Since I'm tired now (being studying 2 days straight, seriously I didn't sleep) I will read your answers tomorrow, and see where my mistakes are.
Yes, the results you got are correct.Is this right though?
This is normal for such a simple circuit. Because in this simple amplifier configuration you do not have too much degree of design freedom. Because Av = gm*Rc = Ic/Vt *Rc = (Ic * Rc)/Vt = V_rc/Vt. And if we set V_rc = Vcc/2 we get:This current "feels" very small.
\[ I_{C_{dc}} = -A_{v}\frac{V_{T}}{R_{C}} = -(-10)\frac{25mV}{2kΩ} = 125μΑ = 0.125mA \]Specs:
Zin = 10kΩ
Zout = 2kΩ
Av = -10
frequency response 20Hz - 20KHz
Vcc = 9V
RE = 4.25V/0.125mA = 34kΩ not 0Ω.Now I think the mistake is somewhere here and especially in what I'm about to do next:
We are saying this because we need to have enough space to "fit" the output voltage swing.You and others are saying that we set Vc = Vcc/2. The graph though is Ic versus Vce, so this is
Εmbarrassing calculations from my side. AnywayRE = 4.25V/0.125mA = 34kΩ not 0Ω.
Yes, I totally get it. So this is why we want Vc = Vcc/2, so it can swing between Vcc | Vc | Ve the maximum way (and equally on both ways so Vpp = 2 times the peak of the output). Of course, Vc can't go below Ve because it will saturate the transistor. This actually raises a question:Also, notice one thing, because now V_rc = 0.25V thus Vc = 8.75V. You almost have no "space" for the output voltage to "swing up". Do you see it?
The voltage at the collector can swing from 8.75V down to Ve. The negative swing is 8.75V - 4.25V = 4.5V. But what about the positive swing?
Vc can go up only by 0.25V before it will hit Vcc rail.
Yes, it can. It is true for every amplifier in the world. If you apply too large an input signal to the amplifier input the amp will start to clip the signal (the signal will be distorted). Of course, this is undesired.Question 1:
Can Vin be big enough so at its negative T/2 (half period) will turn off the transistor by reducing Vb?
As I said earlier this is a very common problem when we are dealing with simple circuits. You demand too much from such a single transistor circuit.Question 2:
So what can I do here? Since I calculated Ic using the gain formula, and Rc=Zout, Vc is already determined to be 8.75V.
So, is the design problematic? Do I need to get rid of Ce and take Re into consideration in the gain formula? And why is the design problematic?
No, you do not bias the transistor completely wrong to set the amount of voltage gain. Roughly, the voltage gain is set by the collector resistor/the emitter resistor if there is no load. If the collector resistor is 2k and the emitter resistor is 180 or 190 then the voltage gain is 10 times. The DC collector voltage should be near half the supply voltage for maximum output level without clipping.I asked for a gain of 10 which the transistor can probably give me, but in order for this to work, the math told me that in order for this gain to work you need Vc=8.75.
Does anything in the math give this away? For example, when you calculate R1 using the Zin=10kΩ, you get a negative resistance, so you know that your assumption will not work.Using proper 1/10th of the collector current in the biasing voltage divider transistors, and using a fairly high hFE 2N3904 transistor, I could not obtain an input impedance of 10k ohms.
R1 is not calculated using Zin.Does anything in the math give this away? For example, when you calculate R1 using the Zin=10kΩ, you get a negative resistance, so you know that your assumption will not work.
I thought of it a little bit differently, and it makes sense. Let's say I figured out the value of R2. Then I probably must calculate R1 using the voltage divider equationR1 is not calculated using Zin.
Zin is R1, R2 and the input impedance of the transistor all in parallel. The datasheet for the 2N3904 transistor shows a typical hFE of 200 then with a 180 ohms emitter resistor, its input resistance is typically 36k ohms.
With a 2mA collector current in the 2k load, so that the collector idles at 5V, R1 should have a voltage across it of 8V and a current in it of 0.2mA, calculating to 40k,but 39k is close. Then R2 must have the remaining 1V across it and with the 0.2mA in it, must be 5k, but 5.1k is close.
Then the input resistance of the circuit is 36k//39k//5.1k= 4k ohms but 10k is needed.
When I used a 2N5089 transistor that has a much higher hFE then its input resistance is higher and its much lower base current allows the resistances of R1 and R2 to be higher.
Why do we want IR1 = Ic / 10 ? Where does this condition come from?5) You want the current in R1 to be 1/10th the collector current then the current in R1 is 0.2mA and the voltage across it is 9V - 1.01V= 7.99V.
To see why try to calculate the Zin of a CE amplifier with an RE resistor and without CE.Lastly, I don't understand how a transistor with bigger Hfe will increase Zin. You didn't use it anywhere in
your calculations.
It is the same condition. In general, we want a "stiff voltage divider" and to achieve this the divider current should be larger than the base current. Typically ten times larger than the base current.Why do we want IR1 = Ic / 10 ? Where does this condition come from?
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