Understanding BJT Voltages and Operating Point Analysis

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The discussion focuses on analyzing BJT voltages and determining the operating point of a transistor circuit. Key points include identifying the common element of the transistor, which is typically the emitter connected to ground, and calculating voltages like VBE, VCE, and VCB based on circuit parameters. Participants clarify that the voltage applied to the circuit is indeed the 12V mentioned, and they discuss how to derive the collector current and other terminal voltages. The operating point is defined in terms of the biasing conditions of the junctions, with four possible regions: active, cutoff, saturation, and inverted. Understanding these concepts is crucial for effective transistor analysis and design.
JasonHathaway
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Hi everyone,


I'm new to transistors analyzing, and I need a way to correctly identify the voltages.

Let's take this circuit for example:

20140618_112601.jpg


Assuming that the Emitter is the common element (The one with arrow which is connected to the ground), then VBE would be:

22222.jpg


My questions are:

1- If I want to know the common element of the transistor, shall I look for the element that is directly connected to the ground?
2- For this circuit, I've VBE - VCE + VCB =0, On which basis I can get the these voltages?
 
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Is the "12" at the top right the voltage applied to the circuit? Or is that something else?
 
You analyze it like this:

(1) The BE junction is forward biased, so Vbe is about 0.7 V.
(2) The voltage across the base resistor is then 12.0 - 0.7 = 11.3 V.
(3) The current through the base resistor Ib = 11.3 / 376 K = .03 mA = 30 uA.
(4) The colector current Ic is beta times this. Assuming beta = 100, Ic = 3 mA.
(5) Vc = 12 - (3mA * 2 k) = 6.0 V.

Does this make sense?
 
Drakkith:
Yes, it's applied on the circuit.

phyzguy:
That makes sense. Now I want to determine Vcb, which is Vc - Vb. Now I've Vc, is Vb the same as Vbe (0.7)? And what's the operating point of the transistor?
 
JasonHathaway said:
That makes sense. Now I want to determine Vcb, which is Vc - Vb. Now I've Vc, is Vb the same as Vbe (0.7)? And what's the operating point of the transistor?

Since you now know the voltages on all three terminals (Ve, Vb, and Vc) calculating the differences is easy. Vbe = Vb - Ve, Vce = Vc-Ve, etc. What do you think the operating point is? What's the definition of operating point? Are the BE and CE junctions forward biased or reverse biased?
 
Actually, that what I'm asking for, the definition of the operating point. As far as I know, the operating point should be one of four regions: 1- Active (forward biased diode and reversed biased one) 2- Cut off (two reversed biased diodes) - 3- Saturation (two forward biased diodes) 4- Inverted (inverse of active).

I know that BE resamples forward biased diode, but I'm not sure about CE
 
In your second pic, those red + and ─ signs are wrong way 'round.
 
NascentOxygen

Thanks :)
 
JasonHathaway said:
Actually, that what I'm asking for, the definition of the operating point. As far as I know, the operating point should be one of four regions: 1- Active (forward biased diode and reversed biased one) 2- Cut off (two reversed biased diodes) - 3- Saturation (two forward biased diodes) 4- Inverted (inverse of active).

I know that BE resamples forward biased diode, but I'm not sure about CE

Well, what is Vc? What is Vb? Is the CB junction therefore forward biased or reverse biased? Given that it is an NPN transistor, you should be able to figure this out.
 
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