For example, assume you have this transistor with a grounded emitter:

enter image description here

Since we are dealing with a saturation mode, \$v_{ce} = 0.2 \$ making the drop across the resistor 9.8 v. The current going through the collector is \$ I_c = 9.8/10 k\Omega = 9.8 mA \$ and assuming \$ \beta = 100 \$ then \$I_b = 0.098 mA \$. But \$v_c = I_c \times R_c = 9.8\$ volts. For a transistor to be considered in the saturation mode \$v_c < v_b \$

My textbook has some contradictory statements about saturation mode. In the beginning of the course, we learned that when a transistor is in saturation mode, the collector should read close to zero volts. The book now says that when a HIGH (> 2 volts) is applied to the input supplied to the transistor's base, it will cause the transistor to conduct and the collector voltage drops to a LOW. But in the calculation from above (assuming that it was done correctly), I can clearly see that the voltage on the collector is 9.8 volts, close to 10 volts and not anywhere close to zero or LOW.

EDIT: Am I suppose to assume that \$v_c = v_{cc} - I_c \times R_c = 0.2\$. Since the emitter is grounded, \$v_c = v_{ce}\$

  • \$\begingroup\$ Vc = 10V - Ic*Rc. \$\endgroup\$
    – user16324
    Commented Dec 1, 2014 at 10:41

3 Answers 3


You cannot make both assumptions about \$V_{ce} = 0.2V\$ and \$\beta = 100\$ at the same time; they are for two different modes of operation for the transistor (there's also a third mode where the transistor is in cutoff, and a fourth less common one in reverse active).

You start out assuming one of the possible states. For example, suppose I assume the transistor is saturated. Then \$V_{ce} = 0.2V\$ and \$V_{be} = 0.7V\$. You can then solve for base and collector currents using these assumptions for the BJT and only these assumptions.

The last step is to check your assumptions. For example, to make sure the transistor isn't actually in the linear active region must check that \$\beta I_{b} \gg I_{c}\$. So let's say in your case you get \$I_{c} = 9.8 mA\$ and \$I_{b} = 9.8 \mu A\$. Well clearly our assumption about saturation was bad because \$\beta I_{b} = 980 \mu A\$, which is less than our calculated \$I_{c}\$. We must then start over with new assumptions and re-solve the problem.

Instead, suppose \$I_b = 2 mA\$. Then \$\beta I_b = 200mA\$, which is much greater than \$I_c\$, so now the saturation assumption is correct.

Note: \$\beta I_b \gg I_c\$ is kind of a vague limit. Typically we use at least 10 times bigger for saturation.

For solving assuming the BJT is forward active, you would assume \$\beta I_b = I_c\$ and only this assumption. To check it (against saturation), simply make sure \$0.2V < V_{ce}\$.


Your calculation of the collector current is correct. So let's focus on calculating the base current.

Since we are dealing with a saturation mode, vce=0.2 making the drop across the resistor 9.8 v. The current going through the collector is Ic=9.8/10kΩ=9.8mA and assuming β=100 then Ib=0.098mA.

β is a characteristic of the transistor in forward-active mode, and doesn't apply in saturation.

Simply use the forward drop of the be junction, \$V_{be}\approx{}0.7\ \mathrm{V}\$. Then the base resistor determines the current, \$I_b=\dfrac{V_{in}-0.7\ \mathrm{V}}{10\ \mathrm{k\Omega}}\$.

I can clearly see that the voltage on the collector is 9.8 volts

You just got mixed up here. The voltage across the collector resistor was 9.8 V. The voltage across the transistor from emitter to collector is 0.2 V. In fact, you got the 9.8 V number by assuming \$v_{ce}=0.2\ \mathrm{V}\$.


VC does NOT equal 9.8V. It is 10 - 9.8 = 0.2 as in your assumption.

The 0.2V is only an approximation - depending upon the transistor it may be as low as a handful of millivolts.


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