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Old 3rd November 2007, 01:44 AM   (permalink)
Default gain of LM565 VCO?

I'm trying to calculate the gain factor of a VCO in the LM565. Is there a general equation for this using values given on the LM565 datasheet?
notwist is offline   Reply With Quote
Old 3rd November 2007, 02:15 AM   (permalink)
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Derive from "demodulated output voltage" at 10% deviation.
mneary is offline   Reply With Quote
Old 3rd November 2007, 03:23 AM   (permalink)
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You can calculate it from the datasheet, at least for (Vcc-Vee)=12V.

VCO gain at Fo=10k is Ko=(6600*2*pi) rad/sec/V = 4.15*10^4 (see attachment).
Since Ko is proportional to Fo (see attachment),
Ko=6600*2*pi*Fo/10000
Ko=4.15*Fo
Therefore, if, for example, Fo=3kHz, Ko=4.15*3000 = 1.244*10^4 rad/sec/V.
You can see that Ko is also inversely proportional to Vc. Kd will vary slightly with Vc, but the phase detector has internal diode clamps, so you could probably calculate Ko fairly accurately at other supply voltages also.
Attached Images
File Type: png LM565 VCO sensitivity.PNG (3.4 KB, 2 views)
File Type: png LM565 loop gain equations.PNG (30.6 KB, 4 views)
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