The signal generator output amplitude is first measured at about 20,000 kHz using an oscilloscope. The unknown inductor is then connected across the signal generator output and the frequency

On a digital oscilloscope I prefer to take rms readings as they are more stable than peak to peak readings.

The equation is quite simple:

If

The result is

For best accuracy remove the inductor when the signal is at half-amplitude. Adjust the signal generator to the previous (convenient) full-amplitude value. Replace the inductor and trim the frequency until the half-amplitude setting is found. Use this frequency to calculate

The best measurement for

Name | Description | |
---|---|---|

A | Calculate L from entered R and f |

Name | Description | |
---|---|---|

1 | Inductor DC resistance R | |

2 | Frequency f |

Line | Display | Key Sequence | Line | Display | Key Sequence | ||
---|---|---|---|---|---|---|---|

000 | 016 | 40 | + | ||||

001 | 42,21,11 | f LBL A | 017 | 48 | . | ||

002 | 44 2 | STO 2 | 018 | 0 | 0 | ||

003 | 34 | x↔y | 019 | 2 | 2 | ||

004 | 44 1 | STO 1 | 020 | 5 | 5 | ||

005 | 2 | 2 | 021 | 45 1 | RCL 1 | ||

006 | 1 | 1 | 022 | 43 11 | g x² | ||

007 | 48 | . | 023 | 20 | × | ||

008 | 1 | 1 | 024 | 30 | − | ||

009 | 1 | 1 | 025 | 11 | √x̅ | ||

010 | 36 | ENTER | 026 | 26 | EEX | ||

011 | 48 | . | 027 | 3 | 3 | ||

012 | 8 | 8 | 028 | 20 | × | ||

013 | 4 | 4 | 029 | 45 2 | RCL 2 | ||

014 | 45 1 | RCL 1 | 030 | 10 | ÷ | ||

015 | 20 | × | 031 | 43 32 | g RTN |