Berkeley example circuits
Circuit 1: Differential Pair
The following deck determines the dc operating point of a simple differential pair. In addition, the ac small-signal response is computed over the frequency range 1Hz to 100MEGHz.
SIMPLE DIFFERENTIAL PAIR VCC 7 0 12 VEE 8 0 -12 VIN 1 0 AC 1 RS1 1 2 1K RS2 6 0 1K Q1 3 2 4 MOD1 Q2 5 6 4 MOD1 RC1 7 3 10K RC2 7 5 10K RE 4 8 10K .MODEL MOD1 NPN BF=50 VAF=50 IS=1.E-12 RB=100 CJC=.5PF TF=.6NS .TF V(5) VIN .AC DEC 10 1 100MEG .END
Circuit 2: MOSFET Characterization
The following deck computes the output characteristics of a MOSFET device over the range 0-10V for VDS and 0-5V for VGS.
MOS OUTPUT CHARACTERISTICS .OPTIONS NODE NOPAGE VDS 3 0 VGS 2 0 M1 1 2 0 0 MOD1 L=4U W=6U AD=10P AS=10P * VIDS MEASURES ID, WE COULD HAVE USED VDS, BUT ID WOULD BE NEGATIVE VIDS 3 1 .MODEL MOD1 NMOS VTO=-2 NSUB=1.0E15 UO=550 .DC VDS 0 10 .5 VGS 0 5 1 .END
Circuit 3: RTL Inverter
The following deck determines the dc transfer curve and the transient pulse response of a simple RTL inverter. The input is a pulse from 0 to 5 Volts with delay, rise, and fall times of 2ns and a pulse width of 30ns. The transient interval is 0 to 100ns, with printing to be done every nanosecond.
SIMPLE RTL INVERTER VCC 4 0 5 VIN 1 0 PULSE 0 5 2NS 2NS 2NS 30NS RB 1 2 10K Q1 3 2 0 Q1 RC 3 4 1K .MODEL Q1 NPN BF 20 RB 100 TF .1NS CJC 2PF .DC VIN 0 5 0.1 .TRAN 1NS 100NS .END
Circuit 4: Four-Bit Binary Adder
The following deck simulates a four-bit binary adder, using several subcircuits to describe various pieces of the overall circuit.
ADDER - 4 BIT ALL-NAND-GATE BINARY ADDER
*** SUBCIRCUIT DEFINITIONS .SUBCKT NAND 1 2 3 4 * NODES: INPUT(2), OUTPUT, VCC Q1 9 5 1 QMOD D1CLAMP 0 1 DMOD Q2 9 5 2 QMOD D2CLAMP 0 2 DMOD RB 4 5 4K R1 4 6 1.6K Q3 6 9 8 QMOD R2 8 0 1K RC 4 7 130 Q4 7 6 10 QMOD DVBEDROP 10 3 DMOD Q5 3 8 0 QMOD .ENDS NAND
.SUBCKT ONEBIT 1 2 3 4 5 6 * NODES: INPUT(2), CARRY-IN, OUTPUT, CARRY-OUT, VCC X1 1 2 7 6 NAND X2 1 7 8 6 NAND X3 2 7 9 6 NAND X4 8 9 10 6 NAND X5 3 10 11 6 NAND X6 3 11 12 6 NAND X7 10 11 13 6 NAND X8 12 13 4 6 NAND X9 11 7 5 6 NAND .ENDS ONEBIT
.SUBCKT TWOBIT 1 2 3 4 5 6 7 8 9 * NODES: INPUT - BIT0(2) / BIT1(2), OUTPUT - BIT0 / BIT1, * CARRY-IN, CARRY-OUT, VCC X1 1 2 7 5 10 9 ONEBIT X2 3 4 10 6 8 9 ONEBIT .ENDS TWOBIT
.SUBCKT FOURBIT 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 * NODES: INPUT - BIT0(2) / BIT1(2) / BIT2(2) / BIT3(2), * OUTPUT - BIT0 / BIT1 / BIT2 / BIT3, CARRY-IN, CARRY-OUT, VCC X1 1 2 3 4 9 10 13 16 15 TWOBIT X2 5 6 7 8 11 12 16 14 15 TWOBIT .ENDS FOURBIT
*** DEFINE NOMINAL CIRCUIT .MODEL DMOD D .MODEL QMOD NPN(BF=75 RB=100 CJE=1PF CJC=3PF) VCC 99 0 DC 5V VIN1A 1 0 PULSE(0 3 0 10NS 10NS 10NS 50NS) VIN1B 2 0 PULSE(0 3 0 10NS 10NS 20NS 100NS) VIN2A 3 0 PULSE(0 3 0 10NS 10NS 40NS 200NS) VIN2B 4 0 PULSE(0 3 0 10NS 10NS 80NS 400NS) VIN3A 5 0 PULSE(0 3 0 10NS 10NS 160NS 800NS) VIN3B 6 0 PULSE(0 3 0 10NS 10NS 320NS 1600NS) VIN4A 7 0 PULSE(0 3 0 10NS 10NS 640NS 3200NS) VIN4B 8 0 PULSE(0 3 0 10NS 10NS 1280NS 6400NS) X1 1 2 3 4 5 6 7 8 9 10 11 12 0 13 99 FOURBIT RBIT0 9 0 1K RBIT1 10 0 1K RBIT2 11 0 1K RBIT3 12 0 1K RCOUT 13 0 1K
*** (FOR THOSE WITH MONEY (AND MEMORY) TO BURN) .TRAN 1NS 6400NS .END
Circuit 5: Transmission-Line Inverter
The following deck simulates a transmission-line inverter. Two transmission-line elements are required since two propagation modes are excited. In the case of a coaxial line, the first line (T1) models the inner conductor with respect to the shield, and the second line (T2) models the shield with respect to the outside world.
TRANSMISSION-LINE INVERTER V1 1 0 PULSE(0 1 0 0.1N) R1 1 2 50 X1 2 0 0 4 TLINE R2 4 0 50
.SUBCKT TLINE 1 2 3 4 T1 1 2 3 4 Z0=50 TD=1.5NS T2 2 0 4 0 Z0=100 TD=1NS .ENDS TLINE
.TRAN 0.1NS 20NS .END