Evaluation Programs for SPICE Op Amp Models Application Note Introduction There is no consistent method for evaluating SPICE models in the industry, so it is hard to reproduce a specific manufacturer’s results or to compare models between manufacturers. Furthermore, many of the SPICE models available from vendors do not correlate the vendor’s data sheet for the corresponding op amp; hence, there is confusion about the validity of the model, the evaluation program, and/or the data sheet. This paper includes a collection of SPICE programs that have been used to evaluate some of the latest Intersil Corporation current feedback op amps. The programs illustrated here will be used to evaluate new Intersil op amp designs, both current and voltage feedback, so they will serve as a standard until modified by common agreement. These programs have several advantages: they are written, they have been debugged, macros exist to eliminate the plotting effort, they cover the pertinent parameters, and they contain equations to normalize the output for “Bode” type plots. January 1996 AN9523 Unless the output is normalized the vertical scale will have to be large enough to accommodate the difference in gains, so small effects such as peaking may be hard to discern or measure. The program configures the load resistor as a voltage divider, and the output is taken at the voltage divider output. If the op amp gain is 10 the load resistor gain will be 0.1. If a load resistor is not required by the data sheet enter a large value such as 1000GΩ; the large resistor will not affect the circuit operation while the normalization feature is retained. Now the three curves will plot on top of each other similar to the GBW curves shown in most data books. 11 RF11 RG11 IN VIN 12 + RA11 OUT1 CL111 RB11 21 Six programs which cover 13 parameters including: inverting gain, non-inverting gain, positive power supply current, negative power supply current, positive input bias current, negative input bias current,offset current, positive offset voltage, negative offset voltage, differential offset voltage, non-inverting common-mode voltage, transient response, and enable/disable response are used to evaluate the op amp. These thirteen parameters are displayed in eight plots which have seventeen curves. Printed copies of these programs are given here, and electronic copies are available on the “Intersil CorporationAnalog SPICE Macromodels” disk dated January 1996 or later. AC Transfer Function For An Inverting Op Amp RG21 RF21 - 22 RA21 OUT2 + CL21 RB21 31 RG31 RF31 - 32 RA31 OUT3 + CL31 RB31 FIGURE 1. SCHEMATIC OF INVERTING OP AMP The first program (see Figures 1, 2, and 3) is named cfaig.cir, and it simulates the AC transfer function for an inverting op amp. This program uses three op amps so it computes the transfer function for three different gains in one pass. The program requires the user to supply the feedback resistance values for each gain, the gain settings, the load resistance, the load capacitance, and the power supply voltage in volts. The program assumes that the op amp is run off two power supplies of equal and opposite polarity each of which is referenced to ground, so it applies the entered voltage to the op amp as a positive and negative supply with equal magnitudes. If a single supply op amp is to be evaluated with these programs just enter half the power supply voltage, and the analysis will be equivalent. 1 1-888-INTERSIL or 321-724-7143 | Copyright © Intersil Corporation 1999 Application Note 9523 *This program simulates the AC transfer function for an inverting op amp. *It has three op amps; each with a gain that is specified by the *user with a .param statement. The user must specify the load RL, CL, the *feedback resistors RF1, RF2, RF3 and the corresponding gains G1, G2, G3. *The power supply voltage is set by the parameter “vsupply”. The load resistors *are automatically split into voltage dividers to normalize the gain plot, *and the gains can be plotted in dB by calling the macros G1, G2, and G3. *The inputs are tied together, and the outputs are called OUT1, OUT2, and *OUT3 corresponding to the respective gains. The op amp model is entered with *a .lib statement. The model in the subcircuit call (x statement) must correspond to the model *called in the .lib statement (3 times). .param CL=10pf .param RL=400 .param RF1=750 .param RF2=750 .param RF3=750 .param G1=1 .param G2=2 .param G3=3 .param vsupply=5 .lib b:ha502x.cir x1 0 11 3 4 12 3 ha502x x2 0 21 3 4 22 3 ha502x x3 0 31 3 4 32 3 ha502x VIN in 0 ac 1 RF11 11 12 {RF1} RF21 21 22 {RF2} RF31 31 32 {RF3} RG11 in 11 {RF1/abs(G1) } RG21 in 21 {RF2/abs(G2) } RG31 in 31 {RF3/abs(G3) } RA11 12 OUT1 {RL* (abs(G1)-.99999) /abs(G1) } RA21 22 OUT2 {RL* (abs(G2)-.99999) /abs(G2) } RA31 32 OUT3 {RL* (abs(G3)-.99999) /abs(G3) } RB11 0 OUT1 {RL/abs(G1) } RB21 0 OUT2 {RL/abs(G2) } RB31 0 OUT3 {RL/abs(G3) } CL11 0 12 {CL} CL21 0 22 {CL} CL31 0 32 {CL} VCC 3 0 {vsupply} VEE 4 0 {-1*vsupply} .ac dec 50 1meg 3000meg .probe .end FIGURE 2. INVERTING OP AMP AC TRANSFER FUNCTION PROGRAM 2 Application Note 9523 VCC 6.0 + - NORMALIZED GAIN IN dB 4.0 RA11 3 OUT1 4 RB11 VEE 2.0 11 RG11 VIN 0 RF11 12 CL11 VCC -2.0 IN G1 G2 G3 -4.0 - 3 10 30 FREQUENCY (MHz) RA21 3 OUT2 4 RB21 VEE -6.0 1 + 100 21 300 RG21 RF21 22 CL21 FIGURE 3. INVERTING OP AMP AC TRANSFER FUNCTION PLOT VCC AC Transfer Function for a Non-Inverting Op Amp Unless the output is normalized the vertical scale will have to be large enough to accommodate the difference in gains, so small effects such as peaking may be hard to discern or measure. The program configures the load resistor as a voltage divider, and the output is taken at the voltage divider output. If the op amp gain is 10, the load resistor gain will be 0.1. If a load resistor is not required by the data sheet enter a large value such as 1000GΩ; the large resistor will not affect the circuit operation while the normalization feature is retained. Now the three curves will plot on top of each other similar to the GBW curves shown in most data books. 3 - 3 RA31 OUT3 4 RB31 VEE 31 RG31 32 RF31 CL31 FIGURE 4. NON-INVERTING OP AMP SCHEMATIC 6.0 4.0 NORMALIZED GAIN IN dB The second program (see Figures 4, 5, and 6) is named cfanig.cir, and it simulates the AC transfer function for a noninverting op amp. This program uses three op amps so it can compute the transfer function for three different gains in one pass. The program requires the user to supply the feedback resistance values for each gain, the gain settings, the load resistance, the load capacitance, and the power supply voltage in volts. The program assumes that the op amp is run off two power supplies of equal and opposite polarity each of which is referenced to ground, so it applies the entered voltage to the op amp as a positive and negative supply with equal magnitudes. If a single supply op amp needs to be evaluated just enter half the power supply voltage, and the analysis will be equivalent. + 2.0 0 G1 -2.0 G2 G3 -4.0 -6.0 1 3 10 30 FREQUENCY (MHz) 100 FIGURE 5. NON-INVERTING OP AMP AC TRANSFER FUNCTION PLOT 300 Application Note 9523 *This program simulates the transfer function for a non-inverting op amp. *It has three op amps; each with a gain that is specified by the user *with a .param statement. The user must specify the load RL, CL, the *feedback resistors RF1, RF2, RF3 and the corresponding gains G1, G2, G3. *The power supply voltage is set by the parameter “vsupply”. The load resistors *are automatically split into voltage dividers to normalize the gain plot, *and they can be plotted in dB by calling the macros G1, G2 and G3. *The inputs are tied together, and the outputs are called OUT1, OUT2, and *OUT3 corresponding to the respective gains. The op amp model is entered with *a .lib statement. The model in the subcircuit call (x statement) must correspond to the model *called in the .lib statement (3 times). .param CL=10pf .param RL=400 .param RF1=1000 .param RF2=681 .param RF3=383 .param G1=1 .param G2=2 .param G3=10 .param vsupply=5 .lib b:ha502x.cir x1 in 11 3 4 12 3 ha502x x2 in 21 3 4 22 3 ha502x x3 in 31 3 4 32 3 has02x vin in 0 ac 1 RF11 11 12 {RF1} RF21 21 22 {RF2} RF31 31 32 {RF3} RG11 0 11 {RF1/(G1-.99999) } RG21 0 21 {RF2/(G2-.99999) } RG31 0 31 {RF3/(G3-.99999) } RA11 12 OUT1 {RL*(G1-.99999)/G1} RA21 22 OUT2 {RL*(G2-.99999)/G2} RA31 32 OUT3 {RL*(G3-.99999)/G3} RB11 0 OUT1 {RL-(RL*(G1-.99999)/G1) } RB21 0 OUT2 {RL-(RL*(G2-.99999)/G2) } RB31 0 OUT3 {RL-(RL*(G3-.99999)/G3) } CL11 0 12 {CL} CL21 0 22 {CL} CL31 0 32 {CL} VCC 3 0 {vsupply} VEE 4 0 {-1*vsupply} .ac dec 50 1meg 3000meg .probe .end FIGURE 6. NON-INVERTING OP AMP AC TRANSFER FUNCTION PROGRAM 4 Application Note 9523 DC Parameters For a Non-Inverting Op Amp The input signal to the op amp is a DC sweep. The sweep input enables a data analysis at VIN = 0V which is often a data book point, and the parameters can be evaluated at various other points of interest. The input currents can be examined by plotting the currents through the feedback resistor, RF, and the input resistor, RI . The difference between these currents is the input offset current. When the voltage is swept through zero the offset voltage for zero input voltage can be calculated. Either input offset voltage can be plotted by selecting the correct node voltage, or the differential input voltage can be plotted be selecting V(11)-V(13). The supply currents are plotted by selecting I(VEE) or I(VCC) for the negative and positive power supplies respectively. The third program (see Figures 7 through 11) is named cfadc.cir, and it simulates the salient DC parameters for a non-inverting op amp. The program requires the user to supply the feedback resistance values, the load resistance, and the power supply voltage in volts. The program assumes that the op amp is run off two power supplies of equal and opposite polarity each of which is referenced to ground, so it applies the entered voltage to the op amp as a positive and negative supply with equal magnitudes. If a single supply op amp needs to be evaluated just enter half the power supply voltage, and the analysis will be equivalent. INPUT CURRENT (µA) 10 RF 11 - OUT 13 + 4 3 RI IN VIN 5 I(RF) - I(RI) I(RF) 0 I(RI) -5 RL VEE VCC -10 -10 FIGURE 7. NON-INVERTING OP AMP SCHEMATIC (DC) 0 VIN (mV) 5 10 FIGURE 8. NON-INVERTING OP AMP INPUT CURRENT PLOT 10 10 5 SUPPLY CURRENT (mA) INPUT VOLTAGE (mV) -5 V(11) V(13) 0 V(11) - V(13) -5 I(VEE) 5 0 -5 I(VCC) -10 -10 -10 -5 0 VIN (mV) 5 FIGURE 9. NON-INVERTING OP AMP INPUT OFFSET VOLTAGE PLOT 5 10 -10 -5 0 VIN (mV) 5 FIGURE 10. NON-INVERTING OP AMP POWER SUPPLY CURRENT PLOT 10 Application Note 9523 *This program simulates the salient DC parameters for a non-inverting op amp. *The user must specify the feedback and load resistance with a .param *statement. The power supply voltage is set by the parameter “vsupply”. Input *currents are measured as IRI, the non-inverting input current, IRF, the *inverting input current, and (IRF-IRI) the offset current. The offset *voltage is calculated with the equation Vos=v(13)-v(11). The power supply *currents can be determined by looking at the parameter ICC and IEE. The *model is entered with a .lib statement. The model in the subcircuit call (x statement) *must correspond to the model called in the .lib statement. .param RF = 1000 .param RL = 400 .param vsupply = 5 .lib b:ha502x.cir x1 13 11 3 4 out 3 ha502x VIN in 0 RF1 11 out {RF} RI in 13 {RF} RL1 0 out {RL} VCC 3 0 {vsupply} VEE 4 0 {-1*vsupply} .dc VIN -.1 .1 .001 .probe .end FIGURE 11. NON-INVERTING OP AMP DC TRANSFER FUNCTION PROGRAM CMRR For A Non-Inverting Op Amp The fourth program (see Figures 12, 13, and 14) is called cfacmrr.cir, and it simulates the common mode rejection ratio for a non-inverting op amp. The program uses two identical non-inverting op amps to implement the equation CMRR = change in input offset voltage divided by the common mode input voltage change. The CMRR equation is shown in circuit parameters in Figure 12. Referring to Figure 13, it is seen that the common mode difference voltage is 2V. The input is a square wave so the measurement should be made in an area which has settled out. Notice that the worse common-mode input voltage is negative so that value of -1.3mV was used in the calculation. Using a square wave rather than a DC signal enables the inspection of both quadrants prior to calculating the CMRR. The program requires the user to supply the feedback resistance value, the load resistance, and the power supply voltage in volts. RF1 = 1K RF2 = 1K VCC 11 - + IN1 3 21 OUT1 4 VEE VIN1 VCC RL1 IN2 VIN2 - + 3 OUT2 4 VEE RL2 ∆V IO ( V ( IN1 ) – V ( 11 ) ) – ( V ( IN2 ) – V ( 21 ) ) CMMR = DB -------------------------------------------------------------------------------------------------- = DB ---------------( V ( IN2 ) – V ( IN1 ) ) ∆V CM FIGURE 12. SCHEMATIC AND EQUATION FOR COMMON-MODE REJECTION CIRCUIT 6 Application Note 9523 *This program simulates the common-mode rejection ratio for a non-inverting op *amp. The equation recommended for the calculation is *CMRR=dB((V(IN1)-V(11)) -(V(IN2)-V(21)))/(V(IN2)-(VIN1)) and this program uses *two identical op amps to obtain the calculation data. The user *must specify the feedback resistance, RF, and the load resistance, RL. *The power supply voltage is set by the parameter “vsupply”. The op amp model *is entered with a .lib statement. The model in the subcircuit call (x statement) must *correspond to the model in the .lib statement (2 times). .param RF=1K .param RL=400 .param vsupply=5 .lib b:ha502x.cir x1 IN1 11 3 4 OUT1 3 ha502x x2 IN2 21 3 4 OUT2 3 ha502x VIN1 IN1 0 pulse (0 1m .1ns .1ns .1ns 100ns 200ns) VIN2 IN2 0 pulse (0 2.001 .1ns .1ns.1ns 100ns 200ns) RF1 11 OUT1 {RF} RF2 21 OUT2 {RF} RI1 OUT1 0 {RL} RL2 OUT2 0 {RL} VCC 3 0 {vsupply} VEE 4 0 {-1*vsupply .tran 20ns 420ns .probe .end FIGURE 13. COMMON-MODE REJECTION PROGRAM 3.0 OUTPUT VOLTAGE (mV) 2.0 1.0 0 -1.0 -1.3mV -2.0 –3 – 1.3 CMMR = 20LOG --------------10 = 63.7dB 2 -3.0 0 50 100 TIME (ns) 150 200 FIGURE 14. PLOT OF COMMON-MODE REJECTION PROGRAM OUTPUT 7 Application Note 9523 Transient Response For A Non-Inverting Op Amp The fifth program (see Figures 15, 16, and 17) is called cfatran.cir, and it simulates the transient response for a noninverting op amp. If the input signal is small, about 100mv as shown in Figure 16, the analysis will be small signal. Larger input signals will yield a large signal analysis. The program requires the user to supply the feedback resistance, the load resistance, the input resistance, and the power supply voltage in volts. The program assumes that the op amp is run off two power supplies of equal and opposite polarity each of which is referenced to ground, so it applies the entered voltage to the op amp as a positive and negative supply with equal magnitudes. If a single supply op amp needs to be evaluated just enter half the power supply voltage, and the analysis will be equivalent. The rise time, fall time, slew rate, and delay time can be read off the plot shown in Figure 16. 120 80 VCC IN 11 RI1 50 40 + OUT - VEE RF1 RL1 0 VIN -40 1K VOUT -80 -120 0 50 100 150 TIME (ns) FIGURE 15. TRANSIENT RESPONSE CIRCUIT SCHEMATIC FIGURE 16. TRANSIENT RESPONSE PROGRAM OUTPUT PLOT *This program simulates the time domain response for a non-inverting op amp. *The response will be small signal or large signal depending on the amplitude *of the input signal. The user must specify the load resistance, RL, the *feedback resistance, RF, and the input resistance, RI. The power supply *voltage is set by the parameter “vsupply”. The op amp model is entered with *a .lib statement. The model in the x statement must correspond to the *model called in the .lib statement. .param RI = 50 .param RF = 1K .param RL = 100 .param vsupply=5 .lib b:ha502x.cir x1 in 11 3 4 out 3 ha502x VIN1 in 0 PULSE (-.1 .1 .1ns .1ns .1ns 100ns 200ns) RF1 11 out {RF} RL1 out 0 {RL} RI1 0 in {RI} VCC 3 0 {vsupply} VEE 4 0 [-1*vsupply} .tran 20ns 420ns .probe .end FIGURE 17. TRANSIENT RESPONSE PROGRAM 8 Application Note 9523 Enable Response For A Non-Inverting Op Amp prior to modification. The program requires the user to supply the feedback resistance, the load resistance, the input resistance, and the power supply voltage in volts. The program assumes that the op amp is run off two power supplies of equal and opposite polarity each of which is referenced to ground, so it applies the entered voltage to the op amp as a positive and negative supply with equal magnitudes. If a single supply op amp needs to be evaluated just enter half the power supply voltage, and the analysis will be equivalent. The enable response times can be read off of Figure 19. The Sixth program (see Figures 18, 19, and 20) is called cfaenabl.cir, and it computes the response of a non-inverting op amp to an enable control signal. A 2VDC excitation is applied to the positive op amp input, and a square wave is applied to the enable input. The enable signal swings from ground to the positive power supply rail, thus simulating an open collector driver. This signal can be modified as required, but it must be disconnected from the supply voltage by removing the {vsupply} term 5.0 4.0 RF (V) 3.0 VCC - 3 + ENA 4 11 IN VIN RI 2.0 OUT ENABLE/ DISABLE 1.0 RL VEE 0 0 FIGURE 18. ENABLE RESPONSE CIRCUIT SCHEMATIC 0.5 1.0 1.5 3.0 3.5 4.0 4.5 FIGURE 19. ENABLE RESPONSE PROGRAM OUTPUT PLOT *This program simulates the time response of the enable/disable function of *the non-inverting op amp. The response is obtained with a 2VDC *input signal, while the voltage on the enable pin swings from VCC to ground. *The user must specify the load resistance, RL, the feedback resistance, RF, *and the input resistance, RI. The power supply voltage is set by a parameter *“vsupply”. The op amp model is entered with a .lib statement. The model in the subcircuit call *(x statement) must correspond to the model called in the .lib statement. If *the rise and fall times of the enable signal are too fast the program may *not converge (10ns is optimal). .param RI = 50 .param RF = 1K .param RL = 100 .param vsupply=5 .lib b:ha502x.cir x1 in 11 3 4 out ena ha502x VIN in 0 2 vena ena 0 PULSE ({vsupply} 0 0ns 10ns 10ns 2000ns 4000ns) RF1 11 out {RF} RL1 out 0 {RL} RI1 0 in {RI} VCC 3 0 {vsupply} VEE 4 0 {-1*vsupply} .tran 20ns 4020ns .probe .end FIGURE 20. ENABLE RESPONSE PROGRAM 9 2.0 2.5 TIME (µs) Application Note 9523 Summary Six programs which compute and plot the response of an op amp are described here. These programs make it easy to complete the SPICE analysis, and by using them the results may be compared to new Intersil Corporation data sheets. This approach gives the minimum data required to evaluate the model, but the actual evaluation must be made by the design engineer. If the model indicates 2dB peaking at the -3dB point while the data sheet shows 3dB peaking at the same point, which does the engineer believe? Since the model is an approximation, the data sheet should be more correct than the model, but because the data sheet is based on “typicals” their curves are sometimes hard to reproduce. This vague area must be resolved through the experience of the design engineer coupled with laboratory data. There is no substitute for accurate measurements! When the data sheet curves, the model curves and the lab curves all fit within a reasonable tolerance the design engineer can begin to trust the models. Keep testing though, because models have been known to be very unpredictable. Any comments, deletions or additions should be directed to one of the authors for inclusion in revisions on this document. All Intersil semiconductor products are manufactured, assembled and tested under ISO9000 quality systems certification. Intersil semiconductor products are sold by description only. Intersil Corporation reserves the right to make changes in circuit design and/or specifications at any time without notice. Accordingly, the reader is cautioned to verify that data sheets are current before placing orders. Information furnished by Intersil is believed to be accurate and reliable. However, no responsibility is assumed by Intersil or its subsidiaries for its use; nor for any infringements of patents or other rights of third parties which may result from its use. No license is granted by implication or otherwise under any patent or patent rights of Intersil or its subsidiaries. For information regarding Intersil Corporation and its products, see web site www.intersil.com Sales Office Headquarters NORTH AMERICA Intersil Corporation P. O. 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