Op-Amp Gain and Offset Design with the HP-41C Programmable Calculator

May 26, 2009

This is a program I originally wrote for the HP-67 calculator, and then ported to the HP-41C series. This program is for designing offset-and-gain stages using a single operational amplifier. Such stages are often necessary to convert an input signal covering one range of voltages (e.g., 0.1V to 0.2V from a sensor) to an output signal covering a different range (e.g., 1.0 to 4.0V into an A/D converter).

Mathematically, such a stage performs a linear transformation on the input voltage,

V_OUT = m V_IN + b

where m is the slope and b is the intercept or offset.

Given an input voltage range, V_IL to V_IH, and an output voltage range, V_OL to V_OH, the slope and offset are given by:

There are four main cases to consider, with different circuits for each, that are addressed by this program:

1. Positive slope and offset (m > 0 and b > 0)
2. Positive slope and negative offset (m > 0 and b < 0)
3. Negative slope and positive offset (m < 0 and b > 0)
4. Negative slope and offset (m < 0 and b < 0)

Positive Slope and Offset

A positive slope and offset stage is implemented by the following circuit:

The designer must select values for R₁ and R_F, and then appropriate values for R₂ and R_G can be calculated using the following formulae:

After determining theoretically ideal values for R₂ and R_G, real-world values can be chosen and the following formula applied to V_IL and V_IH to see the resulting values of V_OL and V_OH respectively:

This case is also used to handle the special case where b = 0 (postive gain with no offset). In such a case, the formula for R₂ would result in dividing by zero, which means R₂ is infinite. In other words, R₂ and V_REF are not needed. The value of R₁ won’t matter then either, and can be replaced by a direct connection. The circuit reduces to:

Positive Slope and Negative Offset

The following circuit implements a positive slope and negative offset stage:

After choosing values for R₁ and R_F, values for R₂ and R_G can be be calculated using these formulae:

The following formula can then be used to determine the effect of real-world values of R₂ and R_G on the transformation:

Negative Slope and Positive Offset

A negative slope and postive offset stage is implemented by the following circuit:

Given values for R₁ and R_F, values for R₂ and R_G can be be calculated as follows:

The following formula can then be used to determine the effect of substituting real-world values of R₂ and R_G:

This case is also used to handle the special case where b = 0 (negative gain with no offset). As in case 1, the formula for R₂ would involve division by zero, so R₂ and V_REF are not needed. The non-inverting input of the op-amp can be connected directly to ground, giving the following circuit:

Negative Slope and Offset

The following circuit implements a negative slope and offset stage:

This circuit has no R₁, so it is only necessary to choose a value for R_F, after which R₂ and R_G are given by:

The effect of using real-world values of R₂ and R_G can then be tested using this formula:

Limitations

Theoretically, the formulae presented here work perfectly well for gains between -1 and 1 (i.e. |m| < 1). However, many real-world op-amps are unstable in such cases. Instead, it will usually be necessary to design for a higher gain (|m| ≥ 1), with an attenuator on the input side. The design procedures for this are described on Texas Instruments’ Op Amp Gain and Offset Page.

Using the Program

First type in the program and save it, or read it from a previously recorded magnetic card. The card should be labelled as follows:

OP-AMP GAIN AND OFFSET STAGE DESIGN
VREF	VIL,VIH	VOL,VOH
→m,b	→CASE	R1,RF→R2,RG	R2,RG→VOL,VOH

Forward Solution: Finding R₂ and R_G

Consider the following example. A sensor has an output ranging from 0.5V to 0.7V, and we want to interface it to an A/D converter that is expecting an input between 1V and 4V. There is no reference voltage available other than the well regulated 5V supply voltage of the circuit. Use a 10kΩ resistor for R₁, and 100kΩ for R_F.

Follow these steps to solve the problem:

Description	Keystrokes	Display
Use engineering notation	ENG 2	0.00  00
Enter VREF	5       a	5.00  00
Enter VIL and VIH	0.5 ENTER 0.7       b	500. -03
Enter VOL and VOH	1 ENTER 4       c	1.00  00
Compute slope m	A	15.0  00
Compute offset b	R/S	-6.50  00
Determine case number	B	2.00  00
Enter R1 and RF, compute R2	10 EEx 3 ENTER 100 EEx 3   C	1.02  03
Compute RG	R/S	6.21  03

Notes

It is not necessary to compute the slope and offset (by pressing A) before determining the case number. Likewise, it isn’t necessary to determine the case number (by pressing B) before computing resistor values (although you’ll want to know the case number in order to know which circuit to build). The program keeps track of which information is up to date, and will (re)compute anything that it needs that hasn’t already been computed.

Reverse Solution: Finding the Effect of R₂ and R_G on V_OL and V_OH

The closest available 5% resistor values for R₂ and R_G are 1kΩ and 6.2kΩ respectively. What effect does using these have on the solution? Follow these steps to find out:

Description	Keystrokes	Display
Enter R2 and RG, compute VOL	1 EEx 3 ENTER 6.2 EEx 3   D	1.13  00
Compute VOH	R/S	4.15  00

This is within the A/D’s input range at the lower bound, but outside the range at the upper bound. What happens if we use the next available value for R_G, 6.8kΩ, instead?

Description	Keystrokes	Display
Enter R2 and RG, compute VOL	1 EEx 3 ENTER 6.8 EEx 3   D	1.09  00
Compute VOH	R/S	3.88  00

This is almost centered within the desired output range, and covers 93% of it.

The only remaining concern is how component tolerances might affect the solution. This can be analyzed by trying different combinations of R₁, R₂, R_F, and R_G representing resistors that are maximally out of tolerance (±5%) in each direction.

For example, to test the case where R₁ and R_F are 5% low and R₂ and R_G are 5% high, follow these steps:

Description	Keystrokes	Display
Enter low R1 and RF, ignore computed R2	0.95 EEx 3 ENTER 95 EEx 3   C	920.  00
Enter high R2 and RG, compute VOL	1.05 EEx 3 ENTER 7.14 EEx 3   D	808. -03
Compute VOH	R/S	3.48  00

The results show that in this case, the lower limit of the output is out of range, meaning either a redesign is necessary, or tighter tolerance components are needed.

Cases where b = 0

In cases where the offset, b, is zero, the program will instead use b = 10^-9 as the offset. This will avoid any division-by-zero errors. The program will then use case 1 (if m > 0) or case 3 (if m < 0) to compute the solution. In both cases, the computed value for R₂ will be very large, typically around 10⁹ times the value specified for R₁. This indicates that R₂ and V_REF can be omitted, and that R₁ can be replaced by a direct connection.

Program Listing

Line	Instruction	Comments
01♦	LBL “GOFF”
02♦	LBL a	Enter available reference voltage
03	STO 09
04	RTN
05♦	LBL b	Enter VIL and VIH
06	STO 06	Store VIH
07	x↔y
08	STO 05	Store VIL
09	CF 00	Must recompute m and b
10	RTN
11♦	LBL c	Enter VOL and VOH
12	STO 08	Store VOH
13	x↔y
14	STO 07	Store VOL
15	CF 00	Must recompute m and b
16	RTN
17♦	LBL A	Compute transfer function slope (m) and intercept (b)
18	FS? 00	Are m and b already up to date?
19	GTO 05
20	RCL 08	Compute and store m = (VOH – VOL) ÷ (VIH – VIL)
21	RCL 07
22	−
23	RCL 06
24	RCL 05
25	−
26	÷
27	STO 20
28	RCL 05	Compute and store b = VOH – m VIL
29	×
30	RCL 07
31	−
32	CHS
33	x≠0?
34	GTO 07
35	1 E-9	Make sure b is never exactly 0;
36♦	LBL 07
37	STO 21
38	SF 00	Stored m and b are now up to date
39	CF 01	Must recompute case number now
40♦	LBL 05	Display computed or already up-to-date m and b
41	RCL 20
42	RTN	Return, leaving m on stack
43	RCL 21	Display b if user pressed R/S after return
44	RTN
45♦	LBL B	Compute case number and store in I
46	XEQ A	Make sure m and b are up to date
47	FS? 01	Is case number already up to date
48	GTO 08
49	x<0?
50	GTO 06	Handle cases where m < 0
51	1
52	STO 25
53	RCL 21
54	x<0?
55	ISZ 25	Case 2: m positive and b negative
56	GTO 08	Case 1: m positive and b positive
57♦	LBL 06	Cases where m is negative
58	3
59	STO 25
60	RCL 21
61	x<0?
62	ISZ 25	Case 4: m negative and b negative
63♦	LBL 08	Case 3: m negative and b positive
64	SF 01	Case number is now up to date (or was already up to date)
65	RCL 25	Display case number and return
66	RTN
67♦	LBL C	Compute solution using appropriate case
68	STO 03	Store RF
69	x↔y
70	STO 01	Store R1
71	XEQ B	Get case number
72	GTO IND 25	Branch to appropriate case
73♦	LBL 01	Positive m and positive b
74	RCL 01	Compute and store R2 = VREF R1 m ÷ b
75	RCL 09
76	×
77	RCL 20
78	×
79	RCL 21
80	÷
81	STO 02
82	R/S	Display R2; 9.99e99 means “open circuit”
83	RCL 09	Compute and store RG = VREF RF ÷ (VREF (m – 1) + b)
84	RCL 03
85	×
86	XEQ 09	Compute VREF (m – 1) + b
87	GTO 05	Divide, store RG, and return, leaving RG on stack
88♦	LBL 02	Positive m and negative b
89	RCL 01	Compute and store R2 = –R1 b ÷ (VREF (m – 1) + b)
90	RCL 21
91	CHS
92	×
93	XEQ 09	Compute VREF (m – 1) + b, leaving VREF (m – 1) in register 0
94	÷
95	STO 02
96	R/S	Display R2
97	RCL 01	Compute and store RG = (R1 b + VREF RF) ÷ (VREF (m – 1))
98	RCL 21
99	×
100	RCL 09
101	RCL 03
102	×
103	+
104	RCL 00	Subroutine 9 left VREF (m – 1) in register 0 for us
105	GTO 05	Divide, store RG, and return, leaving RG on stack
106♦	LBL 03	Negative m and positive b
107	RCL 01	Compute and store R2 = R1 (VREF (m – 1) + b) ÷ –b
108	XEQ 09	Compute VREF (m – 1) + b
109	×
110	RCL 21
111	CHS
112	÷
113	STO 02
114	R/S	Display R2
115	RCL 03	Compute and store RG = –RF ÷ m
116	CHS
117	RCL 20
118	GTO 05	Divide, store RG, and return, leaving RG on stack
119♦	LBL 04	Negative m and negative b
120	RCL 09	Compute and store R2 = VREF RF ÷ –b
121	RCL 03
122	×
123	RCL 21
124	CHS
125	÷
126	STO 02
127	R/S	Display R2
128	RCL 03	Compute and store RG = RF ÷ m
129	RCL 20
130	CHS
131♦	LBL 05	Common code to compute y ÷ x and store in RG
132	÷
133	STO 04
134	RTN	Return, leaving RG on stack
135♦	LBL 09	Compute VREF (m – 1) + b, leaving VREF (m – 1) in register 0
136	RCL 20
137	1
138	−
139	RCL 09
140	×
141	STO 00	Save VREF (m – 1) in register 0 for later
142	RCL 21
143	+
144	RTN
145♦	LBL D	Compute VOL,VOH using supplied R2,RG
146	STO 04	Store RG
147	x↔y
148	STO 02	Store R2
149	XEQ B	Get case number
150	RCL 05
151	XEQ IND 25	Compute VOL from VIL
152	R/S	Display computed VOL
153	RCL 06
154	GTO IND 25	Compute VOH from VIH
155♦	LBL 01	Compute VOUT for positive m and positive b
156	RCL 02
157	×
158♦	LBL 00	Entry point to compute (x + VREF R1) (1 + RF ÷ RG) ÷ (R1 + R2)
159	RCL 09
160	RCL 01
161	×
162	+
163	RCL 03
164	RCL 04
165	÷
166	1
167	+
168	×
169	RCL 01
170	RCL 02
171	+
172	÷
173	RTN
174♦	LBL 02	Compute VOUT for positive m and negative b
175	RCL 01
176	1/x
177	RCL 02
178	1/x
179	+
180	1/x
181	RCL 04
182	+
183	STO 00	Store RG + R1 R2 ÷ (R1 + R2) for later
184	RCL 03
185	+
186	×
187	RCL 00
188	÷
189	RCL 09
190	RCL 02
191	×
192	RCL 03
193	×
194	RCL 01
195	RCL 02
196	+
197	RCL 00
198	×
199	÷
200	−
201	RTN
202♦	LBL 03	Compute VOUT for negative m and positive b
203	0
204	XEQ 00	Compute (VREF R1) (1 + RF ÷ RG) ÷ (R1 + R2)
205	GTO 09	Subtract VIN RF ÷ RG
206♦	LBL 04	Compute VOUT for negative m and negative b
207	RCL 09
208	RCL 03
209	×
210	RCL 02
211	÷
212	CHS
213♦	LBL 09	Entry point to compute x – y RF ÷ RG
214	x↔y
215	RCL 03
216	×
217	RCL 04
218	÷
219	−
220	RTN

Two interesting aspects of this program are its use of indirect addressing for branching to the forward and reverse solution subroutines for each of the four cases, and its use of repeated labels.

The forward solution for each of the four slope/offset cases is implemented by a sequence of instructions starting with the label corresponding to the case number (1 to 4). When the user presses C, the case number is computed if necessary, and a GTO (i) instruction then branches to the appropriate case.

Similarly, the reverse solution for each case is also labeled according to the case number. When the user presses D, V_IL is recalled to the stack, after which a GSB (i) instruction causes V_OL to be calculated. Then V_IH is recalled, and GTO (i) is used to calculate V_OH and then return.

So, there are two each of LBL 1 through LBL 4. This works because the HP-41C/CV/CX (and other vintage HP calculators) search forwards from the current step for the matching label. Thus the GTO (i) in step 73 will branch to the appropriate forward solution, and the GSB (i) in step 152 and GTO (i) in step 155 will branch to the appropriate reverse solution.

This program also makes use of many small subroutines to compute sub-expressions common to multiple solutions. Since there were not enough labels for all the subroutines needed, the same labels were used more than once. Had this not been done, the same sequence of steps would have been repeated several times, and the program would not have fit into the calculator’s 224 step memory.

Registers and Flags

Register	Use
00	Temporary register
01	R1 (Ohms)
02	R2 (Ohms)
03	RF (Ohms)
04	RG (Ohms)
05,06	VIL and VIH, input voltage range
07,08	VOL and VOH, output voltage range
09	VREF
11	m, slope of transfer function
12	b, intercept of transfer function
10	Case number (1,2,3,4)

Flag	Meaning
00	m and b are up to date
01	Case number is up to date

Revision History

2009-May-26 — Initial release.

                    

Related Articles

If you've found this article useful, you may also be interested in:

• Op-Amp Oscillator Design with the HP-41C Programmable Calculator
• Low-Sensitivity Sallen-Key Filter Design with the HP-41C Programmable Calculator
• Low-Sensitivity Sallen-Key Filter Design with the HP-67 Programmable Calculator
• Op-Amp Gain and Offset Design with the HP-67 Programmable Calculator
• Op-Amp Oscillator Design with the HP-67 Programmable Calculator
• Resistor Network Solver for the HP-67 Programmable Calculator
• A Matrix Multi-Tool for the HP 35s Programmable Calculator
• Curve Fitting for the HP 35s Programmable Calculator