Resistor Network Solver for the HP-67 Programmable Calculator

November 18, 2008

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:

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 V_node:

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	i→Ii	i→Pi
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  c	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  d	-124. -06
Look up P2 again	2    f  e	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 (V_node1) is connected by resistors R_1ab, R_2a, R_3a, and R_4a to voltages V_node2, V_2a, V_3a, and ground respectively.

The rightmost node also has four connections, three resistors and a constant current input of 0.1mA. The resistors R_1ab, R_2b, and R_3b connect to voltages V_node1, V_2b, V_3b 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  c	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  b	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 V_node until the values stop changing:

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

The two V_node values stop changing when V_node1 = 1.90V and V_node2 = 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 I1 … IN
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	−	(Vi–Vnode 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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