Transcript CH14

CH14 BIPOLAR DIGITAL CIRCUITS
The Ideal BJT Transistor Switch
The switch terminals
are the collector and
emitter.
The transistor switch is:
 OFF when VBE is zero or negative
 ON when VBE is positive
(These conditions are approximated with a practical transistor
switch.)
The Practical BJT Transistor Switch
Transistor State:
ON – Operating point Is in the Saturation Region
VCE = saturation value typically 0.2 volts
VCE = 0 for the ideal case
(device is said to be operating in the saturation region when VCE
is less than about 0.7 volts.)
OFF – Operating point is in the cutoff region
(the cutoff region exists below the level of IB = 0.)
Note that when IB = 0, IC is not zero. Instead a small leakage
current called ICBO (or ICO) flows (typically1mA).
VCE = VCC – ICBO x RL
VCE = VCC for the ideal case
Transistor Switching Times
Because of internal capacitive
effects, transistors do not
switch in zero time.
Time relationships between iC and vi for the simple inverter
Transistor Switching Times
Delay Time – t d – the time required for IC to reach 10%
of its final level after IB has commenced. It is due to
mainly the time needed to charge the EBJ depletion
capacitance to the forward-bias voltage VBE.
Rise Time – t r – the time it takes for IC to go from 10%
to 90% of its maximum level. (Occurs after the delay
time.)
Turn-On Time – t on = t d + t r
Storage Time – t s – the time between IB switch off and
IC falling to 90% of its maximum level – due to the fact
that the CBJ is FWD biased in saturation. – excess
minority charge carriers, stored in the depletion region
must be withdrawn or recombine before begins IC to
fall.
Fall Time – t f – the time it takes for IC to fall from 90%
to 10% of its maximum.
Turn-Off Time – t off = t s + t f
CH14 INTRODUCTION TO TTL
Conceptual Circuit (Inverter)
Analysis for Case:
Input = logic HI (e.g. 5volts)
In this case Q1 operates in the inverse active mode with a
very low bR  0.02
Base-Collector junction of Q1 is forward biased
Base-Emitter junction of Q1 is reverse biased
(see currents marked on Fig. 14.10)
Gate input current = bR I  0.02 I – very small
Q3 base current = (bR + 1) I  I
R is chosen so that I is large enough to drive Q3 into
saturation with LO output voltage  0.2 volts
CH14 INTRODUCTION TO TTL
Conceptual Circuit (Inverter)
 5V
+0.3V
Analysis for Case:
Input = logic LO (e.g. 0.2 volts)
In this case Q1 operates in the normal active mode with a
bF  large value
Base-Emitter junction of Q1 is forward biased and base
voltage = 0.2 + 0.7 = 0.9 volts
Collector current of Q1 = bF I (large value) which rapidly
discharges the base of Q3 (initially still at 0.7 volts and
saturated immediately after switching) rapidly driving it into
cutoff with a reduced base voltage  0.3volts.
This reduced base voltage is the collector voltage of Q1 so
Q1 saturates with VCE sat  0.1v and negligibly small
collector current. Its base voltage is then  0.3v  the base
voltage of Q3, keeping Q3 in cutoff and output level HI.
Actual Complete Circuit TTL Gate (NAND)
BASIC FEATURES: (Analysis for case of inverter done later)
Input Stage based on Q1 – also known as multi-emitter input.
(Protection diodes shown on actual circuit do not affect logic
function.)
Driver Stage based on Q2 – also known as phase-splitter.
--Causes either Q3 or Q4 to turn on while other is off.
Output Stage based on Q3 and Q4 -- also known as
totem-pole output. – Provides active pull-up through Q3.
Analysis when Input is HI
Circled numbers in Fig. give order of analysis.
(Our analysis relies on prior understanding gained from the
conceptual circuit analysis already completed.)
1. Q3 is on and has 0.7v at its base
2. Q2 is on supplying Q3 with sufficient base current to drive it into
saturation.
3. Q1 is operating in the inverse active mode with BC junction FWD
biased (0.7v), so its base voltage is 1.4 + 0.7 = 2.1v
4. Ohm’s law gives current through 4k resistor.
5. Gate input current IIH = Q1’s inverse mode emitter current =bR I
6. Q2’s base current = Q1’s inverse mode collector current
= (bR +1) I  0.73mA
7. Q2’s collector voltage = VBE3 + VCE sat  0.7 + 0.2 = 0.9v
8. Ohm’s law gives current in 1.6k resistor = 2.6 mA.
9. The 0.9v at Q4’s base cannot turn it on due to diode D, so IB4 = 0
10. For Q2, IE = IC + IB = 3.3mA
Analysis when Input is HI
11. Current through 1k resistor by Ohm’s law
12. Current into base of Q3 by KCL
13. Output voltage of gate is VCE SAT of Q3
In the LO output state Q3 can sink a load current
I L <= b x 2.6mA
If this value is exceeded – no longer in saturation and output
logic LO voltage level not maintained within specified limits.
(Fig. shows Q3 leaving saturation at higher load current).
There is thus a limit on allowable load current that is directly
related to the gate’s maximum fan-out.
The Fig. is simply the vCE vs. iC characteristic curve of Q3
at the base current it has when output is LO.
Analysis when Input is LO
1. BE junction of Q1 is FWD biased and base voltage  0.2+0.7 = 0.9v
2. Ohm’s law gives current in 4k resistor.
4. 0.9v is insufficient to FWD bias the series combination of CB
junction of Q1 and BE junction of Q2, so Q2 is cutoff and Q1 has
zero collector current.
3. For Q1, IE = IB + IC = IB + 0 = I  1mA.
5. For Q1, VC = VCE SAT + VI  0.1 + 0.2 = 0.3v
6. For Q2 (cutoff), VE = IE x 1k = 0
7. Q3 is also cutoff with base current = 0 mA
Analysis when Input is LO
Q4 supplies the load current when input is LO (output HI)
1.
Gate output terminal open:
v0 = 5 – (IB4 X 1.6k) – 0.65 – 0.65
 5 – 0.65 – 0.65 = 3.7v (since IB4  0)
2.
Q4 supplying load current large enough to cause it to
saturate:
v0 = VCC – iL x 130 – VCE SAT4 – VD
3.
Q4 supplying load current below value causing
saturation while in active mode:
v0 = VCC – iL /(b +1) x 1.6k – VBE4 – VD
Function of 130 Ohm Resistance
Function is to limit the current through the pull-up
transistor Q4.
Limit current due to:
1. Accidental short-circuit to ground.
2. Current pulse due to Q3 and Q4 briefly on at the
same time during HI to LO transition.
(Q3 needs extra time to turn off (discharging the
base through the 1k) after Q2 has caused
Q4 to turn on.)
TTL Supply Bypassing
The current pulses due to the brief time when both output
transistors of the totem pole turn on, are current spikes
which cause corresponding voltage spikes superimposed
on VCC that can be coupled to other components of the
digital system.
Thus bypassing capacitors should be connected
between
The VCC rail and ground at frequent locations.
(Usually at the VCC pins of the IC’s).
Check IC manufacturer’s application notes for
Recommended practice.