3 The TTL NAND Gate
3.1
TTL NAND Gate Circuit Structure
The circuit structure is identical to the previous TTL inverter
circuit except for the multiple emitter input transistor. This is used to
implement a diode switching structure in active transistor form using
parallel junction diffusions for several emitters.
⇒
Fig. 3.1 Multiple Input Emitter Structure of TTL
If any input is low, the corresponding base-emitter junction
becomes forward-biased and the transistor conducts. The other
characteristics of the circuit and its transfer characteristic are identical
to those of the inverter circuit.
3.2
Logical Operation
A table of conduction states can be drawn up showing the state of
each transistor in the circuit for all possible input conditions as before
to verify the logic function performed. The direction of conduction of T1
can be in the forward or reverse mode so this should also be noted in
the table. It can be seen from the table that the output goes LO only
when both inputs are HI which verifies the NAND function.
IN1
IN2
T1
T2
T3
T4
D
OUPUT
LO
LO
ONfor
OFF
OFF
ON
ON
HI
LO
HI
ONfor
OFF
OFF
ON
ON
HI
HI
LO
ONfor
OFF
OFF
ON
ON
HI
HI
HI
ONrev
ON
ON
OFF
OFF
LO
1
VCC
R3
R1
130Ω
1. 6 k Ω
N5
RB
N3
4kΩ
N1
Input 1
Input 2
T4
N6
N2
T2
T1
N7
N4
Output
(no load)
T3
R2
1kΩ
Fig. 3.2
Circuit Diagram of a Standard 2-input TTL NAND Gate
2
3.3. Circuit Analysis
It is of interest to examine the conditions for the different logic states
of the NAND Gate circuit, particularly with regard to estimating the
power consumption in each state. This can be done by first
establishing the voltages at each of the nodes N1 – N7 in the circuit
and then finding the total current drawn from the power supply.
(a) At Least One Input LO – Output HI
To aid in the analysis, the NAND Gate circuit can be redrawn with the
transistors which are non-conducting or OFF removed from the circuit
as shown in Fig. 3.3. Then the potentials, relative to ground, can be
determined for each of the nodes in turn. Under this condition, T1 is ON
in forward mode, T2 is OFF, T3 is OFF, while T4 is ON at the point of cutin and therefore T2 and T3 have been removed from the circuit.
VCC
I1
IB
I3
R3
130Ω
R1
1.6kΩ
RB
N5
N3
4kΩ
N1
Input 1
Input 2
0.1 V
T4
N6
N2
N7
T1
Output
(no load)
N4
R2
1kΩ
Fig. 3.3
NAND Gate Circuit Redrawn with at least One Input LO
3
(i) T1 ON in forward mode and is operating in saturation as there is
only a leakage current from T2 available as collector current, i.e. T1
operates with a large base current and negligible collector current
where IC MAX = 0. The input logic LO voltage is taken as 0.1V. Then:
Node N1 : VN1 = Vi + VBE1 SAT = 0.1 + 0.8 = 0.9V
(ii) With T1 operating in saturation, its collector-emitter voltage is VCE
SAT = 0.1V so that:
Node N2 : VN2 = Vi + VCE1 SAT = 0.1 + 0.1 = 0.2V
(iii) With T4 operating at the point of cut-in its base current and hence
its collector current can
be taken as zero. This means that there is no voltage drop across
either resistor R1 or R3
and so the potential at both sides of these resistors is equal to
the supply voltage VCC giving:
Node N3, Node N5 : VN3 = VN5 = VCC = 5V
(iv) Node N4 is pulled low by the resistor R2 which has no current
flowing through it so that:
Node N4 :
(v)
VN4 = 0V
Finally, with T4 operating at the point of cut-in:
Node N6 : VN6 = VN3 − VBE4 CUT-IN = 5 − 0.6 = 4.4V
and with the diode at cut-in also:
Node N7 : VN7 = VN6 − VD CUT -IN = 4.4 − 0.4 = 4.0V
4
The current drawn from the supply can then be obtained as:
IB =
VCC − VN1 5 − 0.9V
=
= 1.025mA
RB
4kΩ
with I1 and I3 = 0 since negligible current flows into the base or
collector of T4 while at the point of cut-in.
The power consumption of the gate with the output in the logic Hi
state can then be obtained as:
POH = VCC x IB = 5V x 1.025mA = 5.125mW
(b) Both Inputs HI – Output LO
Under this condition T1 is ON in the reverse mode, T2 is ON, T3 is ON
and T4 is OFF. Fig. 3.4 shows the NAND gate circuit redrawn with T4
removed. Potentials must be determined in a different order this time.
VCC
I1
I3
IB
130Ω
N5
R1
1.6kΩ
RB
N3
4kΩ
N1
Input 1
Input 2
5V
5V
R3
N6
N2
T2
T1
N7
N4
Output
(no load)
T3
R2
1kΩ
Fig. 3.4
NAND Gate Circuit Redrawn with Both Inputs HI
5
(i) With T3 ON and operating in saturation:
Node N4 : VN4 = VBE3 SAT = 0.8V
(ii) With T2 also ON and in saturation:
Node N2 :
VN2 = VN4 + VBE2 SAT = 0.8 + 0.8 = 1.6V
(iii) Since T1 is ON in the reverse mode, the base-collector voltage in
this mode can be taken as the same as the base-emitter voltage of
a transistor operating in the forward active mode so that:
Node N1 :
VN1 = VN2 + VBC1 ON REV = 1.6 + 0.7 = 2.3V
(iv) With T2 operating in saturation, its collector emitter voltage will be
VCE SAT = 0.1V so that:
Node N3 :
VN3 = VN4 + VCE2 sat = 0.8 + 0.1 = 0.9V
(v) With T4 OFF no current will flow through resistor R3 and
consequently Node N5 will be pulled up to the supply rail voltage:
Node N5 :
VN5 = VCC = 5V
(vi) With T3 ON and in saturation, its collector-base voltage will be at a
saturation value so that the output voltage at Node N7 is simply:
Node N7 :
VN7 = VCE3 SAT = 0.1V
(vii) With T4 and the diode non-conducting, the potential at Node N6 is
somewhat ill-defined and depends on the resistances of the nonconducting junctions of these devices but will lie somewhere between
that of Nodes N3 and N7, i.e. between 0.1 and 0.9V. However, this
voltage is not significant.
6
The current drawn from the supply this time is given by the sum of IB
and I1 with I3 = 0:
Then:
IB =
VCC − VN1 5 − 2.3V
=
= 0.675mA
RB
4kΩ
and
I1 =
VCC − VN3
5 − 0.9V
=
= 2.56mA
R1
1.6kΩ
The power consumption when the output is LO is then given as:
POL = VCC x (IB + I1 ) = 5 x ( 0.675 + 2.56) = 16.175mW
If the NAND gate is assumed to spend half of its time in each logic
state then the average power consumption can be expressed as:
PAVE =
POH + POL
5.125 + 16.175
=
= 10.56mW
2
2
7