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Can you build a CS amplifier that his drain voltage (VD) is equal to 0.5Vdd or 2/3Vdd ??
In which region MOSFET is operating? In saturation ? So how it is possible if MOS work as a VCCS?
Also notice that in your LDO regulator your MOS (series pass element) work as a CE amplifier.
You made a mistake in your Zout calculation.
In this circuit Ib is not equal to 0A. To find Zout we "short" transistor base terminal to ground. But this "short" is only for AC signal. And this means the in real life we "short" transistor base to GND via large capacitor. So transistor is ON and...
You as a IC designer need to design a circuit that give us as much output voltage swing
as possible. Also notice that as long as your amp will work in closed loop and and in linear region M6 will stay in saturation region.
Do we need to worry about this in the firs place ? What will happen if M6 enter triode region?
The output voltage will be clipped. So M6 will limit the maximum positive output voltage.
And all amplifiers has specified output voltage swing. The maximum positive or negative peak output voltage...
No, this is not why we connect D and G of M3 together. M3 together with M4 form a current mirror circuit.
Please read this
https://forum.allaboutcircuits.com/showthread.php?p=678790#post678790 (post 33 and 39)
You don't need to worry about M6, the negative feedback will take care of it and...
In this case RB2 is simply pull-down resistor.
But in some cases the low input voltage is greater than 0.6V so we need voltage divider to ensure BJT cut-off.
Yes, I made mistake.
This is what I mean Ib > Ic/Hfe_min
The most datasheet show Vce_sat for Ic/Ib = 10
See the example...
Normally the only information we have is that if we "force" the base current to be equal Ib = Ic/10 then the BJT for sure will be in saturation region.
Most small-signal BJTs saturation specs are defined saturation when Ic/Ib (called forced beta) = 10...
When I start learning about BJT and FET I notice the big difference between saturation in FET and in BJT.
The saturation for the FET's transistor is what we describe in BJT as a "active-region".
LvW do you have any info what are those "historical reasons" for such a big difference?
Every real life "active device" to work properly as an amplifier supplied from a single source need proper bias circuit.
When you use BJT as a CE amplifier, you use a voltage divider to bias the active device somewhere in the "linear region".
In case of single supply op amp you have to do the...
Yes you right, I use R1 = 1K so this is why "my gain" is equal to 15V/V (and Fo=5.1Mhz) instead of 1.5V/V
And to find Fo i use "brute force" (nodal analysis).
For node A
\[\frac{Vin - Va}{R1} = \frac{Va-Vb}{R2}+\frac{Va-Vout}{Rf} \]
And for node B
\[\frac{Va-Vb}{R2} =...
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