Op-Amp Oscillator Design with the HP-67 Programmable Calculator

November 26, 2008

This is my second HP-67 program, which I wrote to help me with the design of a project that is currently (Nov. 2008) in the planning stages. This program selects component values for an op-amp based relaxation oscillator, given the desired frequency and output wave form peak voltages. It can also solve the inverse problem, finding the frequency and voltages resulting from given component values. The following is the schematic for such an oscillator:

R₁ and R₂ form a voltage divider, with an additional input from the op-amp output through R₃. When the op-amp output is at a high-level, the voltage at the non-inverting input of the op-amp is higher than when the op-amp output is at a low level. When the output is high, capacitor C₁ also charges through R₄ until the voltage across it (which is applied to the op-amp’s inverting input) reaches the voltage at the non-inverting input. At that time, the op-amp output goes low, and the capacitor begins to discharge through R₄ until the voltage once again reaches the (now lower) voltage at the non-inverting input.

If one were to monitor the op-amp output, it would alternate between a high level (V_OH, generally close to positive supply voltage, V_POS) and a low level (V_OL, generally close to negative supply voltage, V_NEG). The duty cycle of this square wave depends on the relative time it takes to charge and discharge C₁ through R₄, which in turn depends on the low (V_PL) and high (V_PH) peak voltages that C₁ cycles between (which in turn depend on R₁, R₂, and R₃). Monitoring the voltage across C₁ shows a triangle wave.

With this program, you can select components for such an oscillator to achieve a desired frequency, and if it matters to your design, desired low and high triangle peaks (V_PL and V_PH). After the program computes the required component values, you can modify these values to match those actually available in the real world. The program will then compute what effect these changes have on the frequency and triangle peak voltages.

The following equations describe the operation of the oscillator:

You will first need to choose values for C₁ and R₁ arbitrarily, since for any desired frequency and given C₁ and R₁, it will be possible to find (possibly impractical) values for R₂, R₃, and R₄. A good choice for R₁ is generally somewhere around 10kΩ to 100kΩ. The choice of C₁ depends on the frequency, and a readily available value near (50/f) μF is usually suitable.

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 OSCILLATOR DESIGN
VNEG,VPOS	VOL,VOH	C1	R1	VPL,VPH
f	→%DC	→R2→	→R3→	→R4→

Forward Solution: Finding R₂, R₃, and R₄

Consider the following example: It is desired to find values for R₂, R₃, and R₄ to produce an oscillator of about 2500Hz, with a triangle waveform that oscillates between 1.2V and 1.4V. The op-amp is to operate from a single 5V supply, and the op-amp’s output is capable of a low of 0.3V and a high of 5V. Use a 0.022μF capacitor for C₁, and a 22kΩ resistor for R₁.

Follow these steps to solve the problem:

Description	Keystrokes	Display
Select engineering notation	h  ENG   DSP 2	0.00  00
Enter power supply voltages	0 ENTER 5    f  a	0.00  00
Enter low and high level output voltages	0.3 ENTER 5    f  b	300. -03
Enter C1 (Farads)	0.022 EEx CHS 6    f  c	22.0 -09
Enter R1 (Ohms)	22 EEx 3    f  d	22.0  03
Enter desired triangle lower and upper voltage peaks	1.2 ENTER 1.4    f  e	1.20  00
Enter desired frequency (Hz)	2500   A	2.50  03
Compute square wave duty cycle	B	788. -03
Compute value of R2 (Ohms)	C	7.26  03
Compute value of R3 (Ohms)	D	123.  03
Compute value of R4 (Ohms)	E	71.4  03

Notes

Specifying V_OL and V_OH is optional. If this step is omitted, the program will assume the op-amp output can span the entire negative and positive supply voltage range.

If the triangle wave form peak voltages don’t matter to your design (because you’re only using the square wave output), you don’t need to specify them. The program will assume peak voltages ranging from V_OL+(V_OH–V_OL)/3 to V_OH-(V_OH–V_OL)/3, which is the middle third of the op-amp output voltage range. This also happens to result in a 50% duty cycle.

To achive a low duty cycle square wave, choose V_PL and V_PH close to the bottom of the op-amp output range (V_OL). Likewise for a high duty cycle, choose V_PL and V_PH close to the top of the range (V_OH).

For the most stable oscillation frequency, choose V_PL and V_PH far away from the op-amp output voltage limits, V_OL and V_OH (to keep the triangle edges steep), and far away from each other (to keep any variations a small percentage of the overall voltage swing). Since these two goals are at odds with one another, a good compromise is to select V_PL and V_PH to span the middle of the V_OL to V_OH range (which is the default if V_PL and V_PH aren’t specified).

Reverse Solution: Finding V_PL, V_PH, Frequency, and Duty Cycle

After finding the above ideal solution, we’ll want to use real-world component values to build the physical circuit. The closest E-24 resistor values to those computed are: R₂ = 7.5kΩ, R₃ = 120kΩ, and R₄ = 68kΩ. What effect will using these values have on the frequency, V_PL, and V_PH?

These are the steps to find out:

Description	Keystrokes	Display
Enter new value for R2	7.5 EEx 3   C	7.50 03
Enter new value for R3	120 EEx 3   D	120. 03
Enter new value for R4	68 EEx 3   E	68.0 03
Compute resulting value for VPL	f  e	1.23 00
Compute resulting value for VPH	R/S	1.44 00
Compute resulting frequency	A	2.57 03

Other Uses for this Program

The oscillator design facilitated by this program consists of two parts, a comparator with hysteresis, and a capacitor being charged and discharged by the comparator output through a resistor. The equations describing the comparator aspect of the circuit are not affected by those describing the behaviour of the resistor-capacitor network, so the program can be used to design such comparators for other applications.

Here is a brief example of using this program to design a comparator: Assume we want to design a comparator operating from a +/-12V supply, whose output goes low when the voltage exceeds +2V, and goes high when the voltage subsequently drops below -3V. Assume the op amp used has an output that can swing to within 0.7V of the voltage limits. Use a 10kΩ resistor for R₁. Follow these steps to solve the problem:

Description	Keystrokes	Display
Enter power supply voltages	12 CHS ENTER 12    f  a	-12.0 00
Enter low and high level output voltages	11.3 CHS ENTER 11.3    f  b	-11.3 00
Enter R1	10 EEx 3    f  d	10.0 03
Enter desired lower and upper switching points	3 CHS ENTER 2    f  e	-3.00 00
Compute value of R2	C	8.98 03
Compute value of R3	D	16.7 03

Now select the closest real-world resistor values for R₂ and R₃ and determine how that affects the switching points:

Description	Keystrokes	Display
Enter new value for R2	9.1 EEx 3   C	9.10 03
Enter new value for R3	16 EEx 3   D	16.0 03
Compute resulting lower switching point	f  e	-3.03 00
Compute resulting upper switching point	R/S	2.16 00

Additional Real-World Considerations

The mathematical model used as the basis of this program assumes that V_OL and V_OH are constant, regardless of load. For sufficiently low current, this is close enough to true to be ignored. Thus it is important to use fairly high resistor values for R₃ and R₄ (10kΩ or bigger to be on the safe side). If the value of R₃ computed by the program is too low, start with a higher value for R₁. Similarly, if the value computed for R₄ is too low, use a lower value for C₁.

Some op-amps have an open-collector output. This means that when the output is low, it is pulled low through an output transistor, but when the output is high, it is simply floating. Thus, a pull-up resistor is needed to pull the output high. The chosen pull-up resistor must meet two requirements:

1. It must have a high-enough resistance that the output transistor can overcome the pull-up current when the output is low.

2. It must have a low-enough resistance that it is not so large a percentage of the resistance of R₃ or R₄ that it throws off the solution.

For the LM339 comparator that I often use in my designs, I’ve found that a 1kΩ resistor works well, together with R₃ and R₄ values about 100 times as much. As described above, use a higher value for R₁ to achieve a higher value for R₃, and use a lower value for C₁ to achieve a higher R₄.

Program Listing

Line	Instruction	Comments
001♦	LBL a	Store VNEG and VPOS
002	STO 6	VPOS
003	x↔y
004	STO 5	VNEG
005	x↔y	Fall through and initialize VOL and VOH to VNEG and VPOS
006♦	LBL b	Store VOL and VOH
007	CF 3
008	STO 8	VOH
009	x↔y
010	STO 7	VOL
011	−	Initialize VPL and VPH
012	3
013	÷	(VOH–VOL)/3
014	RCL 7
015	x↔y
016	+
017	STO A	Set VPL = VOL + (VOH–VOL)/3
018	RCL 8
019	LSTx
020	−
021	STO B	Set VPH = VOH – (VOH–VOL)/3
022	CF 1
023	RCL 8	Leave VOL and VPH on stack as feedback to user
024	RCL 7
025	RTN
026♦	LBL c	Store C1
027	CF 3
028	STO C
029	RTN
030♦	LBL d	Store R1
031	CF 3
032	STO 1
033	RTN
034♦	LBL e	Store or compute (if necessary) VPL and VPH
035	F? 3	If data entered, store new VPL and VPH
036	GTO 9
037♦	LBL 1	Otherwise, compute VPL and VPH if necessary
038	F? 1	Need to compute VPL and VPH?
039	GTO 8
040	RCL B	Recall already-up-to-date VPL and VPH
041	RCL A
042	RTN
043	RCL B	If user presses R/S after seeing VPL, display VPH
044	RTN
045♦	LBL 8	Recompute VPL and VPH
046	RCL 1
047	RCL 3
048	×
049	STO D
050	RCL 2
051	RCL 3
052	×
053	ST I
054	+
055	RCL 2
056	RCL 1
057	×
058	STO 9
059	+
060	1/x
061	STO E
062	RCL 5
063	×
064	RCL D
065	×
066	RCL E
067	RCL 6
068	×
069	RC I
070	×
071	+
072	STO D	Partial result common to VPL and VPH
073	RCL E
074	RCL 9
075	×
076	STO E	End of computation common to VPL and VPH
077	RCL 7	Compute VPL
078	×
079	+	End of computation of VPL
080	RCL E	Compute VPH
081	RCL 8
082	×
083	RCL D
084	+	End of computation of VPH; VPL is in Y-register
085♦	LBL 9	Store entered or computed VPH and VPL
086	STO B	Store VPH
087	x↔y
088	STO A	Store VPL
089	CF 1	VPL and VPH are now up to date
090	RTN
091	RCL B	If user presses R/S after seeing VPL, display VPH
092	RTN
093♦	LBL B	Compute duty cycle
094	CF 3
095	GSB 7	Get numerator and denominator (also used for computing R4 or f)
096	LSTx	Numerator
097	x↔y
098	÷
099	RTN
100♦	LBL 7	Compute denominator of duty cycle, leaving numerator in LSTx
101	GSB 1	Recompute VPL and VPH if necessary; leaves VPH in X, VPL in Y
102	RCL 8	Compute first half of denominator
103	−
104	RCL B
105	RCL 8
106	−
107	÷
108	LN
109	RCL B	Compute second half of denominator (which is also the numerator)
110	RCL 7
111	−
112	RCL A
113	RCL 7
114	−
115	÷
116	LN
117	+	Combine two halves, leaving numerator in LSTx
118	RTN
119♦	LBL A	Store or compute f
120	F? 3
121	GTO 0
122	GSB 7	Get denominator (also used for computing R4 and duty cycle)
123	RCL 4	Multiply by R4 and C1
124	×
125	RCL C
126	×
127	1/x
128♦	LBL 0	Store entered or computed f
129	STO 0
130	RTN
131♦	LBL C	Store or compute R2
132	F? 3
133	GTO 2
134	GSB 5	Compute numerator of R2
135	RCL 6	Compute denominator of R2
136	GSB 6
137	÷
138	STO 2	Store computed R2
139	RTN
140♦	LBL 2	Store R2 and invalidate VPL and VPH
141	STO 2
142	SF 1	Must recompute VPL and VPH for user-defined R2
143	RTN
144♦	LBL D	Store or compute R3
145	F? 3
146	GTO 3
147	GSB 5	Compute numerator of R3 (same as R2)
148	RCL 6	Compute denominator of R3
149	RCL 5
150	−
151	÷
152	RCL A
153	RCL B
154	−
155	÷
156	STO 3	Store computed R3
157	RTN
158♦	LBL 3	Store R3 and invalidate VPLh and VPH
159	STO 3
160	SF 1	Must recompute VPL and VPH for user-defined R3
161	RTN
162♦	LBL E	Store or compute R4
163	F? 3
164	GTO 4
165	GSB 7	Get denominator (also used for computing f or duty cycle)
166	RCL 0
167	×
168	RCL C
169	×
170	1/x
171♦	LBL 4	Store entered or computed R4
172	STO 4
173	RTN
174♦	LBL 5	Compute numerator common to R2 and R3
175	GSB 1	Recompute VPL and VPH if necessary
176	RCL 5
177	GSB 6
178	RCL 1
179	×
180	CHS
181	RTN
182♦	LBL 6	Compute (VOL – VPL – VOH + VPH) * X + VOH * VPL – VPH * VOL
183	RCL 7
184	RCL A
185	−
186	RCL 8
187	−
188	RCL B
189	+
190	×	Multiply (VOL – VPL – VOH + VPH) by VPOS or VNEG (now in Y)
191	RCL 8
192	RCL A
193	×
194	+
195	RCL B
196	RCL 7
197	×
198	−
199	RTN

Registers and Flags

Register	Use
0	Frequency (Hz)
1,2,3,4	Resistors R1, R2, R3, and R4 (Ohms)
5,6	VNEG and VPOS (Volts)
7,8	VOL and VOH (Volts)
A,B	VPL and VPH (Volts)
C	Capacitor C1 (Farads)
9,D,E,I	Temporary registers

Flag	Meaning
1	VPL and VPH need to be recomputed
3	User supplied input

Revision History

2008-Nov-26 — Initial release.

2009-May-25 — Fixed a bug that caused an error if one did not specify V_PL and V_PH. Changing R₁ no longer forces V_PL and V_PH to be recomputed.

2009-Jun-11 — Fixed a bug where the data entry flag sometimes wasn’t cleared even though there had been no data entry.

2015-Jun-23 — Fixed a bug in the subroutine for computing the denominator of the duty cycle, wherein V_PL and V_PH were accidentally interchanged. Bug was in web-site listing only, not in original program.

                    

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