How to Find Ibq and Vceq in a Common Base Amplifier?

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SUMMARY

The discussion focuses on calculating the DC base current (Ibq) and the DC collector-emitter voltage (Vceq) in a common base amplifier circuit with given parameters: β=100, Vcc=15V, VEE=4V, RE=3.3KΩ, and Rc=7.1KΩ. The user successfully derived Ibq using the formula Ibq=IEq/(1+β), resulting in approximately 0.0099mA, and found Vceq to be 8.6V. The calculations confirm that both Ibq and Vceq are determined using the DC equivalent of the circuit, clarifying the relationship between the currents and voltages in the common base configuration.

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Rampart
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Homework Statement


In the following common base circuit,β=100.If Vcc=15V, VEE=4V,RE=3.3KΩ, Rc=7,1KΩ find the
Ibq and Vceq(they are related to the dc values of the Q point)

R9x3Xff.png


Homework Equations

The Attempt at a Solution


Well ah...i mainly want to ask some things.However knowing that i must show my current work,check the pdf.I want to ask at first 2 things:
a)I believe Ibq refers to the dc base current.Now how is it possible to have Base current without a Base source.

I found Ic(=IE) but can i say that Ibq=Icq/β even if i don't have any base source?

b)Also can you please describe me the current's route?I mean Vin is in the area of the emitting diode.The ac voltage goes to The diode E.How does it move pass him,when the diode is reverse biased?

so i would like to get these things cleared out.Thanks in advance.
 

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Rampart said:

The Attempt at a Solution


Well ah...i mainly want to ask some things.However knowing that i must show my current work,check the pdf.I want to ask at first 2 things:
a)I believe Ibq refers to the dc base current.Now how is it possible to have Base current without a Base source.

I found Ic(=IE) but can i say that Ibq=Icq/β even if i don't have any base source?
Ground is a perfectly good 0 V source :) Remember that the emitter supply is negative.
b)Also can you please describe me the current's route?I mean Vin is in the area of the emitting diode.The ac voltage goes to The diode E.How does it move pass him,when the diode is reverse biased?
The AC voltage modulates the voltage at the emitter. As you might imagine, this will cause a big modulation in the base current since the base-emitter junction behaves like a typical diode where forward current changes drastically with forward bias voltage. The collector current change is even larger since it's β times the base current.
so i would like to get these things cleared out.Thanks in advance.
Hope that helps.
 
gneill said:
Hope that helps.
Of course it helped sir.First thing in the morning,tomorrow,i will retry.So for now,have a good night.
 
gneill said:
Ground is a perfectly good 0 V source :) Remember that the emitter supply is negative.

Ok so there is Base current.So i found it by using Ibq=IEq/(1+β).And it is around 0.0099mA.Vceq as found in the pdf is 8.6V.Was this the whole exercise?I mean i just used the DC equivalent.Am i missing something?Ibq isn't the dc base current & vceq the dc voltage?Won't they be found by using only the DC equivalent?
 
Rampart said:
Ok so there is Base current.So i found it by using Ibq=IEq/(1+β).And it is around 0.0099mA.Vceq as found in the pdf is 8.6V.Was this the whole exercise?I mean i just used the DC equivalent.Am i missing something?Ibq isn't the dc base current & vceq the dc voltage?Won't they be found by using only the DC equivalent?
If the question refers to the currents and voltages using upper-case letters, I and V, e.g., IBQ and VCEQ, then you are being asked to find the Q-point DC values. (Lower-case letters, i and v are, by convention, reserved for AC components due to that AC signal capacitively-coupled to the emitter.)

Your calculations match mine. :)
 
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