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Find the collector current in Fig 2.

Assume:

  • The base current is negligible
  • Si diode.

enter image description here


enter image description here

I have solved it, but I don't know if it's right or wrong.

Furthermore I have considered Ic=Ie+Ib which in wrongly written.

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    \$\begingroup\$ You have the current thru the 10k resistors. now figure out the base voltage. then the emitter voltage. then you will have the emitter current. then finally the collector current \$\endgroup\$ – Pete W Mar 18 at 18:55
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    \$\begingroup\$ @BornMad compute Vb starting from the fact that Ib=0 \$\endgroup\$ – Sredni Vashtar Mar 18 at 19:00
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    \$\begingroup\$ why would that be? they have similar voltages but more than 2x smaller R \$\endgroup\$ – Pete W Mar 18 at 19:02
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    \$\begingroup\$ Ok I get it I can take Vbe = .7 and apply KVL to find base voltage \$\endgroup\$ – BornMad Mar 18 at 19:04
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    \$\begingroup\$ @BornMad Ignoring base recombination current makes this a very simple problem. The two resistors at the base, combined with the two diodes, means that the midpoint between the two diodes will be exactly -10 V. Since the base-emitter junction drop is the same as a diode drop, this means the emitter of the BJT will also be exactly -10 V. The rest just quickly falls out. \$\endgroup\$ – jonk Mar 18 at 19:15
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Assume that the collector is connected to a potential such that the transistor is in active mode (eg. grounded). Assume junctions have 0.7V across them.

The current through the 10K resistors and diodes is (20V-1.4V)/20K = 0.93mA (so far so good)

So the base voltage is 10.7V above the -20V rail and therefore the emitter is at 10.0V above the -20V rail so the emitter current is ~10.0V/4.7K = 2.1mA.

Since base current is said to be negligible, collector current is also 2.1mA. Doing a quick simulation with real parts:

enter image description here

We can see that that's quite accurate.

If the -20V rail was much lower (say -3V) the exact magnitude of the Vbe and diode drops would be far more important, but here most of the voltage is dropped across the 10K resistors.

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20V / (10k+10k) minus diode drops is approx 1mA . This means all Si diodes will be 0.6V not 0.7 @1mA

Therefore the two 10k voltage divider gives;
Vb= -10V + 2x0.6. = -8.6V which controls the emitter voltage.

Thus Ve = -8.6 - 0.6 = - 9.2V.

Now you can compute Ie which = Ic within 1% for hFE >100.

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Given problem shows no connection to the collector line. If the diagram is correct current in emitter is 1 mA

This can be taken as base current. If β is(Ic/Ib) given we can find Ic. Say β=150 then Ic= 150 mA. enter image description here

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  • \$\begingroup\$ It is customary to draw parts of a circuit this way. The dashed line means that there is a current going into the collector coming from a part of the circuit that is not shown. Ideally if the BJT is in active region collector current does not depend on what is attached to the collector (within reason) \$\endgroup\$ – Sredni Vashtar Mar 20 at 13:46
  • \$\begingroup\$ If there is a collector current , to find it( in active reagion) Beata(Ic/Ib) Should be given. \$\endgroup\$ – upali Mar 21 at 5:10
  • \$\begingroup\$ there is a collector current and the assumption that ib=0 means it is equal to the emitter current. It's a common simplifying assumption. \$\endgroup\$ – Sredni Vashtar Mar 21 at 5:13

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