12/3/2004 section 5_4 BJT Circuits at DC 1/1 Jim Stiles The Univ. of Kansas Dept. of EECS Section 5.4 – BJT Circuits at DCReading Assignment:pp. 421-436To analyze a BJT circuit, we follow the same boring procedure as always: ASSUME, ENFORCE, ANALYZE and CHECK. HO: Steps for D.C. Analysis of BJT Circuits HO: Hints for BJT Circuit Analysis For example: Example: D.C. Analysis of a BJT Circuit Example: An Analysis of apnpBJT Circuit Example: Another DC Analysis of a BJT Circuit Example: A BJT Circuit in Saturation
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12/3/2004 Steps for DC Analysis of BJT Circuits 1/11
Jim Stiles The Univ. of Kansas Dept. of EECS
Steps for D.C. Analysis of
BJT CircuitsTo analyze BJT circuit with D.C. sources, we must follow thesefive steps :
1. ASSUME an operating mode
2. ENFORCE the equality conditions of that mode.
3. ANALYZE the circuit with the enforced conditions.
4. CHECK the inequality conditions of the mode forconsistency with original assumption. If consistent, theanalysis is complete; if inconsistent, go to step 5.
5. MODIFY your original assumption and repeat all steps.
12/3/2004 Steps for DC Analysis of BJT Circuits 5/11
Jim Stiles The Univ. of Kansas Dept. of EECS
This of course is a consequence of KCL, and is trueregardless of the BJT mode.
But think about what this means! We have two currentequations and three currents (i.e., , , B C E i i i )—we only needto determine one current and we can then immediately findthe other two!
Q: Which current do we need to find?
A: Doesn’t matter! For a BJT operating in the activeregion, if we know one current, we know them all!
b) In addition to 0 7 ( 0 7). .BE EB V V = = , we have a second
useful relationship:
(npn)
(pnp)
CB CE BE
BC EC EB
V V V
V V V
= +
= +
This of course is a consequence of KVL, and is trueregardless of the BJT mode.
12/3/2004 Steps for DC Analysis of BJT Circuits 7/11
Jim Stiles The Univ. of Kansas Dept. of EECS
Thus, for an analysis of circuit with a BJT in cutoff, we need tofind any two of the three quantities:
( )
( )
CB BE CE
BC EB EC
V , V , V npn
V , V , V pnp
We can then use KVLto find the third.
4. CHECK
You do not know if your D.C. analysis is correct unless youCHECK to see if it is consistent with your original assumption!
WARNING!-Failure to CHECK the original assumption willresult in a SIGNIFICANT REDUCTION in credit on exams,
regardless of the accuracy of the analysis !!!
Q: What exactly do we CHECK?
A: We ENFORCED the mode equalities , we CHECK the modeinequalities.
Active
We must CHECK two separate inequalities after analyzing acircuit with a BJT that we ASSUMED to be operating in active mode. One inequality involves BJT voltages , the other BJT currents .
12/3/2004 Steps for DC Analysis of BJT Circuits 8/11
Jim Stiles The Univ. of Kansas Dept. of EECS
a) In the active region, the Collector-Base Junction is“off” (i.e., reverse biased). Therefore, we must CHECK ouranalysis results to see if they are consistent with:
0 (npn)
0 (pnp)
CB
BC
V
V
>
>
Since 0 7.CB CE V V = + , we find that an equivalent inequalityis:
0 7 (npn)
0 7 (pnp)
.
.
CE
EC
V
V
>
>
We need to check only one of these two inequalities ( notboth !).
b) In the active region, the Base-Emitter Junction is “on”(i.e., forward biased). Therefore, we must CHECK theresults of our analysis to see if they are consistent with:
12/3/2004 Steps for DC Analysis of BJT Circuits 10/11
Jim Stiles The Univ. of Kansas Dept. of EECS
Cutoff
For cutoff we must CHECK two BJT voltages .
a) Since the EBJ is reverse biased , we CHECK:
( )
( )
0
0
BE
EB
V npn
V pnp
<
<
b) Likewise, since the CBJ is also reverse biased , weCHECK:
( )
( )
0
0
CB
BC
V npn
V pnp
>
>
If the results of our analysis are consistent with each of theseinequalities, then we have made the correct assumption! Thenumeric results of our analysis are then likewise correct. Wecan stop working!
However, if even one of the results of our analysis isinconsistent with active mode (e.g., currents are negative, or
0 7.CE V < ), then we have made the wrong assumption! Time tomove to step 5.
12/3/2004 Steps for DC Analysis of BJT Circuits 11/11
Jim Stiles The Univ. of Kansas Dept. of EECS
5. MODIFY
If one or more of the BJTs are not in the active mode, then itmust be in either cutoff or saturation . We must change ourassumption and start completely over!
In general, all of the results of our previous analysis areincorrect, and thus must be completely scraped!
5. Forget about what the problem is asking for! Just startby determining any and all the circuit quantities that youcan. If you end up solving the entire circuit, the answerwill in there somewhere!
6. If you get stuck, try working the problem backward ! Forexample, to find a resistor value, you must find the voltageacross it and the current through it.
7. Make sure you are using all the information provided inthe problem!
12/3/2004 Example Another BJT Circuit Analysis 2/3
Q: Now, how do we ANALYZE the circuit ??
A: This seems to be a problem ! We cannot easily solve theemitter base KVL, as i 1 is NOT EQUAL to i E1 (make sure youunderstand this !). Instead, we find:
1 B2E1i = i + iSo, what do we do ?
First, ask the question: What do we know ??
Look closely at the circuit, it is apparent that VB1 = 5.3 V andVE2 = 7.7 V .
See! Base current i B = 23.8 µ A, just like before.Therefore collector current and emitter current are again i C = 99i B = 2.356 mA and i E = 100 i B = 2.380 mA. Right ?!
Well maybe , but we still need to CHECK to seeif our assumption is correct!
We know that i B = 23.8 µA > 0 a , but what about V CE ?
From collector-emitter KVL we get:
10.7 – 10 iC – VCE – 2 iE =0Therefore,
VCE = 10.7 – 10(2.36) – 2(2.38) = -17.66 V < 0.7 V X
Our assumption is wrong ! The BJT is not in active mode.
In the previous example, the collector resistor was 1K ,whereas in this example the collector resistor is 10K.
Thus, there is 10X the voltage drop across the collectorresistor, which lowers the collector voltage so much thatthe BJT cannot remain in the active mode.