hi David,
There is full PDF, its +12meg in size.
Search for ltspice_4_e2.pdf

https://www.google.com/search?q=ltspice_4_e2.pdf&ie=utf-8&oe=utf-8&client=firefox-b

Thats a number links for LTS.

Eric

Thanks Eric

 

Maybe this video lectures about BJT's

Thanks Jony

 

Because you still forget that the BJT is a very nonlinear device. And that the voltage gain is not constant but will change with the input signal exponentially.

Av = Vout/Vin = - (Rc * Is)/Vt * exp^(Vin/Vt) (with CE capacitor across RE ressitor)

And this is why single stage BJT amplifier produces alot of distortion. We get about 1%THD per 1mV at the base (10% for Vin =10mV) .
https://www.researchgate.net/public...C_DISTORTION_IN_LOW_FREQUENCY_POWER_AMPLIFIER

http://www.kevinaylward.co.uk/ee/bipolardesign2/bipolardesign2.xht

You can improve the distortion by adding the emitter resistor (without CE capacitor). But to get "low THD" you need to pick RE >> re.
But it is impossible to get high voltage gain and low THD at low load resistance in such a simple circuit.
As you can see you're expecting too much from this simple circuit. And this is why no one uses this circuit as "low THD" amplifier.

I have been watching some of Professor Razavi's video lectures. How do you
find the value of Is for a device?

Ic = Is * (exp vbe/vt - 1)

Thanks,

David

 

Welcome to the real world, Isat current is not very constant and highly dependent on device temperature additional it also varies a lot from device to device even the same type or batch. This is why exact calculations of a BJT circuits are impossible to make.

But if you whant Isat value the for small signal BJTs, expect it to be in the area of about 10^-14 A ... 10^-15 A .
Or use the LTspice value.
Or you can calculate it. If you know the Vbe1 voltage for a given collector current Ic1 you can solve for Isat:

Isat = Ic1 * exp ( -Vbe1/Vt )

Or if you know the what Vbe1 is at some collector current Ic1 , then you can compute Vbex for any other collector current (Icx) without knowing the Isat current:

Vbex = Vbe1 + Vt * ln( Ic1/Icx)

And remember that the Vbe will go up 60mV or down for every upward /down change in Ic by a factor of ten (decade change).
For example if Vbe is 0.62V for Ic = 1mA if now Ic increases to 10mA the Vbe voltage will rise to 0.62V + 60mV = 0.68V.
Or if Ic current drops to 0.1mA the Vbe will also drop to 0.56V.

Good bedtime reading

https://forum.allaboutcircuits.com/blog/fun-with-the-diode-equation.589/#comment-1283

 

Welcome to the real world, Isat current is not very constant and highly dependent on device temperature additional it also varies a lot from device to device even the same type or batch. This is why exact calculations of a BJT circuits are impossible to make.

But if you whant Isat value the for small signal BJTs, expect it to be in the area of about 10^-14 A ... 10^-15 A .
Or use the LTspice value.
Or you can calculate it. If you know the Vbe1 voltage for a given collector current Ic1 you can solve for Isat:

Isat = Ic1 * exp ( Vbe1/Vt )

Or if you know the what Vbe1 is at some collector current Ic1 , then you can compute Vbex for any other collector current (Icx) without knowing the Isat current:

Vbex = Vbe1 + Vt * ln( Ic1/Icx)

And remember that the Vbe will go up 60mV or down for every upward /down change in Ic by a factor of ten (decade change).
For example if Vbe is 0.62V for Ic = 1mA if now Ic increases to 10mA the Vbe voltage will rise to 0.62V + 60mV = 0.68V.
Or if Ic current drops to 0.1mA the Vbe will also drop to 0.56V.

Good bedtime reading

https://forum.allaboutcircuits.com/blog/fun-with-the-diode-equation.589/#comment-1283

Thanks Jony.....David