ReactanceP

Description

J E Patterson - jepspectro.com - 20180610


hp-15c Program Index and documentation


This program calculates solutions to common reactance problems and also allows an inductor to be specified. Solver is not needed to do reactance calculations. The equations can be solved directly. The advantage with using Solver is that programs can be shorter. An example is here.

The equations which follow are all used in this program. An unknown variable C, L or f can be calculated by entering it as zero.

Usage:

C ENTER L ENTER f GSB A
set unknown to zero. Its value will be calculated.
R/S
The reactances are now in registers 4 and 5. At resonance the values are the same. New values of C, L and f can be stored in registers 1, 2 and 3.
GSB 4 for capacitive reactance Xc and GSB 5 for inductive reactance Xl.

If you wish you can enter C, L and f directly in registers 1, 2 and 3. Store 0 for the unknown variable.
Press GSB B to calculate the value for the unknown variable.

You can calculate an approximate number of turns for an inductor using the Wheeler formula. See also the ARRL Handbook.
Turns = 5040*√((18*d +40*l)L)/d where L is in henrys, d is the diameter of the winding in mm, l is the length of the winding in mm. The factor 5040 in front of the standard Wheeler formula corrects for henrys instead of microhenrys (1000) and mm instead of inches (5.04).
d ENTER l ENTER GSB 8

An approximation or rough check for the Wheeler formula is T = 40√(aL)/d where a is the average of the coil length l and diameter d. L is in microhenries. For a 10 microhenry, 20 mm diameter, 30 mm long coil the Wheeler formula shows 31.5 turns and the approximation shows 31.6 turns. With more extreme size ratios the error is greater.

For coils with a 2:3 diameter to length ratio an even simpler formula is T = 45 √(L/d). This formula shows 31.8 turns.
Using the inductance stored in register 2, enter the coil diameter d GSB 9 to test this simplified formula. In this program the factor is 45000 since L in register 2 is in henrys.

Example:

C is unknown, L = 1E-6 henrys, the frequency f is 1 MHz.
0 ENTER 1 EEX CHS 6 ENTER 1 EEX 6 GSB A


Result:

C = 25.33 nanofarads
R/S
RCL 4 Xc = 6.28 ohms
RCL 5 Xl = 6.28 ohms

1 EEX 7 STO 3 - The frequency is increased 10 times to 10 Mhz.
GSB 4 Xc = 0.63 ohms
GSB 5 Xl = 62.83 ohms

0 ENTER RCL 2 RCL 3 GSB A
For resonance at 10 MHz the capacitor should now be 253 picofarads.

For a coil, diameter d = 20 mm and length l = 20 mm:
2 0 ENTER 2 0 GSB 8
8.6 turns are required. This inductor is loaded with a lot of capacitance. Reducing capacitance to 15 picofarads requires a 17 microhenry inductor of 35 turns.

Program Resources

Labels

Name Description Name Description
 A Enter capacitance, Inductance and frequency. Unknown value is entered as zero.  2 * RCL3, x^2, 1/x code fragment
 B Calculate unknown  4 Calculate capacitive reactance Xc
 C Get capacitance  5 Calculate inductive reactance Xl
 D Get inductance  8 Calculate number of turns using the Wheeler formula
 E Get frequency  9 Simple turns calculator using 45000 √(L/d), L is in henrys
 1 1/(2π)^2 code fragment

Storage Registers

Name Description
 1 Capacitance C farads
 2 Inductance L henrys
 3 Frequency f hertz
 4 Capacitive reactance Xc ohms
 5 Inductive reactance Xl ohms
 6 2π temporary storage
 7 Inductor length l mm
 8 Inductor diameter d mm

Program

Line Display Key Sequence Line Display Key Sequence Line Display Key Sequence
000 035 43 32 g RTN 070 44 5 STO 5
001 42,21,11 f LBL A 036 42,21,15 f LBL E 071 43 32 g RTN
002 44 3 STO 3 037 32 1 GSB 1 072 42,21, 8 f LBL 8
003 33 R⬇ 038 45,20, 2 RCL × 2 073 44 7 STO 7
004 44 2 STO 2 039 45,20, 1 RCL × 1 074 33 R⬇
005 33 R⬇ 040 11 √x̅ 075 44 8 STO 8
006 44 1 STO 1 041 15 1/x 076 1 1
007 32 12 GSB B 042 44 3 STO 3 077 8 8
008 43 32 g RTN 043 31 R/S 078 20 ×
009 42,21,12 f LBL B 044 43 32 g RTN 079 45 7 RCL 7
010 45 3 RCL 3 045 42,21, 1 f LBL 1 080 4 4
011 43 20 g x=0 046 2 2 081 0 0
012 32 15 GSB E 047 43 26 g π 082 20 ×
013 45 2 RCL 2 048 20 × 083 40 +
014 43 20 g x=0 049 44 6 STO 6 084 45,20, 2 RCL × 2
015 32 14 GSB D 050 43 11 g 085 11 √x̅
016 45 1 RCL 1 051 43 32 g RTN 086 45,10, 8 RCL ÷ 8
017 43 20 g x=0 052 42,21, 2 f LBL 2 087 5 5
018 32 13 GSB C 053 20 × 088 0 0
019 32 4 GSB 4 054 45 3 RCL 3 089 4 4
020 32 5 GSB 5 055 43 11 g 090 0 0
021 43 32 g RTN 056 20 × 091 20 ×
022 42,21,13 f LBL C 057 15 1/x 092 43 32 g RTN
023 32 1 GSB 1 058 43 32 g RTN 093 42,21, 9 f LBL 9
024 45 2 RCL 2 059 42,21, 4 f LBL 4 094 45 2 RCL 2
025 32 2 GSB 2 060 45 6 RCL 6 095 34 x↔y
026 44 1 STO 1 061 45,20, 3 RCL × 3 096 10 ÷
027 31 R/S 062 45,20, 1 RCL × 1 097 11 √x̅
028 43 32 g RTN 063 15 1/x 098 4 4
029 42,21,14 f LBL D 064 44 4 STO 4 099 5 5
030 32 1 GSB 1 065 43 32 g RTN 100 26 EEX
031 45 1 RCL 1 066 42,21, 5 f LBL 5 101 3 3
032 32 2 GSB 2 067 45 6 RCL 6 102 20 ×
033 44 2 STO 2 068 45,20, 3 RCL × 3 103 43 32 g RTN
034 31 R/S 069 45,20, 2 RCL × 2