# Solving a DC Circuit: Find I_B, V_C, V_{CE}

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In summary, we had a circuit with a DC equivalent circuit that was easy to draw. We then computed ##I_B, V_C, V_{CE}## using the given equations and obtained the relationships: $$I_C = \frac{V_{CC} - V_C}{R_C} = \frac{21 - V_C}{2k}$$ $$I_B = \frac{V_C - V_{BE}}{R_1 + R_2} = \frac{V_C - 0.7}{81k}$$ $$I_E = \frac{V_E}{R_E} = \frac{V_E}{1k}$$ Next, we related ##I_C## and ##I_B## using
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## Homework Statement

For the circuit above, draw the DC equivalent circuit.

Then compute ##I_B, V_C, V_{CE}##.

## The Attempt at a Solution

So the DC equivalent circuit was easy to draw. My concern is about the final answer. The answer given does not match the answer I've obtained and I'm wondering who is correct.

After drawing the DC circuit, I obtained the following relationships:

$$I_C = \frac{V_{CC} - V_C}{R_C} = \frac{21 - V_C}{2k}$$
$$I_B = \frac{V_C - V_{BE}}{R_1 + R_2} = \frac{V_C - 0.7}{81k}$$
$$I_E = \frac{V_E}{R_E} = \frac{V_E}{1k}$$

Now we know ##I_C = \beta I_B##, so the first and second equations can be related:

$$40I_B = \frac{21 - V_C}{2k}$$
$$I_B = \frac{V_C - 0.7}{81k}$$

Simplifying we get:

$$I_B = \frac{21 - V_C}{80k}$$
$$I_B = \frac{V_C - 0.7}{81k}$$

Solving we get:

$$\frac{21 - V_C}{80k} = \frac{V_C - 0.7}{81k}$$
$$V_C = 10.9V$$

Going back to the original equations, we see:

$$I_C = \frac{21 - V_C}{2k} = \frac{21 - 10.9}{2k} = 5.05 mA$$
$$I_B = \frac{V_C - 0.7}{81k} = \frac{10.9 - 0.7}{81k} = 0. 126 mA$$

Then we know:

$$I_E = I_C + I_B = 5.05 mA + 0.126 mA = 5.176 mA$$

From which we find:

$$I_E = \frac{V_E}{1k}$$
$$V_E = (5.176 mA)(1k) = 5.176 V$$

Hence we know:

$$V_{CE} = V_C - V_E = 10.9V - 5.176V = 5.72V$$

Now all of my analysis seems reasonable to me, but for some reason the answer is given:

$$I_B = 0.0995 mA$$
$$V_C = 12.8 V$$
$$V_{CE} = 8.76 V$$

This answer does not match mine whatsoever. Have I done something wrong?

EDIT: Here is a picture of the analysis given in the answer:

This doesn't look right:
Zondrina said:
$$I_B = \frac{V_C - V_{BE}}{R_1 + R_2} = \frac{V_C - 0.7}{81k}$$

I didn't check the rest. I think the solution you were given is correct.

You have the two equations in three unknowns ##I_C = \beta I_B,I_C + I_B = I_E##, so find a third using KVL and your knowledge of ##V_\mathrm{BE}##.

When you calculated IB you did not take into account the fact that the emitter is not at ground potential, but has some potential VE due to the emitter resistor carrying current (IB and IC).Edit: Oops! Beaten to the punch by milesyoung!

So I have the two equations:

$$I_C = \beta I_B$$
$$I_E = I_C + I_B$$

That connection from the collector to the base is what's causing a little bit of confusion. Why do they write ##I_C + I_B = I_E## in the spot where usually they write ##I_C##? In other words, why are they saying ##I_E## flows through ##R_C##? If I understood this, then everything else is clear.

Zondrina said:
Why do they write ##I_C + I_B = I_E## in the spot where usually they write ##I_C##?
Consider where the base current comes from in this circuit compared to the ones you're used to.

Maybe think of it this way: The emitter current flows down through ##R_C##, splits into ##I_B## and ##I_C##, and then forms up as ##I_E## again through ##R_E##.

Maybe a picture to help?

I see how I'm supposed to think about the current "breaking up" now. Pictures are worth a million words.

Thank you so much to both of you. That one connection threw me off my usual thinking. So:

$$-21V + I_ER_C + I_B(R_1 + R_2) + 0.7V + I_ER_E = 0$$
$$I_ER_C + I_B(R_1 + R_2) + I_ER_E = 20.3V$$

Then using ##I_E = (\beta + 1)I_B##:

$$I_B(R_1 + R_2) + (\beta + 1)I_B(R_C + R_E) = 20.3V$$

The rest of the analysis follows.

## 1. How do I calculate the base current (IB) in a DC circuit?

To calculate the base current in a DC circuit, you can use Ohm's Law (I = V/R) or Kirchhoff's Current Law (sum of currents entering and leaving a node is equal to zero). Alternatively, you can use the formula IB = (IC/β), where β is the current gain of the transistor.

## 2. What is the significance of VC in a DC circuit?

VC (collector voltage) is important in a DC circuit as it determines the amount of voltage dropped across the collector resistor and the output voltage of the circuit. It also affects the operating point of the transistor and can help determine its amplification capabilities.

## 3. How do I find the voltage across the collector-emitter (VCE) in a DC circuit?

The voltage across the collector-emitter can be calculated by subtracting the voltage at the base (VB) from the voltage at the collector (VC). Alternatively, you can use Ohm's Law (V = IR) or Kirchhoff's Voltage Law (sum of voltages in a closed loop is equal to zero) to find VCE.

## 4. What factors affect the values of IB, VC, and VCE in a DC circuit?

The values of IB, VC, and VCE are affected by various factors such as the values of the resistors in the circuit, the voltage and current sources, the type of transistor used, and the circuit topology. Temperature and other external conditions can also affect these values.

## 5. Can I use the same method to solve for IB, VC, and VCE in all types of DC circuits?

While the general principles and formulas for calculating IB, VC, and VCE apply to all DC circuits, the specific calculations may vary depending on the complexity and components of the circuit. It is important to carefully analyze the circuit and use the appropriate equations and techniques to accurately solve for these values.

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