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 (VOH, generally close to positive supply voltage, VPOS) and a low level (VOL, generally close to negative supply voltage, VNEG). 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 (VPL) and high (VPH) 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 (VPL and VPH). 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:

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 
 0.00  00 
Enter low and high level output voltages  0.3 ENTER 5 
 300. -03 
Enter C1 (Farads)  0.022 EEx CHS 6 
 22.0 -09 
Enter R1 (Ohms)  22 EEx 3 
 22.0  03 
Enter desired triangle lower and upper voltage peaks  1.2 ENTER 1.4 
 1.20  00 
Enter desired frequency (Hz)  2500 
 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 


Specifying VOL and VOH 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 VOL+(VOHVOL)/3 to VOH-(VOHVOL)/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 VPL and VPH close to the bottom of the op-amp output range (VOL). Likewise for a high duty cycle, choose VPL and VPH close to the top of the range (VOH).

For the most stable oscillation frequency, choose VPL and VPH far away from the op-amp output voltage limits, VOL and VOH (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 VPL and VPH to span the middle of the VOL to VOH range (which is the default if VPL and VPH aren’t specified).

Reverse Solution: Finding VPL, VPH, 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, VPL, and VPH?

These are the steps to find out:

Description Keystrokes Display
Enter new value for R2  7.5 EEx 3 
 7.50 03 
Enter new value for R3  120 EEx 3 
 120. 03 
Enter new value for R4  68 EEx 3 
 68.0 03 
Compute resulting value for VPL   f   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 
-12.0 00 
Enter low and high level output voltages  11.3 CHS ENTER 11.3 
-11.3 00 
Enter R1  10 EEx 3 
  10.0 03 
Enter desired lower and upper switching points  3 CHS ENTER 2 
-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 
  9.10 03 
Enter new value for R3  16 EEx 3 
  16.0 03 
Compute resulting lower switching point   f  -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 VOL and VOH 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  ÷  (VOHVOL)/3
014  RCL 7   
015  x↔y   
016  +   
017  STO A  Set VPL = VOL + (VOHVOL)/3
018  RCL 8   
019  LSTx   
020  −   
021  STO B  Set VPH = VOH – (VOHVOL)/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 (VOLVPLVOH + VPH) * X + VOH * VPLVPH * VOL
183  RCL 7   
184  RCL A   
185  −   
186  RCL 8   
187  −   
188  RCL B   
189  +   
190  ×  Multiply (VOLVPLVOH + 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 VPL and VPH. Changing R1 no longer forces VPL and VPH 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 VPL and VPH were accidentally interchanged. Bug was in web-site listing only, not in original program.

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1 Comment

  1. Willy Kunz
    May 16, 2013

    The program as shown in the listing gives wrong answers for duty cycle and resistor R4. To fix it:
    Step 109: change RCL A to RCL B
    Step 112: change RCL B to RCL A
    Step 116: insert x↔y after LN
    With these changes, the duty cycle is shown as 212.-03 which is equivalent to 788.-03 (=1-0.212).

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