Op-Amp Gain and Offset Design with the HP-67 Programmable Calculator

December 4, 2008

This is the third in an ongoing series of electronic design programs that I’m writing for the HP-67. 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	h  ENG   DSP 2	0.00  00
Enter VREF	5    f  a	5.00  00
Enter VIL and VIH	0.5 ENTER 0.7    f  b	500. -03
Enter VOL and VOH	1 ENTER 4    f  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
001♦	LBL a	Enter available reference voltage
002	STO 9
003	RTN
004♦	LBL b	Enter VIL and VIH
005	STO 6	Store VIH
006	x↔y
007	STO 5	Store VIL
008	CF 0	Must recompute m and b
009	RTN
010♦	LBL c	Enter VOL and VOH
011	STO 8	Store VOH
012	x↔y
013	STO 7	Store VOL
014	CF 0	Must recompute m and b
015	RTN
016♦	LBL A	Compute transfer function slope (m) and intercept (b)
017	F? 0	Are m and b already up to date?
018	GTO 5
019	RCL 8	Compute and store m = (VOH – VOL) ÷ (VIH – VIL)
020	RCL 7
021	−
022	RCL 6
023	RCL 5
024	−
025	÷
026	STO A
027	RCL 5	Compute and store b = VOH – m VIL
028	×
029	RCL 7
030	−
031	CHS
032	x≠0?
033	GTO 7
034	EEx	Make sure b is never exactly 0
035	CHS
036	9
037♦	LBL 7
038	STO B
039	SF 0	Stored m and b are now up to date
040	CF 1	Must recompute case number now
041♦	LBL 5	Display computed or already up-to-date m and b
042	RCL A
043	RTN	Return, leaving m on stack
044	RCL B	Display b if user pressed R/S after return
045	RTN
046♦	LBL B	Compute case number and store in I
047	GSB A	Make sure m and b are up to date
048	F? 1	Is case number already up to date
049	GTO 8
050	x<0?
051	GTO 6	Handle cases where m < 0
052	1
053	ST I
054	RCL B
055	x<0?
056	ISZ I	Case 2: m positive and b negative
057	GTO 8	Case 1: m positive and b positive
058♦	LBL 6	Cases where m is negative
059	3
060	ST I
061	RCL B
062	x<0?
063	ISZ I	Case 4: m negative and b negative
064♦	LBL 8	Case 3: m negative and b positive
065	SF 1	Case number is now up to date (or was already up to date)
066	RC I	Display case number and return
067	RTN
068♦	LBL C	Compute solution using appropriate case
069	STO 3	Store RF
070	x↔y
071	STO 1	Store R1
072	GSB B	Get case number
073	GTO (i)	Branch to appropriate case
074♦	LBL 1	Positive m and positive b
075	RCL 1	Compute and store R2 = VREF R1 m ÷ b
076	RCL 9
077	×
078	RCL A
079	×
080	RCL B
081	÷
082	STO 2
083	R/S	Display R2; 9.99e99 means “open circuit”
084	RCL 9	Compute and store RG = VREF RF ÷ (VREF (m – 1) + b)
085	RCL 3
086	×
087	GSB 9	Compute VREF (m – 1) + b
088	GTO 5	Divide, store RG, and return, leaving RG on stack
089♦	LBL 2	Positive m and negative b
090	RCL 1	Compute and store R2 = –R1 b ÷ (VREF (m – 1) + b)
091	RCL B
092	CHS
093	×
094	GSB 9	Compute VREF (m – 1) + b, leaving VREF (m – 1) in register 0
095	÷
096	STO 2
097	R/S	Display R2
098	RCL 1	Compute and store RG = (R1 b + VREF RF) ÷ (VREF (m – 1))
099	RCL B
100	×
101	RCL 9
102	RCL 3
103	×
104	+
105	RCL 0	Subroutine 9 left VREF (m – 1) in register 0 for us
106	GTO 5	Divide, store RG, and return, leaving RG on stack
107♦	LBL 3	Negative m and positive b
108	RCL 1	Compute and store R2 = R1 (VREF (m – 1) + b) ÷ –b
109	GSB 9	Compute VREF (m – 1) + b
110	×
111	RCL B
112	CHS
113	÷
114	STO 2
115	R/S	Display R2
116	RCL 3	Compute and store RG = –RF ÷ m
117	CHS
118	RCL A
119	GTO 5	Divide, store RG, and return, leaving RG on stack
120♦	LBL 4	Negative m and negative b
121	RCL 9	Compute and store R2 = VREF RF ÷ –b
122	RCL 3
123	×
124	RCL B
125	CHS
126	÷
127	STO 2
128	R/S	Display R2
129	RCL 3	Compute and store RG = RF ÷ m
130	RCL A
131	CHS
132♦	LBL 5	Common code to compute y ÷ x and store in RG
133	÷
134	STO 4
135	RTN	Return, leaving RG on stack
136♦	LBL 9	Compute VREF (m – 1) + b, leaving VREF (m – 1) in register 0
137	RCL A
138	1
139	−
140	RCL 9
141	×
142	STO 0	Save VREF (m – 1) in register 0 for later
143	RCL B
144	+
145	RTN
146♦	LBL D	Compute VOL,VOH using supplied R2,RG
147	STO 4	Store RG
148	x↔y
149	STO 2	Store R2
150	GSB B	Get case number
151	RCL 5
152	GSB (i)	Compute VOL from VIL
153	R/S	Display computed VOL
154	RCL 6
155	GTO (i)	Compute VOH from VIH
156♦	LBL 1	Compute VOUT for positive m and positive b
157	RCL 2
158	×
159♦	LBL 0	Entry point to compute (x + VREF R1) (1 + RF ÷ RG) ÷ (R1 + R2)
160	RCL 9
161	RCL 1
162	×
163	+
164	RCL 3
165	RCL 4
166	÷
167	1
168	+
169	×
170	RCL 1
171	RCL 2
172	+
173	÷
174	RTN
175♦	LBL 2	Compute VOUT for positive m and negative b
176	RCL 1
177	1/x
178	RCL 2
179	1/x
180	+
181	1/x
182	RCL 4
183	+
184	STO 0	Store RG + R1 R2 ÷ (R1 + R2) for later
185	RCL 3
186	+
187	×
188	RCL 0
189	÷
190	RCL 9
191	RCL 2
192	×
193	RCL 3
194	×
195	RCL 1
196	RCL 2
197	+
198	RCL 0
199	×
200	÷
201	−
202	RTN
203♦	LBL 3	Compute VOUT for negative m and positive b
204	0
205	GSB 0	Compute (VREF R1) (1 + RF ÷ RG) ÷ (R1 + R2)
206	GTO 9	Subtract VIN RF ÷ RG
207♦	LBL 4	Compute VOUT for negative m and negative b
208	RCL 9
209	RCL 3
210	×
211	RCL 2
212	÷
213	CHS
214♦	LBL 9	Entry point to compute x – y RF ÷ RG
215	x↔y
216	RCL 3
217	×
218	RCL 4
219	÷
220	−
221	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-67 (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
0	Temporary register
1	R1 (Ohms)
2	R2 (Ohms)
3	RF (Ohms)
4	RG (Ohms)
5,6	VIL and VIH, input voltage range
7,8	VOL and VOH, output voltage range
9	VREF
A	m, slope of transfer function
B	b, intercept of transfer function
I	Case number (1,2,3,4)

Flag	Meaning
0	m and b are up to date
1	Case number is up to date

Revision History

2008-Dec-04 — Initial release.

                    

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