[] Output current of transistor

AI Thread Summary
The discussion centers on the behavior of current in a transistor circuit, specifically regarding why the output current (Io) flows through resistor Rc but not through capacitor C2, which is open-circuited. It is clarified that Io does not pass through C2 because it has no connection on the other side, making it effectively an open circuit. The presence of capacitor C1 is noted as blocking DC gain, while Rc biases the transistor for proper operation. The conversation also touches on the conditions for calculating input and output impedance, emphasizing the importance of having either input or output current equal to zero for accurate analysis. Overall, the circuit is designed to amplify AC signals, with components affecting the gain and behavior of the current flow.
cuongbui1702
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Warning: Homework template must be used for all homework questions
My teacher gave me this picture below. My question is why Io go through Rc but does not go through C2
Mtb5V_zpskj3obfmb.png
 
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Presumably C2 is open circuited in this case (no connection at Vo).
 
gneill said:
Presumably C2 is open circuited in this case (no connection at Vo).
But why Ii is not equal 0 because of C1
 
cuongbui1702 said:
But why Ii is not equal 0 because of C1
##I_i## would be assumed to be some injected input current.

I realize that it may seem to present contradictory notions, but this is a just a convention that's often used to set up the scenario for analysis of a circuit: Some signal is assumed to be present at the input and the output is unloaded.
 
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cuongbui1702 said:
My teacher gave me this picture below. My question is why Io go through Rc but does not go through C2
You could say Rc represents the load. The capacitor provides a handy point where you can imagine monitoring the output voltage.

When you connect a resistive load between the capacitor and ground, this can be accounted for by just reducing Rc in your AC voltage-gain analysis.
 
do not forgot that with capacitors i=(1/C)Cdv/dt. so you could say say that there will be a Ii when Vi changes.

io does not go through the capacitor because it is an open circuit on the other side, so the current has nowhere to flow.

This is not important for you solving the problem but it might help prospective. the purpose of circuits like this are to amplify an AC signal. The C1 is in there to block DC gain from entering the amplifier. Rb, Re and Rc will bias your transistor so it is operating in the correct region. C2 will prevent a DC voltage from leaving the amplifier. Since we are dealing with AC gains, all components will affect the gain. also note that in many cases circuits like this are attached to a high impedance circuit on the output, so it can be modeled as an op circuit.
 
So almost is assuming, and Io = ic( AC of IC, IC is DC current going through RC). That makes me messing up. When I take a Rload, where is Io, does it continue going through Rc?
 
is this for a homework assignment. If it is then the knowledge that current cannot flow through an open current should be enough.

If this isn't for a homework assignment, then when you add a load the current will only flow through the capacitor when the voltage across the cap changes, which will be due to a change in input voltage. The Io signal will have both a DC and AC component to it while the circuit is operating.
 
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In the schematic shown, Io is defined as the AC current through Rc.

If there were a load, RL, separate from Rc, then the load current would be IRL.
 
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Thank for your answers. And one more question, my teacher said me that: when I calculate Zin or Zout, I must have 1 condition such as: Iin =0, or Iout = 0. Why I must have this condition?
 
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Of course, there is a possiblity that Io is incorrectly marked in your schematic. It may have been the intention that it be the current through C2 to a load not shown.
 
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cuongbui1702 said:
My teacher gave me this picture below. My question is why Io go through Rc but does not go through C2
Mtb5V_zpskj3obfmb.png
This is a dc problem. DC does not go thru a capacitor.
Plus the fact that there is no connection with one of C's terminals so AC current would not go through it either.
 
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