# Resistor Network Solver for the HP-67 Programmable Calculator

November 18, 2008

My HP-67 magnetic card programmable calculator, manufactured in 1978.

I acquired my HP-67 at the beginning of 2008, and it sat on the shelf until November when I finally found the time to rebuild the battery pack and overhaul the card reader. I then set out to enter a program I’d written for the HP-67 earlier in the year (and debugged on an HP-41C, after hand-translating it).

I wrote this program because I often need to determine voltages at the nodes of simple resistor networks (for example, while designing my USB Powered AA Battery Charger). This program can find node voltages in networks of one or two nodes, with up to five resistors and/or constant current sources per node. The following is an example of such a network:

A simple network with one node, two resistors, two voltage sources, and a current source.

The program uses Kirchoff’s Current Law together with Ohm’s Law to compute the voltage at the node(s). Kirchoff’s law states that the sum of all the currents flowing into a node must equal zero. In other words, for every Amp of current that flows in, an Amp of current must flow out. Ohm’s Law states that the current flowing through a resistor equals the voltage across the resistor divided by the resistance.

For the single node case, finding the node voltage for a network with N resistors and M current sources is a matter of solving the following equation for Vnode:

Fortunately, the solution is rather simple, and is well suited to being computed by a programmable calculator such as the HP-67:

Solving a problem with two nodes is an iterative process of alternately solving two problems with one node each. First we solve the problem for one of the nodes, treating the the other node as a voltage source and guessing its voltage (any reasonable guess will do). This gives an estimate of the voltage of the first node. We can then solve the second node the same way, treating the first node as a voltage source having the voltage we just estimated, and giving an estimate for the voltage of the second node. Now we can go back and solve the first node again and we’ll get a better estimate of its voltage. We can continue like this, going back and forth, getting ever closer estimates of the node voltages. When the estimates stop changing at the level of precision we need, we’re done.

## 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:

RESISTOR NETWORKS OF 1 OR 2 NODES
DROP 1↔2 CLEAR iIi iPi
Vi,Ri Ii Vnode I1,I2,… P1,P2,…

### Single Node Problems

Consider a single node network consisting of two resistors and a constant current source as shown:

Solve the problem using the following steps:

Description Keystrokes Display
Select engineering notation   h  ENG
DSP 2
0.00  00
Clear any existing data   f   0.00  00
Enter V1 and R1  5 ENTER
56 EEX 3
A
1.00  00
Enter V2 and R2  0.3 ENTER
27 EEX 3
A
2.00  00
Enter I3  0.1 EEX 3 CHS
B
3.00  00
Compute Vnode (Volts)  C   3.65  00
Compute I1 (Amps)  D   24.1 -06
Compute I2  R/S  -124. -06
Compute (recall) I3  R/S   100. -06
Compute P1 (Watts)  E   32.5 -06
Compute P2  R/S   416. -06
Compute P3 (not applicable)  R/S   0.00  00
Look up I2 again  2
f
-124. -06
Look up P2 again  2
f
416. -06

Notice that the computed value of I2 is negative, meaning that current is flowing out of the node through R2.

Since I3 was entered as a fixed current as opposed to a voltage or resistor, computing I3 just returns the entered value, and computing P3 returns zero.

If you make a mistake when entering an input to a node (a voltage/resistance pair or a current), you can remove the most recently entered input using the DROP command (press f a).

### Two Node Problems

Consider the following two node network:

Each node has four connections. The leftmost node (Vnode1) is connected by resistors R1ab, R2a, R3a, and R4a to voltages Vnode2, V2a, V3a, and ground respectively.

The rightmost node also has four connections, three resistors and a constant current input of 0.1mA. The resistors R1ab, R2b, and R3b connect to voltages Vnode1, V2b, V3b respectively.

To solve this problem, first consider the leftmost node in isolation, and partially solve it using the following steps:

Description Keystrokes Display
Clear any existing data   f   0.00  00
Enter guess for Vnode2, and R1ab  0 ENTER
68 EEX 3
A
1.00  00
Enter V2a and R2a  5 ENTER
56 EEX 3
A
2.00  00
Enter V3a and R3a  0.3 ENTER
27 EEX 3
A
3.00  00
Enter V4a (ground = 0V) and R4a  0 ENTER
47 EEX 3
A
4.00  00
Estimate Vnode1 (Volts) assuming Vnode2 = 0  C   1.10  00

Now solve the rightmost node, using the estimated voltage of the left node:

Description Keystrokes Display
Switch to node 2. Display of 1.00 indicates that R1 is already entered.   f   1.00  00
Enter V2b and R2b  6 ENTER
33 EEX 3
A
2.00  00
Enter V3b and R3b  3 ENTER
22 EEX 3
A
3.00  00
Enter I4b  0.1 EEX 3 CHS
B
4.00  00
Estimate Vnode2 assuming Vnode1 = 1.10  C   4.80  00

Finally, alternate between the two nodes, recomputing Vnode until the values stop changing:

Description Keystrokes Display
Switch to node 1   f   4.00  00
Estimate Vnode1 (Volts) assuming Vnode2 = 4.80  C   1.88  00
Switch to node 2   f   4.00  00
Estimate Vnode2 assuming Vnode1 = 1.88  C   4.93  00
Switch to node 1   f   4.00  00
Estimate Vnode1 (Volts) assuming Vnode2 = 4.93  C   1.90  00
Switch to node 2   f   4.00  00
Estimate Vnode2 assuming Vnode1 = 1.90  C   4.93  00
Switch to node 1   f   4.00  00
Estimate Vnode1 (Volts) assuming Vnode2 = 4.93  C   1.90  00

The two Vnode values stop changing when Vnode1 = 1.90V and Vnode2 = 4.93V, so that is the final solution.

Just as with the single-node solution, you can view and review the currents and power dissipation through each resistor connected to the currently selected node.

## Program Listing

Line Instruction Comments
001♦   LBL a  Back up by one input
002  RC I
003  2
004  x>y?  Don’t back up if we’re already at the beginning
005  RTN
006  −
007  ST I
008  SF 1  Must recompute Vnode
009  GTO 1  Display number of inputs entered
010♦   LBL B  Enter a current as Vi=Ii, Ri=0
011  0  (0 Ii)
012♦   LBL A  Enter a voltage,resistance pair, with X=Ri, Y=Vi
013  RC I  Test that we’re not out of space
014  2
015  0
016  −
017  x=0?
018  1/x  Force an error if we’re out of space
019  R↓  (Ri Vi)
020  x↔y  (Vi Ri)
021  STO (i)  Store Vi in R0, R2, R4, …
022  ISZ I
023  x↔y  (Ri Vi)
024  STO (i)  Store Ri in R1, R3, R5, …
025  ISZ I
026  SF 1  Must recompute Vnode
027♦   LBL 1  Display number of inputs entered
028  RC I
029  2
030  ÷
031  RTN
032♦   LBL C  Compute (if necessary) and display Vnode
033  F? 1  Need to recompute?
034  GTO 5
035♦   LBL 3  Compute numerator / denominator
036  RCL B  Σ Vi/Ri + Σ Ii
037  RCL C  Σ 1/Ri
038  ÷  Vnode
039  STO D  Save for later use
040  CF 1  Record that we’ve computed it
041  RTN
042♦   LBL 5  Compute numerator and denominator
043  RC I
044  STO A  Save I for re-use
045  CLx  Clear numerator and denominator
046  STO B
047  STO C
048  ST I  Clear I (original value was saved in A)
049♦   LBL 2  Loop to sum numerator and denominator
050  RC I
051  RCL A
052  x=y?  Are we done?
053  GTO 3
054  RCL (i)  (Vi)
055  ISZ I
056  RCL (i)  (Ri Vi)
057  ISZ I
058  x≠0?  Is this a voltage,resistance pair?
059  GTO 4
060  R↓  X=0 means a current is in Y
061  RCL B  Add current to numerator
062  +  HP-67/97 can’t do STO+ B
063  STO B
064  GTO 2  Next input
065♦   LBL 4  Process voltage,resistance pair
066  ENTER  (Ri Ri Vi)
067  1/x  (1/Ri Ri Vi)
068  RCL C  Add 1/Ri to denominator
069  +
070  STO C
071  R↓  (Ri Vi)
072  ÷  (Vi/Ri)
073  RCL B  Add Vi/Ri to numerator
074  +
075  STO B
076  GTO 2  Next input
077♦   LBL D  Display I1IN
078  SF 0  Indicates we want Ii, not Pi
079  CLx
080  STO E  We’re using E as an index in this loop
081  GTO 0  Use same loop as subroutine E
082♦   LBL E  Display P1 .. PN
083  CF 0  Indicates we want Pi, not Ii
084  CLx
085  STO E
086  LBL 0  Entry point for subroutine E to call us
087  RCL E
088  RC I
089  x=y?  Exit if we’re done
090  GTO 1  Display number of inputs
091  F? 0  If called from subroutine D, set flag 2
092  SF 2
093  GSB 8  Compute Ii (or Pi) for input specified in register E
094  R/S  Stop to display result
095  RCL E  Point to next voltage,resistance pair
096  2
097  +
098  STO E
099  GTO 0  Process next pair
100♦   LBL d  Compute Ii for i=X
101  SF 2
102♦   LBL e  Compute Ii (if flag 2 set) or Pi for i=X
103  1
104  −  i-1
105  2
106  ×  2(i-1)
107  STO E  Set E to point to Vi,Ri pair
108♦   LBL 8  Compute Ii based on index in E (also leaves Ri in Y)
109  F? 1  Do we need to recompute Vnode?
110  GSB 5  Recompute denominator of solution
111  RCL E
112  x↔I
113  STO A
114  RCL (i)  (Vi)
115  ISZ I
116  RCL (i)  (Ri Vi)
117  x=0?  Is this a current instead of a voltage,resistance pair?
118  GTO 6
119  x↔y  (Vi Ri)
120  RCL D  (Vnode Vi Ri)
121  0  (0 Vnode Vi Ri)
122  +  (Vnode Vi Ri Ri)
123  −  (ViVnode Ri Ri Ri)
124  x↔y
125  ÷  (Ii Ri Ri Ri)
126  GTO 7
127♦   LBL 6  This input is a fixed current
128  x↔y  (Ii 0)
129♦   LBL 7  Restore saved value of I
130  RCL A
131  ST I
132  R↓  (Ii Ri)
133  F? 2  Do we only want Ii? (This test clears the flag)
134  RTN
135  x2  (Ii^2 Ri)
136  ×  Pi
137  RTN
138♦   LBL c  Clear all data
139  CLREG
140  P↔S
141  CLREG
142  GTO 1  Display number of inputs so far (zero)
143♦   LBL b  Prepare to swap nodes
144  9
145  RC I
146  −
147  x<0?  Throw an error if more than 5 inputs
148  √x
149  RC I  Save number of inputs in V1 of active node
150  STO 0
151  RCL 1  Get R1 to the stack before swapping
152  P↔S  Swap networks
153  STO 1  Set R1 equal to R1 from the other node
154  2
155  RCL 0  Retrieve number of inputs of other node
156  x=0?  New node?
157  +  Skip R1,V1 since they are copied from other node
158  ST I
159  RCL D  Set V1 equal to Vnode from the other node
160  STO 0
161  SF 1  Force recomputation
162  GTO 1  Display number of inputs to this node

An interesting feature of this program is that it takes advantages of the HP-67’s primary and secondary register sets when solving problems with two nodes. Most of the work of switching from one node to the other is done by a P↔S instruction (additional instructions are used to keep R1 consistent between the two nodes, and to save the number of inputs of the inactive node in its V1 register).

## Registers and Flags

Register Use
0, 2, 4, …, 18  Vi or Ii
1, 3, 5, …, 19  Ri or 0
A  Holds N during computation of Vnode
B  Accumulator for numerator of solution
C  Accumulator for denominator of solution
D  Computed value of Vnode
E  Loop index
Flag Meaning
0  Selects between computing Ii or Pi
1  Indicates that Vnode needs to be recomputed
2  Temporarily selects between computing Ii or Pi

## Revision History

2008-Nov-18 — Initial release.

## Possible Improvements

The process for solving a network with two nodes is a bit tedious, because the user must manually switch back and forth between the two nodes (by pressing f b) and compute the node voltage (by pressing C). It is up to the user to decide when the solution has stabilized. This sort of thing is easy for a programmable calculator to do, but the program uses all the calculator’s memory registers already, so none are left to perform the necessary book-keeping to automatically control such an iteration. Hopefully, I’ll think of some clever “trick” to get around this problem. If any one reading this has a suggestion, please leave a comment below.

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## 1 Comment

1. July 22, 2013

Dear Dr. Stefan Vorkoetter:

Im an Agronomist researcher working for the Argentine government.
Since 1976 I become an HP calculator user, from HP 25C (1976), HP-67 (later) and HP 35s (at present time). Also, some thing else in common, my relation with aviation is by pesticide application

I enjoy very much your high quality programming techniques, and would like to contact you by email

Congratulations for your excellent job

Sincerely, Daniel

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