Determinant Zero, Saturated Bipolar Transistor

[mishima]

Homework Statement

I am looking for the base, collector, and emitter voltages in the following circuit:

Homework Equations

KCL
KVL
offset voltage = VB - VE
saturation voltage = VC - VE

The Attempt at a Solution

First I made a matrix without assuming any values for offset or saturation voltages. Doing so gave me this:

1 55.56 14.7 | 678.72 KCL with transistor as node
9/500 1 9/34 | 12.216 KVL right hand mesh
17/250 34/9 1 | 46.149 KVL left hand mesh

The determinant for this matrix is 0, and seems to have infinite solutions. So that was not a good approach here. I checked the circuit with CircuitLab, and sure enough the voltages work with my matrix. Its just that the matrix can't produce them unfortunately... (10.07, 9.539, 9.428 for Vbase, Vcollector, and Vemitter). First, I was wondering if this always happens with a saturated bipolar transistor. Or, this isn't a textbook problem, is that the only reason the matrix didn't work out?

A much easier solution was to just assume that the offset voltage was 0.7 V and saturation voltage (Vce) was 0.2 V. Those 2 equations

Vbase - Vemitter = 0.7
Vcollector - Vemitter = 0.2

along with the first equation in the matrix gave solutions consistent with CircuitLab to within a hundreth of a volt. So second, I'm just kind of wondering what makes the first assumption valid. I was told offset voltage is not a constant. The Vce assumption makes sense just looking at a datasheet for the 2n2222. But why can I say that Vbe is 0.7 with confidence here?

Thanks.

 

[mishima]

I think I see what happened, I used the same equation in rows 1 and 2 of my matrix.

Actually they are all the same...

 

[CWatters]

But why can I say that Vbe is 0.7 with confidence here?

If the transistor is ON then Vbe will be roughly 0.7V. Look at the diode model of a transistor.

It might seem obvious for this problem but it would be best to check the transistor is saturated by comparing the base and collector currents. For example most small signal transistors have a gain >50 so multiply the base current by 50 and if the answer is more than the collector current the transistor will be ON/saturated.

 

[jfgobin]

If you want the super simplified model, you consider the transistor is ON, That V_BE=0.7V and that the current gain is 50 (approximate value for a 2n2222 at I_c=1mA)

That gives you the following matrix:

[itex]
\left ( \begin{array}{ccc}
10680 & 680& 0\\
680 & 860 &1\\
-50 & 1 &0
\end{array} \right ) \cdot \left (
\begin{array}{c}
I_{b} \\
I_{c} \\
V_{CE}\end{array} \right ) = \left ( \begin{array}{c}
11.3\\
12\\0\end{array}\right )[/itex]

If you want a model that's slightly more complex, you can use the hybridparameters. (See here)

(Edit, correct Vbe to Vce)

 

[rezeew]

I can understand your confusion with the initial approach using the matrix method. It is true that with a saturated bipolar transistor, the determinant of the matrix will be 0 and it will have infinite solutions. However, this does not mean that the matrix cannot produce valid solutions. In fact, as you have observed, using the matrix method with some assumptions can lead to accurate results.

As for the assumptions of offset voltage and saturation voltage, they are typically used in circuit analysis to simplify the calculations and make them more manageable. The offset voltage is not a constant and can vary depending on the specific transistor and circuit conditions. However, it is often approximated to be around 0.7V for a bipolar transistor. Similarly, the saturation voltage is also not a constant and can vary depending on the specific transistor and circuit conditions. In this case, assuming a saturation voltage of 0.2V may not be entirely accurate, but it is a reasonable approximation for circuit analysis purposes.

Overall, the assumptions made in circuit analysis are not always exact, but they allow us to solve problems and understand circuit behavior in a more practical and manageable way. It is important to keep in mind that these assumptions may not always hold true in real-world scenarios and further analysis may be necessary.