References: [ +subject:/^(?:^\s*(re|sv|fwd|fw)[\[\]\d]*[:>-]+\s*)*\[AMPS\]\s+More\s+L\-net\/Series\s+to\s+Parallel\s+questions\s*$/: 4 ]

** Total 4 documents matching your query.**

- 1.
**[AMPS] More L-net/Series to Parallel questions**(score: 1) **Author**:*jono@enteract.com (Jon Ogden)***Date**:*Tue, 16 Feb 99 16:09:23 -0600*- OK, more questions on series to parallel conversions and L-nets. Let's try this one on for size: Assume a resonant RLC circuit. The input impedance is 4600 Ohms. C is conected to ground at 7.6 pF. L
- /archives//html/Amps/1999-02/msg00430.html (8,857 bytes)

- 2.
**[AMPS] More L-net/Series to Parallel questions**(score: 1) **Author**:*Peter_Chadwick@mitel.com (Peter Chadwick)***Date**:*Wed, 17 Feb 1999 09:02:56 -0000*- Jon asks: Because the example given isn't resonant. 7.6pF at 28 MHz is -j747.9 ohms. In shunt with 4600 ohms, this gives a series equivalent of -j728.6 in series with 118.5 ohms. 1 microhenry is j 17
- /archives//html/Amps/1999-02/msg00435.html (8,074 bytes)

- 3.
**[AMPS] More L-net/Series to Parallel questions**(score: 1) **Author**:*jono@enteract.com (Jon Ogden)***Date**:*Wed, 17 Feb 99 09:40:27 -0600*- I don't understand. The extra inductor acts as an impedance transformer. It's reactance changes with frequency and so the resulting output impedance (prior to the tank circuit) changes over frequenc
- /archives//html/Amps/1999-02/msg00442.html (8,374 bytes)

- 4.
**[AMPS] More L-net/Series to Parallel questions**(score: 1) **Author**:*Peter_Chadwick@mitel.com (Peter Chadwick)***Date**:*Wed, 17 Feb 1999 16:07:40 -0000*- Bandwidth on 10m. I set up the Smith Chart with the 4600 ohms as the load, and looked in from the 50 ohm end. On 10m, the 2:1 SWR Bandwidth for any particular setting of the L's and C's was less with
- /archives//html/Amps/1999-02/msg00443.html (7,656 bytes)

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