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Here is the example with its solution:

The problem is that when I apply KVL to the circuit (the second circuit) from the 10V source to the ground at the bottom, I get wrong values. Here is what I did:

$$-10 + 4700 \times I_C + V_{CE} + 3300 \times I_E = 0$$

$$I_C = \alpha I_E = \frac{\beta}{\beta+1}I_E = (100/101)I_E$$

$$\rightarrow -10 + 4700 \times \frac{100}{101}I_E + V_{CE} + 3300 \times I_E = 0$$

This gives that \$I_E = 1.23\$ mA which is different from \$1.6\$ mA.

What is wrong ?

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    \$\begingroup\$ Did you notice the comment at the bottom of your page about Beta forced? \$\endgroup\$ – Tut Jan 21 '15 at 21:41
  • \$\begingroup\$ Aha. In saturation I can't use the old Beta. Thanks. \$\endgroup\$ – ammar Jan 21 '15 at 21:45
  • \$\begingroup\$ Just FYI, when you set out to use a BJT as a saturated switch, you deliberately set up the circuit so that the forced beta will be far below the datasheet beta. For example, you might use a forced beta of 10. This helps keep Vce as low as it can be. \$\endgroup\$ – mkeith Jan 22 '15 at 4:30
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Here's the way I'd analyze this circuit:

1) By inspection, the emitter resistor is smaller than the collector resistor. So short C-E and calculate the voltage at that junction. I get about 4.13V.

2) The Base voltage is set at 6V. The voltage from above (4.13V) means that the E-B junction is forward biased. That changes things. Note that you need to do the above calculation to determine if, in fact, the junction is forward-biased.

3) The Emitter voltage is the Base voltage minus Vbe. That is: 6V - (0.7) = 5.3V. The current through the Emitter resistor is that voltage divided by the resistance: 5.3 / 3300 = ~1.61 mA.

4) Assume that the transistor is saturated. We'll do the calculations based on that assumption, then go back and check the assumption once we have some numbers to work with.

5) The collector voltage is about Vemitter plus the saturation voltage. I would normally assume saturation voltage at about 150 mV but your text says 200 mV, so use that. Ve + 0.2 = 5.5V.

6) Calculate the collector current. (10 - 5.5) / 4700 = ~0.957 mA

7) The Base current is the Emitter current minus the Collector current. 1.61 - 0.957 = ~0.649 mA

Now check to see if the transistor is in saturation. Most small-signal transistors have a Hfe of anywhere from 40 to 200. Let's use the worst-case value of 40.

If the transistor was NOT saturated, the collector current would be greater than 0.649 mA * 40 = ~25.9 mA. But we already know the collector current is about 0.957 mA. Therefore, the transistor is saturated and the above calculations hold.

Note that it took far longer to type this out than it did to calculate.

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