Hello, so my task is to design a 2 or 3 stage amplifier that has the following parameters:
Open-circuit (no-load) voltage gain: |
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Hmm okay. So i'm taking another look at it and I'm not quite sure where I should begin changing some values.As already mentioned by PRS you BJT stage amp is not properly biased.
And your Ic current is much large then 0.626mA. Because your BJT is in saturation region.
I have seen your work around these forums while I was "Googling" things and stuff, your help is appreciated, although it seems your away atm, I'll be here untill I figure this out!Broodsliver, your BJT is improperly biased. I'll help you with this problem if you come back.
Start with the output parameters. Exaggerate a little so that you design for 12 volts peak to peak at the collector. This gives you leeway when you use standard value resistors.First just want to apologize when I posted this on the computers at school, the requirements displayed values in case some cannot read them:
Open-circuit (no-load) voltage gain: |Avo|=100 (± 10%)
Number of transistors (stages): no more than 3;
Power supply: +15V relative to the ground;
Output resistance: 4.0 Komhs (±10%);
Input resistance: no less than 50kohms;
Maximum no-load output signal swing: 10V peak to peak (from −5V to +5V).
NPN Bipolar-Junction Transistor (BJT) 2N3904, β = 150,
VBE(on) = 0.7V Vcesat = 0.3 V
n-type enhancement MOSFET included in the chip ALD1106PBL
K=0.55 (mA/V^2), Vt=0.75
Broodsilver, I think I may have found a way to get this done. Your idea of using the MOSFET in the follower mode is good. But the next stage, if it were a differential amplifier, might do the trick. Have you studied differential amplifiers yet? If so, this is what your instructor is looking for --- a FET follower feeding a differential amplifier. That's three transistors given that you use a resistor at the emitters of the differential amplifier. Also use the common mode design, not the single-ended design.This is an open ended lab design. We only have to build and show it to our professor if we wish to get some bonus mark, other wise I just need to do the simulations and have the "pre-lab" work to find the resistor/capacitor values and the stages to use.
Well it is unfortunate, I'm thinking of another way maybe using a CS,CC,then another CS amp. Maybe this will work better? Av1 10, Av2 close to 1 (hopefully at worst 0.9) then maybe 12 for the last gain.
Although it would be more work DC biasing the stages is it ultimately the better way to do things? Or should I not have a capacitor block the DC part from my 1st stage to 2nd stage and let the full AC signal and DC part flow and use that? Like wise from my 2nd to 3rd stage.
Appreciated
Well he did learn in lecture about differential amplifiers, however I don't have any practice at solving their structures, nor do I yet know how they should be linked together (for symmetry), so far they have they just been MOSFET examples however, so I have no idea how their voltage gains should be calculated from a BJT differential amp, even as a base example. I'm guessing you mean BJT since you mentioned to use resistors in their emitters.Broodsilver, I think I may have found a way to get this done. Your idea of using the MOSFET in the follower mode is good. But the next stage, if it were a differential amplifier, might do the trick. Have you studied differential amplifiers yet? If so, this is what your instructor is looking for --- a FET follower feeding a differential amplifier. That's three transistors given that you use a resistor at the emitters of the differential amplifier. Also use the common mode design, not the single-ended design.
I'll look into it and get back with you tomorrow. I used to be pretty good at this but I'm rusty and as they say use it or lose it. But do ask your instructor about the 10-100mV input verses his 10 V peak to peak output problem with respect to gain and make sure he is aware of it. If he is then that's good! It means there's a way to do this. I know there is a way using multiple stages with feedback, but I'm guessing this is not the case here.Well he did learn in lecture about differential amplifiers, however I don't have any practice at solving their structures, nor do I yet know how they should be linked together (for symmetry), so far they have they just been MOSFET examples however, so I have no idea how their voltage gains should be calculated from a BJT differential amp, even as a base example. I'm guessing you mean BJT since you mentioned to use resistors in their emitters.
Also how to calculate it's output resistance.
Ohhh okay I see what you did. That actually helps me b/c I forgot to add in a couple of capacitors to thend of my CS's to increase the gain. But correct me if I'm wrong, If I wanted my output voltage signal swing to be 10V (± 5V) a CE amp can't do the job because it cant take in more than 10mVp-p?Yesterday we and up with this circuit
With Ic = 1.2765mA and Ie = 1.285mA.
And we need Ve = 2.7V so
Re = 2.7V/1.285mA ≈ 2.2K
Now we need to take care of the voltage gain. Without Ce capacitor the gain is equal to
Av_without = Rc/Re = 4.7K/2.2K = 2.1V/V
So we need to decrease Re value for AC signals to increase the voltage gain. We can do this by adding Ce capacitor parallel with Re resistor.
This capacitor will short emitter to ground for AC signals. And thanks to this the gain increase to the new value:
Av = gm*Rc = Rc/re where
re ≈ 1/gm = 25mV/Ie = 19.5Ω
So our gain will reach new value
Av = 4.7KΩ/19.5Ω = 241V/V
But now our gain is way too high. So to fix this problem we need to add another resistor. But this time we need to add resistor in series with Ce capacitor. The value for this resistor can be find quite easy.
Rx = Rc/110 - 19.5 ≈ 22Ω
I use 110 instead of 100 because I want to compensate the gain drop in source follower.
No, if the hole amplifier will have a voltage gain equal to 100V/V.But correct me if I'm wrong, If I wanted my output voltage signal swing to be 10V (± 5V) a CE amp can't do the job because it cant take in more than 10mVp-p?
Because you simply forget about the loading effect. Your first CS stage is loaded by CD stage input impedance.My problem is, when I put them together, staging them with capacitors I get a gain of about 1.02. So my question is, how does the signal work in passing through?
PRS mentioned in a post yesterday that: "you can't put more than 10mV into a common emitter amplifier without creating harmonic distortion. Staging doesn't help. You can partially bypass the emitter resistor but even then you can increase the input voltage only to about 100 mV."No, if the hole amplifier will have a voltage gain equal to 100V/V.
And you want Vout = 10Vp-p the Vin_max = 10/100 = 100mVp-p.
So what is your problem here?
Ohhhh, well that makes things much more interesting. Hmm so, what are some steps I can take to help reduce my input impedances but still keep my overall gain or rather increase them from what I already have? If I change my Vg then I change my Vgs and following that the Vgs changes then so does Id and if that changes so does gm and ultimately Avo.Because you simply forget about the loading effect. Your first CS stage is loaded by CD stage input impedance.
Rin for CD stage is equal to 3.3K||1K = 0.77kΩ.
So the voltage gain of your firs stage will drop
Av = gm * (RD||Rin) ≈ gm*0.77K
And this is why your gain drops.
With the 22 ohm resistor in series with the emitter bypass capacitor 100mV is acceptable. Jony's circuit is probably good enough -- unless your instructor is worried about a little harmonic distortion. You'll do well to discuss the distortion problem in your report. By the way, I breadboarded Jony's circuit and even though the bottom portion is more narrow than the top I can't see doing any better than this with an emitter follower.PRS mentioned in a post yesterday that: "you can't put more than 10mV into a common emitter amplifier without creating harmonic distortion. Staging doesn't help. You can partially bypass the emitter resistor but even then you can increase the input voltage only to about 100 mV."
So if i can't put more than 10mVp-p into I can't get 100mV to get my 10Vp-p. Orrrr am I looking at this wrong?
by Jake Hertz
by Duane Benson
by Duane Benson
by Duane Benson