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In solution of this question they said that current through 3k and 9k resistors are same... Here my confusion is Why Ib of Q3 is zero? Because currents for both resistors are same only when Ib
is zero. Also base voltage of Q3 is 3V given in solution, but i think that it should be voltage across 9k resistor (which is 9V as per voltage divider)

enter image description here

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  • \$\begingroup\$ You shouldn't accept a answer so fast. You don't know what others might say that haven't even had a chance to see the question yet. Give it a day or so. \$\endgroup\$ Commented Apr 27, 2017 at 12:01

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For reference and to protect against edits to the question, here is the schematic being discussed:

they said that current through 3k and 9k resistors are same

That is a legitimate first approximation. It is basically saying that the gain of Q3 is infinite. Since the base current is the collector current divided by the gain, the base current is 0 in this approximation.

Assuming that the transistor gain is infinite is often useful just to get a basic idea what the circuit is doing. Once you solve for the voltages and currents with this simplifying assumption, you get some idea what parts are sensitive to transistor gain. Then you can go back and figure things more accurately in that area.

Another point is that good BJT circuits work with the transistor gains anywhere from the guaranteed minimum to infinite. Therefore analyzing at infinite gain is verifying one of these criteria. Often there is little difference between a gain of, say 100, and infinite. Whether the collector current is 99% of the emitter current or 100% of the emitter current shouldn't make much difference in a good design.

Let's see what we get with infinite gain, then compare that to the fixed gain of 100. With infinite gain, the base current is 0, so there will be 9 V across the 9K resistor, and 3 V across 3K. This leaves about 8.3 V across 2.77K, for a collector and emitter current of 3 mA.

With a gain of 100, the resistance connected to the emitter is reflected to the base times (gain + 1). The base therefore looks like a 280 kΩ load to -11.3 V. The base voltage then comes out to -3.07 V. The voltage across the emitter resistor is 8.23 V, and the emitter current 2.97 mA. Again using a gain of 100, the collector current is therefore 2.94 mA.

This is a good example showing how the infinite gain approximation can lead to useful first pass results. With Q3 having infinite gain, I0 is 3.00 mA. With it having the finite gain of 100, I0 is 2.94 mA. If this transistor has a minimum guaranteed gain of 100 at this operating point, then the rest of the circuit has to be able to tolerate the 2.94 to 3.00 mA variation. That's life with real BJTs. Fortunately, that level of uncertainty should not be a hardship.

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  • \$\begingroup\$ Can you give me some insight about voltage divider confusion? \$\endgroup\$
    – user146551
    Commented Apr 28, 2017 at 5:35
  • \$\begingroup\$ @user: I don't know what you mean. \$\endgroup\$ Commented Apr 28, 2017 at 10:39
  • \$\begingroup\$ In solution they are saying that Base Voltage of Q3 is 3V, but i think that base voltage is voltage across 9K resistor as it connected across base, am i wrong? \$\endgroup\$
    – user146551
    Commented Apr 28, 2017 at 13:17
  • \$\begingroup\$ @user: The base will be at about -3 V relative to ground. Generally when a absolute voltage is given, it is implied to be relative to ground. \$\endgroup\$ Commented Apr 28, 2017 at 13:45
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The solution should have said that the currents are nearly the same. The base current is much less than the 1 milliamp flowing through the 3k and 9k resistors.

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