Why Do Equations Remain Unchanged Across Different Transistor Operating Regions?

AI Thread Summary
The discussion centers on the confusion regarding the equations used in transistor circuit analysis, particularly the assumption of saturation mode while determining the active region. It is clarified that the equations presented are valid for both active and saturation regions, and the condition of the base-collector diode being reverse-biased is implicit in the analysis. The user seeks to understand why the equations do not change despite the operating region, leading to a deeper exploration of the relationships between currents and voltages in the circuit. The importance of using correct notation for subscripts in equations is also emphasized. Ultimately, the discussion highlights the need for a comprehensive approach to solving transistor circuit problems by establishing multiple equations to account for all variables.
noamriemer
Messages
50
Reaction score
0
Hi there! I'm having trouble understanding the transistor circuit analysis. Hope you could help me :)
Capture.PNG


First I need to find the active region of the transistor. What I saw in the solutions was an assumption that the transistor is in saturation region and then:

VBB=IbRb+Vbe+IeRe
Vcc=IcRc+Vce+IeRe

Why did the solver assume the transistor is in saturation mode, when he is supposed to find the active region?

Next, I have to calculate the minimal RE needed for the transistor to be active.
This time, they did refer to it as in active region. But, the equations remained the same. I thought that in Active region, it is supposed to reverse:

VBB=IbRb+Vbe+IeRe
Vce-IcRc-Vcc=IeRe

Why isn't there any change in the equations ? the BC diode is supposed to be in reverse mode, isn't it ?

Thank you!
 
Physics news on Phys.org
noamriemer said:
First I need to find the active region of the transistor. What I saw in the solutions was an assumption that the transistor is in saturation region and then:

VBB=IbRb+Vbe+IeRe
Vcc=IcRc+Vce+IeRe

Why did the solver assume the transistor is in saturation mode, when he is supposed to find the active region?
There is nothing here to indicate an assumption of saturation. Those equations hold for both regions of operation, active and saturated.
Why isn't there any change in the equations ? the BC diode is supposed to be in reverse mode, isn't it ?
Yes, and that condition is implicit in setting VCE = 0. (Explanation: If VCE = 0 and VBE = 0.6, then it follows that VCE = -0.6)

[PLAIN]https://www.physicsforums.com/images/icons/icon2.gifAlso, may I point out that you should not be using lower-case subscripts here, as they mean something different from the capitals subscripts. For DC or average values, use upper-case subscripts.
 
Last edited by a moderator:
noamriemer said:
Hi there! I'm having trouble understanding the transistor circuit analysis. Hope you could help me :)
View attachment 50224

First I need to find the active region of the transistor. What I saw in the solutions was an assumption that the transistor is in saturation region and then:

VBB=IbRb+Vbe+IeRe
Vcc=IcRc+Vce+IeRe

Why did the solver assume the transistor is in saturation mode, when he is supposed to find the active region?

Next, I have to calculate the minimal RE needed for the transistor to be active.
This time, they did refer to it as in active region. But, the equations remained the same. I thought that in Active region, it is supposed to reverse:

VBB=IbRb+Vbe+IeRe
Vce-IcRc-Vcc=IeRe

Why isn't there any change in the equations ? the BC diode is supposed to be in reverse mode, isn't it ?

Thank you!

The first set of 2 equations is correct irrespective of values of the R's or power supply voltages. The second set is garbage.

A quick check shows that this configuration IS in saturation. Approximately,
ie = (3V-0.7V)/0.5K = 4.6 mA which when multiplied by Rc = 3K gives Vcc - Vc = 13.8V, exceeding even the Vcc power supply.

(How did I do that? I just assumed β = ∞, a reasonable approx. with the low value of Rb).

To get your answer, you need 3 more equations:

Obviously,

ie = ic + ib

and ic = βib.

Then you need one more. Hint: what is the current ib thru Rb? And if you know Ve do you automatically also know Vb?

That's 5 independent equations and 5 unknowns: the three currents plus emitter and collector voltages. Set Vc - Ve = 0 and solve for Re.
 
Thank you both for your great help!
 
Back
Top