# Calculation of base current and what decides the current through collector-emitter branch

I have the below circuit. Its not a homework material. I am understanding how to analyze the transistor circuits.

Below are my questions while trying to analyze :

1. If I am given the below circuit, how to determine whether the transistor is operating in active/saturation/cut off region?
2. How is the base current determined in the below circuit when there is not base resistor given? The voltage at the base is calculated to be 1.4V. But how is base current calculated?
3. What determines the current through the collector-emitter branch? Is is the emitter resistor or collector resistor?

To determine the current through the collector-emitter branch, we need to find the region of operation of the transistor, right? How to find Ib and Ic?

Can someone help.

• Are you aware of how to re-cast two resistors tied between a voltage source into a Thevenin equivalent that is just one new voltage source and one new resistor? Are you aware of KVL? If you are, then re-cast your $5\:\text{V}$ source, $R_1$ and $R_2$ into $V_\text{TH}$ and $R_\text{TH}$ and then just solve the KVL starting with $V_\text{TH}$ and ending at ground below the emitter. – jonk Dec 15 '19 at 15:34
• No . Could you just help with out thevenin equivalent circuit? I am aware of KVL. – Newbie Dec 15 '19 at 15:38
• Replace $R_1$ and $R_2$ with a new resistor that is in series with the BJT base. This resistor will be the Thevenin equivalent resistance of $\frac{R_1\cdot R_2}{R_1+R_2}$. The new, open end of this resistor is then connected to the Thevenin voltage of $5\:\text{V}\cdot \frac{R_2}{R_1+R_2}$. This simplifies the circuit, but doesn't change its function. (You should learn about Thevenin and Nortion equivalents, someday.) With this new circuit, KVL can be applied. (You could also just use KCL. But it's a little less intuitive in this case.) – jonk Dec 15 '19 at 15:47
• Ok . But can't we not analyse this without using thevenin? If yes, please help me with that way as I am trying to analyse intuitively. And could you write an answer to my other questions? – Newbie Dec 15 '19 at 15:53
• Also look here, as there is a similar discussion and a lot more detail provided. – jonk Dec 15 '19 at 15:53

If I am given the below circuit, how to determine whether the transistor is operating in active/saturation/cut off region?

We can, for example, assume that the BJT is working in an active region. And do the calculations based on this assumption. Because if our assumption is wrong we get "unreal" results.

How is the base current determined in the below circuit when there is not base resistor given? The voltage at the base is calculated to be 1.4V. But how is base current calculated?

We can do it in multiple ways.

The first way is to write a KCL equation and solve it.

$$\I_1 = I_B + I2 \$$ (1)

And the II Kirchhoff's law we can write:

$$\V_{CC} = I_1R_1 + I_2 R_2\$$ (2)

$$\ I_2 R_2 = V_{BE} + I_E R_E\$$ (3)

We can write

$$\ \large I_B = \frac{I_E}{\beta + 1}\$$ (4)

We can solve this for $$\I_B\$$ current

$$I_B = \frac{R_2V_{CC} - V_{BE}(R_1+R_2)}{(\beta + 1)R_E(R_1+R_2) +(R_1R_2) }$$

But there is also a simpler way to solve this circuit by using Thevenin's theorem.

We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit:

$$V_{TH} = V_{CC} \times \frac{R_2}{R_1+R_2} = 1.4V$$

$$R_{TH} = R_1||R_2 =\frac{R_1 \times R_2}{R_1 + R_2} \approx 2.8k\Omega$$

So, we end up with this circuit:

And base on KVL we can write:

$$\V_{TH} = I_B R_{TH} + V_{BE}+I_E R_E\$$

And we also know that $$\I_E = (\beta +1)I_B\$$

so we end up with

$$\V_{TH} = I_B R_{TH} + V_{BE}+ (\beta +1)I_B R_E\$$

and the base current:

$$I_B = \frac{V_{TH} - V_{BE}}{R_{TH} + (\beta +1)R_E } = \frac{1.4V - 0.7V}{2.8k\Omega + 201*180\Omega} \approx 18 \mu A$$

$$I_C = \beta I_B = 200 \times 18 \mu A = 3.6mA$$

$$I_E = (\beta+1) I_B = 201 \times 18 \mu A = 3.618mA$$

$$V_E = I_E R_E = 0.651V$$

$$V_C = V_{CC} - I_C R_C = 2.552V$$

What determines the current through the collector-emitter branch? Is is the emitter resistor or collector resistor?

If the BJT is on the active region ($$\V_C > V_E\$$) the truth is that the $$\V_{BE}\$$ voltage determines the current through the collector-emitter. Or the base current if we prefer the "current control" point of view. How is possible that with same Ibase there is more than one Vce?

• Thanks for taking up the Thevenin/KVL approach! Together, we've provided a comprehensive response, I think. – jonk Dec 15 '19 at 17:05
• @Jonk no problem – G36 Dec 15 '19 at 17:08

## Using only KCL

I'll completely avoid setting up a Thevenin equivalent, followed by a KVL analsysis. Instead, I'll only use KCL on your circuit:

simulate this circuit – Schematic created using CircuitLab

There are three unknown nodes. Let's call them $$\V_\text{B}\$$, $$\V_\text{E}\$$, and $$\V_\text{C}\$$. (You should have no problem assigning those to their associated circuit nodes.)

So, assuming the BJT is in active mode (and we have to assume that, to start -- we can always disprove that assumption if this analysis doesn't produce reasonable values) and using KCL we can get:

\begin{align*} \frac{V_\text{B}}{R_1}+\frac{V_\text{B}}{R_2}+I_\text{B}&=\frac{V_\text{CC}}{R_1}\\\\ \frac{V_\text{E}}{R_\text{E}}&=I_\text{E} \end{align*}

But we know a few added things, also assuming active mode. For example, $$\I_\text{E}=\left(\beta+1\right)I_\text{B}\$$ and also $$\V_\text{E}=V_\text{B}-V_\text{BE}\$$. So the above can be rewritten as:

\begin{align*} \frac{V_\text{B}}{R_1}+\frac{V_\text{B}}{R_2}+I_\text{B}&=\frac{V_\text{CC}}{R_1}\\\\ \frac{V_\text{B}-V_\text{BE}}{R_\text{E}}&=\left(\beta+1\right)I_\text{B} \end{align*}

We now only have two unknowns and two equations, $$\V_\text{B}\$$ and $$\I_\text{B}\$$. So it's solvable by the usual means.

## Question 1

If I am given the below circuit, how to determine whether the transistor is operating in active/saturation/cut off region?

Start by following through with the above analysis and then compute quantities from it. From there, you can determine $$\I_\text{E}\$$ and thereby $$\I_\text{C}\$$ on the assumption that it is in active mode. If you now examine $$\V_\text{C}=V_\text{CC}-R_\text{C}\cdot I_\text{C}\$$ and compare it with $$\V_\text{E}=R_\text{E}\cdot I_\text{E}\$$ and find the difference value to be below about $$\600\:\text{mV}\$$ in this case, then it is in saturation and not active mode. The lower that the computed $$\V_\text{C}-V_\text{E}\$$ is, the deeper the saturation.

Otherwise, it's in active mode.

## Question 2

How is the base current determined in the below circuit when there is not base resistor given? The voltage at the base is calculated to be 1.4V. But how is base current calculated?

By using the above-mentioned KCL solution process. $$\I_\text{B}\$$ just falls out.

## Question 3

What determines the current through the collector-emitter branch? Is is the emitter resistor or collector resistor?

If it is not in saturation, then the emitter voltage follows the base voltage and this determines the voltage across $$\R_\text{E}\$$ -- which determines its current and therefore the emitter current. So in this case, only the emitter resistor determines the current through the $$\R_\text{C}\$$+$$\V_\text{CE}\$$+$$\R_\text{E}\$$ path. The collector itself acts like a current source that reflects the emitter current.

If it is in saturation, then both resistors determine the current. You take $$\V_\text{CC}\$$, subtract some estimated tiny value for $$\V_\text{CE}\$$ (but obviously non-zero and positive) that should be on the order of a few hundred millivolts or less, and then divide that result by $$\R_\text{C}+R_\text{E}\$$. In this case, the collector acts like a voltage source.

Suppose you assume (and it is an assumption for now) that $$\V_\text{BE}\approx 700\:\text{mV}\$$ and that $$\\beta=200\$$. Then the above calculations with your circuit would find that $$\I_\text{B}\approx 18\:\mu\text{A}\$$, $$\V_\text{B}\approx 1.35\:\text{V}\$$, $$\V_\text{E}\approx 652\:\text{mV}\$$ and $$\V_\text{C}\approx 2.55\:\text{V}\$$. This would imply $$\I_\text{C}\approx 3.6\:\text{mA}\$$, which is consistent with the assumption that $$\V_\text{BE}\approx 700\:\text{mV}\$$. Since $$\V_\text{C}-V_\text{E}\approx 1.9\:\text{V}\$$, the BJT is not saturated.
Feel free to try other values for $$\\beta\$$ or $$\V_\text{BE}\$$ and see how things vary. It's worth the effort.
If things had turned out differently, and the circuit was in fact saturated, then the computations are different. As I pointed out, the current through $$\R_\text{C}\$$ and $$\R_\text{E}\$$ would then be determined by $$\V_\text{CC}\$$, less some assumed small value for $$\V_\text{CE}\$$, divided into by the sum of $$\R_\text{C}\$$ and $$\R_\text{E}\$$. So, you'd have different results in that case.