Find the collector current in Fig 2.
Assume:
- The base current is negligible
- Si diode.
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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2\$\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 WCommented Mar 18, 2021 at 18:55
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2\$\begingroup\$ @BornMad compute Vb starting from the fact that Ib=0 \$\endgroup\$– Sredni VashtarCommented Mar 18, 2021 at 19:00
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2\$\begingroup\$ why would that be? they have similar voltages but more than 2x smaller R \$\endgroup\$– Pete WCommented Mar 18, 2021 at 19:02
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2\$\begingroup\$ Ok I get it I can take Vbe = .7 and apply KVL to find base voltage \$\endgroup\$– BornMadCommented Mar 18, 2021 at 19:04
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2\$\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\$– jonkCommented Mar 18, 2021 at 19:15
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3 Answers
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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:
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.
answered Mar 20, 2021 at 11:50
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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.
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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\$ Commented Mar 20, 2021 at 13:46
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\$\begingroup\$ If there is a collector current , to find it( in active reagion) Beata(Ic/Ib) Should be given. \$\endgroup\$– upaliCommented Mar 21, 2021 at 5:10
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\$\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\$ Commented Mar 21, 2021 at 5:13