Semiconductor Devices - Hour 31
REAL Bipolar Junction Transistors (BJTs)
We also neglected certain strange distortional effects
Up until now, we have considered an idealized bipolar transistor to be ~ always in:
Today we will discuss a complete and realistic picture of the full range of BJT operation:
"Forward-active mode"
NOTE: There is nothing WRONG about our earlier picture
Implying substantial top-to-bottom (emitter to base) DC bias producing:
It is just doesn't cover the full range of possible BJT operation likely to be encountered in real circuits
Mild forward bias on its emitter-base junction (lowering its barrier)
List of REAL effects & considerations follows. Stick with me - there's some neat device physics going on!
Strong reverse bias on its base-collector junction (increasing its "fall")
"Forward-active" NPN BJT:
"Forward-active" PNP BJT:
Start by taking a more realistic look at what happens over wider range of bias and signal conditions
Basic amplifier circuit:
R1 and R2 chosen so that Emitter-Base forward biased:
Vpwr
if Vpwr = 10 volts
RC
choose R2 = 0.7 k:
Vout
R1
Further, to make sure that we stayed in this "forward active" mode, we assumed small applied AC signals
then VBE = 0.67 volts
vin
So that the biases on the junctions (their barriers, depletion regions . . . ) stayed ~ constant
R1 = 10 k:
RB
R2
We also treated these bipolar transistors as simple three layer structures
With all layers having the same width and depth (only their thicknesses and doping were different)
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Using standard circuit convention that upper case V and I refer to DC and lower case v and i refer to AC
vin
RB
IC3 = ⠘ IB3
IC2 = ⠘ IB2
Then, with transistor acting as a perfect current amplifier, at the circuit output:
iC = ⠘ iB
§ ☠RC ·
vout = RC ˜ iC = ¨
˜ vin
¨ RB
©
¹
then:
voltage gain =
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So if ran circuit with three different base currents (where IB1 IB2 IB3 ) might expect following I-V
curves:
IC
Assuming vin is small compared to 0.67 volts, then AC base current is:
iB =
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Changing VCE changes nothing
Only way to change things is to change I B
IC1 = ⠘ IB1
⠘ RC
RB
VCE = total, end-to-end, transistor voltage
HOWEVER, this assumes that changes in I B and VCE do not affect the value of E
Now let's look more closely at the DC biasing. An equivalent DC circuit would be:
Turns out (not surprisingly) that E can change in a number of ways
Where in "forward-active" mode with NPN transistor:
VCE
IB
B
VBE
E
0 ! VBC is our requirement, but given that VXY = VX VY we can show that in terms of our
batteries:
VBE ! 0
IC
And, again assuming transistor = current amplifier
expect:
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1) To continue to act as amplifier, we must maintain a reverse bias on BC junction (so that carriers are collected!)
VBC 0
C
3
0 ! VBC = VCB = VCE VBE
IC = ⠘ IB
(Equation 1)
In order to increase IC (the current through the transistor) we MUST increase V BE (lowering the barrier)
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·
§ q ˜ VBE
¨ k˜T
IC ~ IBE = Isat ˜ © e
1¹
But as VBE increases in, equation above, will reach a point where V BE > VCE then sign flips
At that point (and for higher currents) we no longer have reverse biased collector > lose gain
Forward active:
inverting this:
k ˜ T §¨ Ic ·
VBE =
˜ ln
¨ Isat
q
©
¹
Inserting this into equation 1, above:
§
k ˜ T §¨ Ic · ·
0 ! VBC = ¨ VCE ˜ ln
¨
¨ Isat
q
Not forward active:
©
©
¹¹
=>
This inequality is violated (and transistor stops operating a current amplifier with gain E) when:
By lowering barrier in middle (to get more IC)
We've lost reverse bias on right BC junction
q ˜ VCE
k ˜ T §¨ Ic ·
˜ ln
! VCE
¨ Isat
q
©
or rearranging, will loose normal operation when:
¹
k˜t
Leading to a revised set of I-V curves (each for a different base current):
And it can now start to backward inject!
IC
We can calculate this changeover by analyzing the emitter-base diode
⠘ IB3
§ q ˜ VBE
·
¨ k˜T
IBE = Isat ˜ © e
1¹
For the forward biased E-B diode:
IC ! Isat ˜ e
LEFT of the dashed exponential we have lost the
⠘ IB2
⠘ IB1
reverse bias on the base-collector junction
But for a narrow-base transistor, this current across the E-B junction is ~ the current across the B-C junction:
VCE
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Higher base currents = higher current through the device => more voltage needed on emitter base junction
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IC
IC = ⠘ ( n 1) ˜ ∆IB ⠘ ( n) ˜ ∆IB = ⠘ ∆IB
n 1
n
Then:
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Spacing between curves is D current gain
=> Less of power supply voltage left over for base-collector reverse bias
Constant gain = Uniform spacing:
Decreasing gain = Decreasing spacing:
=> REASON dashed line is farther right for higher currents
IC
IC
Add to this the knowledge that if V CE => 0 expect NO current through the device - and connect the points:
IC
IC3 = ⠘ IB3
VCE
IC2 = ⠘ IB2
IC1 = ⠘ IB1
VCE
What sort of things (far from saturation on left) might lead to changes in current gain?
"Base Width Modulation"
VCE
VBE = E-B forward bias
"Forward-Active" Amplifying Mode
= small value
"Saturation" = So much current (used so much voltage on EB junction)
VCE
IB
that no longer have enough voltage left to hold B-C junction in reverse bias
B
VBE
Machine that produces these curves, typically steps up the base current by equal amounts between curves:
IB1 = ∆IB
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IB2 = 2 ˜ ∆IB
IB3 = 3 ˜ ∆IB
7
C
E
= held ~ constant by VBE (or R1 and R2)
IC
THUS increasing VCE means base-collector
reverse bias must be increasing
...
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Looks like extrapolations of straight line segments might converge to left:
Means B-C depletion width must increase:
IC
VBC = VR
VBE
Emitter
Collector
xB_eff
WBC =
2 ˜ κ ˜ εo ˜ Vbi VR
§ 1 1 ·
˜¨
q
© NB NC ¹
VCE
VEarly
Point of approximate convergence = "Early Voltage" after James Early of Fairchild Semiconductor
More VCE => more voltage across B-C junction => Wider B-C depletion layer => Narrower effective base width
(very nice guy, met him several times)
As effective base width shrinks, less recombination => increase in gain, thus expect:
Ideal transistor (no slopes, as early in this lecture):
VEarly = ∞
IC
Real transistor: VEarly is finite (but the larger the better)
To increase Early voltage (decrease effect of base-width modulation) by making base thicker:
- Depletion makes less of a percentage change in base width
VCE
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- But this decreases the transistor's E!
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B-C Junction Avalanche Breakdown
Why care about that slight slope in the curves (=slight increase in E as VCE increases)?
Similar to the avalanche breakdown of a simple diode - except transistor AMPLIFIES the effect!
"Base-width modulation" / "Early Effect" distorts the amplification of signals!
Base
Emitter
Apply input AC signal (here a triangular wave) to the base of an NPN transistor:
- Negative voltage swing => Increase VBC, increase WBC, increase E
VBC
C
B
- Positive voltage swing => Decrease VBC, decrease WBC, decrease E
vin:
Collector
VBE
E
vout:
Hole drifting across B-C junction loses energy
Energy creates new electron-hole pair
After:
Before:
Then NEW carriers can do the same thing => Avalanche
How does this differ from simple diode avalanche breakdown?
Distortion, harmonics . . .
Large number of avalanche holes produced flow into base
=> Charge base slightly more positive
Other non-ideal effects:
=> Lowers Emitter-Base barrier turning transistor more strongly on!
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Step 1) New hole produced
Step 2) Holes accumulate in base
charging it more positive
Step 3) Lowered E-B barrier
=> massive current
Loss of "Low-Level Injection" - A reason why gain tends to decrease as transistor is driven hard:
Normal "Low-Level Injection" (remember, low-level injection = ~ no change in majority carrier population)
ppo
No change in holes!!!
Base region of NPN:
np
Transistor AMPLIFIES the effect of the beginning avalanche:
npo
Avalanche multiplication is thus important at voltages lower than one expects from simple diode
Breakdown voltage of transistor BC junction ~
Breakdown voltage of comparable diode
Emitter Side
β
Manifestation of B-C avalanche breakdown in I-V curves:
Collector Side
High-Level Injection (transistor really strongly on):
IC
<= Electron buildup pulls in extra holes
= More holes trying to jump into emitter
BVceo = Breakdown voltage common emitter . . .
= These holes must be replaced by IB
Strong accumulation of negative
electrons at emitter side of base
= Less gain in transistor!
VCE
BVceo
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I-V Manifestation (as shown early in lecture):
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At High VCE, depletion layer of BC junction actually CONNECTS with depletion layer of EB junction:
IC
Bands no longer rise as far in base => Massive increase in current ( I-V similar to avalanche breakdown)
Higher drive => Smaller intervals between curves
=> Reduced gain
To these effects (and a couple of others I didn't go into), must add:
VCE
Actual 3D configuration of bipolar transistor (here a big power transistor):
Base Contact
"Punch Through" - Another thing that can go wrong if drive transistor too hard:
High VCE:
Medium VCE:
Three-Dimensional Effects
Emitter Contact
ASSEMBLY SEQUENCE:
Base
1) N - substrate
Emitter
Emitter
2) Big P well diffused in (or implanted)=> Base
3) Smaller N+ well diffused in (or implanted) => Emitter
Collector
N+
WEB xB_eff
P+
P
WBC
WEB WBC
4) Ring of P+ added to exposed base
(improves ohmic contact & current flow around base)
N-
Collector
Collector Contact
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Which gives the following cross-section (with current flows):
2) Base Resistance:
Half cross-section of transistor showing current flowing in to base region from side
Emitter
Base layer is VERY thin (0.01 - 0.2 micron)
Emitter
But emitter can be very wide (microns)
Base
Problem #1) "Current Crowding" at edges
So Base resistor W / L very small
Collector
Collector
=> Big ohmic voltage losses
Problem #2) Resistance of thin base to horizontal current
VBase
1) Current Crowding: Vertical current naturally spreads out - even beyond bounds of emitter
But VEmitter ~ constant across its width so:
x
Most of current thus conducted at or near edges of the emitter
VBE
=> To get transistor to handle more current, make more edges!!!
Base-Emitter bias decreases in from edge!
Transistor is less "on" as move inward!!!!
Simple design
(seen from above):
x
Better "interdigitated"
Emitter
design (more edge):
Solution i) Make emitter narrower, use interdigitation to get more useful edge length
Solution ii): Dope base more heavily => less resistance
Collector
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But this also decreases transistor's E (more holes in base, more to escape to emitter, more I B)
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But, to complete our picture of microelectronics, it is time to turn to the very different field effect transistor
Another solution gained use ~ 1990 based on my personal work: HETEROJUNCTION BIPOLAR TRANSISTOR
Find a DIFFERENT material to make the base out of - one with a smaller bandgap!
∆Ec
For GeSi on Si, 'Ec >'Ev
So when put this into a bipolar transistor:
Silicon 1.1eV
Emitter
Ge0.2Si0.8
Base
0.9eV
Emitter
∆Ev
GeSi
Base
Collector
And then bias up the transistor for gain:
Kill off "reverse injection" of holes
with the higher energy barrier
Small E step
Then are free to increase base doping
Large E step
=> Reduces base series resistance
(higher gain / faster transistor)
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