bjtcircuit analysiscurrent-sourcedcthevenin
I am trying to calculate the high frequency cut off of a common collector BJT circuit. I replaced Cu with a test voltage and open circuited the other capacitors. The circuit I came up with has the test voltage in series with a current source. I am just unsure how to find the test voltage. Here is the work I have done so far. I just can't figure out what an expression for Vt would be or how to relate it to the other variables. I needto find Vt/it
That could only be replaced by an infinite voltage source.
Current sources operate on the basis that they will output their current no matter the load. When you leave it open circuit your open circuit voltage would be infinity. IE, this circuit cannot exist.
The standard replacement will be a current source with a parallel resistance to a voltage source with series resistance. As shown in this wikipedia image.
http://en.wikipedia.org/wiki/Th%C3%A9venin's_theorem">
The MMUN2211LT1 already has a base resistor of 7~13kΩ with an hFE of 35 ~60.
If you apply 2.5V to base resistor it will saturate to 0.2Vmax with 5mA collector current.
added- - I read the above in the product spec. Have you read it? My analysis of this switch is as follows; Output 5mA for input 2.5V/10kΩ nom = 0.25mA which translates to current gain of 20x which is typical for a saturated switch. 5x is common for high current switch.
So apply more base voltage to this device and remove the external base resistor.
Best Answer
First of all your small-signal diagram is wrong. If this is a CC amp then collector should also be at AC ground. And this means that Cu capacitor is connected between the base and the ground. And the resistance seen by this capacitor is simply equal to: $$R_u=R_a||(r_{\pi}+(\beta+1)R_E)$$