable for small-signal amplification.Note that capital letters, e.g.Vc, indicate DC or quiescent (large) values, while small letters, e.g.vcindicate AC or small changes in levels, for either current (Iori) or voltage (Vorv). Single subscripts on voltages indicate with respect to ground.We start the design by choosing DC (quiescent) parametersVcc= DC supply voltage, sayVcc= 12 V.Ic= quiescent collector current, here assumed to equalIe= quiescent emitter current (since beta >> 1), sayIc= 2 mA.Vce= quiescent collector-emitter voltage, usually nearVcc/2 to allow reasonable swing from quiescent point, sayVce= 5 V (to allow forVebelow)Therefore,R4= (12 V - 5 V - 3 V)/(2 mA) = 2.0 k\$\Omega\$.Now, the DC base-emitter voltageVbewill be close to 0.7 V.To reduce the effect of variations inVbe, we set the quiescent emitter voltage (due to current through R4) to be larger, sayVe= 3.0 VForIe= 2 mA, we use R4 = 1.5 k\$\Omega\$ at the emitterTherefore, the DC bias voltage at the base becomesVb= 3.0 + 0.7 = +3.7 V.To get that voltage, we can use small resistors at R1 and R2, so that the Thévenin equivalent resistance will be low andVbwill change little with the actual base currentIb=Ic/ beta.Now, if we want high voltage gain, we connect emitter capacitor C3 across emitter resistor R4, so that the AC voltage at the emitter is negligible. As can be shown, that means that the reactanceXC3must be very low compared withre= 1/gm= 0.013\$\Omega\$, which is a much lower resistance than the external resistor R4, at the lowest frequency of interest.Similarly, we assume that the coupling capacitors C1 and C2 are large enough to be neglected, and we ignore high-frequency effects of any capacitance to ground.In this case, withIc= 2 mA, we havegm=ic/vbe(note AC values) =Ic/(26 mV) = (2 mA)/(26 mV) = 77 mA/V.Therefore, the voltage gain from In to Out is given byA=gmxrload.Here, the AC load impedancerloadis the parallel combination of the external resistors R3 and R5 and the internal output impedance of the transistor collector (which should be much larger than the external resistors), approximately .Now, there will be finite (but small) DC and AC currents into the base due to finite (but large) beta.The DC bias currentIbflowing through R1 and R2 in parallel will give a small change inVbandVe, modifying the quiescent collector currentIc. If beta = 100 for these parameters, the base bias current will be 0.02 mA; if the parallel combination of R1 and R2 is 2.0 k\$\Omega\$, that gives a small change in the voltage across R4 of 0.04 V from the initial value of 3.0 V. If the bias-circuit resistance is higher, you can iterate the calculation for the new value ofVe, and the calculation should converge to within slide-rule accuracy very quickly.Now, it gets interesting if we disconnect C3 to obtain "emitter degeneration".The effective emitter resistance of the transistorre= 1/gmis an internal resistor that can be considered in series with the emitter of an infinite-transconductance ideal transistor.Without C3, the external resistor R4 is in series with that emitter, and the new transconductance for the degenerated transistor isgm' = 1 / (re+R4).Obviously, the external resistorR4>>re, (1500 vs 0.013), sogm' = 1/(1500\$\Omega\$) = 0.67 mA/V and the voltage gain is very low, only about (0.67 mA/V) x (2000\$\Omega\$) = 1.3 V/V.To increase that value, R4 can be split into two resistors in series, with the bypass capacitor C3 across part of it, thus increasing the degenerated transconductance by decreasing the AC resistance in the emitter circuit.Feel free to audit the arithmetic here.

author: Zero999
author_rank: Super Contributor
timestamp: February 13, 2025, 07:16:21 pm
content: Beta theoretically makes no difference to the transconductance of a BJT and therefore the voltage gain of a BJT amplifier. Here's a simulation showing a current mirror made with transistors of different betas, yet the collector currents are identical.Current mirror hFE.ascHowever in this case, beta will affect the voltage gain a bit. Lower beta will pull down the biasing potential divider to the base a little, which will reduce VB, therefore VEand IE, thus lower the voltage gain. The solution is to reduce the impedance of the potential divider, but then that also reduces the input impedance, which can be alleviated with bootstrapping.

author: TimFox
author_rank: Super Contributor
timestamp: February 13, 2025, 07:48:31 pm
content: In my example, the main effect of beta is on the input impedance.For an effective emitter resistanceret=R4+ (26 mV)/Ic, whereR4can be zero (if bypassed completely), the input impedance (after the bias divider R1 and R2) is (beta) xret

author: Zero999
author_rank: Super Contributor
timestamp: February 13, 2025, 09:17:19 pm
content: Quote from: TimFox on February 13, 2025, 07:48:31 pmYes you're right. The base impedance is determined by the beta.I was wrong about bootstrapping. It won't work, with the emitter bypassed with a capacitor, because the base impedance is hFE*26mV/ie, which is much less than the divider.