04-03-2017, 05:34 PM
(This post was last modified: 04-03-2017, 05:35 PM by Hartley118.)
It's certainly not immediately obvious what's going on.
I too was inclined at first to dismiss that RC feedback network (Ca1 & Ra2) as just a bit of frequency response correction. However, on a second look, it seems to me to be central to the whole basis of the circuit because Ca1 at 2200 pf is probably of similar order of capacitance to the crystal pickup, and what's more will be 'amplified' in value by the gain of the stage (Miller effect). Not sure what Ra2 is for because its resistance value only becomes significant compared with the reactance of Ca1 in the tens of kHz. Perhaps it went unstable without it! So forget its presence for a moment and what do we have? - an integrator.
So I look at it as a capacitor with gain. Roughly its value will be 2200pF x 18k/2.2k or 18nF. As Mark has calculated, its voltage gain will be 18k/2.2k, or just over 8.
Why was it designed this way? Well, a crystal pickup has a capacitive output impedance,so loading it with a capacitor reduces its output voltage, but also reduces the output load resistance needed to preserve low frequency performance. Quite a useful way to match it into a bipolar transistor input.
Martin
I too was inclined at first to dismiss that RC feedback network (Ca1 & Ra2) as just a bit of frequency response correction. However, on a second look, it seems to me to be central to the whole basis of the circuit because Ca1 at 2200 pf is probably of similar order of capacitance to the crystal pickup, and what's more will be 'amplified' in value by the gain of the stage (Miller effect). Not sure what Ra2 is for because its resistance value only becomes significant compared with the reactance of Ca1 in the tens of kHz. Perhaps it went unstable without it! So forget its presence for a moment and what do we have? - an integrator.
So I look at it as a capacitor with gain. Roughly its value will be 2200pF x 18k/2.2k or 18nF. As Mark has calculated, its voltage gain will be 18k/2.2k, or just over 8.
Why was it designed this way? Well, a crystal pickup has a capacitive output impedance,so loading it with a capacitor reduces its output voltage, but also reduces the output load resistance needed to preserve low frequency performance. Quite a useful way to match it into a bipolar transistor input.
Martin
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