Timeline for Discrete component FM radio receiver circuit - explanation needed
Current License: CC BY-SA 4.0
16 events
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| Aug 22, 2019 at 17:45 | comment | added | dhrovner | Thank you for your help. The lesson I learned is the base/collector junction does not NEED to be in reverse bias in order for the common base configuration to amplify a signal. I didn't know that. Thank you! | |
| Aug 22, 2019 at 16:41 | comment | added | TimWescott | It can't be hard in saturation, because it is amplifying. Period. | |
| Aug 22, 2019 at 15:57 | comment | added | dhrovner | Last comment/question! If the circuit actually is hovering/in saturation (base collector in forward bias) I am still not 100% clear how the circuit is amplifying the transient noise! If you can try to explain to me again :) | |
| Aug 22, 2019 at 15:53 | comment | added | dhrovner | In the simulations I reduced the value of R2 to exaggerate the results. | |
| Aug 22, 2019 at 15:47 | comment | added | dhrovner | Thanks! I've added a screenshot of my simulation results. | |
| Aug 22, 2019 at 15:45 | comment | added | TimWescott | It could be that the transistor is, indeed, going into saturation rather than cutoff. That would have the effect of varying the gain within a cycle, which is what's necessary for mixing. Having the transistor saturate in an oscillator circuit like that tends to kill the Q of the resonator, which significantly reduces the radio performance, but things ought to still work. | |
| Aug 22, 2019 at 15:21 | comment | added | dhrovner | Sadly when I simulate the autodyne stage (Q1), it indeed amplifies and oscillates and stabilizes as expected, however, I don't see Q1 going into cutoff. I see the opposite... hovering around saturation. So either something is unique about this circuit, or, I am not simulating the circuit correctly. When actually fabricated this circuit works and picks up many FM stations, so I am now just trying to understand the schematic and logic and electronics behind the working circuit. I am sure you can tell I am a novice, perhaps not knowing how to explain myself well... But thank you so much! | |
| Aug 22, 2019 at 15:16 | history | edited | Voltage Spike♦ | CC BY-SA 4.0 |
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| Aug 22, 2019 at 15:09 | history | edited | TimWescott | CC BY-SA 4.0 |
added 9 characters in body
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| Aug 22, 2019 at 15:09 | comment | added | TimWescott | Yes, class C implies that the transistor is going into cutoff -- but that's a likely thing to happen in an oscillator. Oscillation grows until something in the circuit acts to reduce the gain of the active element. In the case of a circuit like that, usually what happens is that the base capacitor gets "pumped down" from injected base current, and holds the base at a lower voltage than it would be in linear operation. That reduces the conduction angle of the amplifier, making it class C (or B, which would have the same effect -- I'm editing my answer). | |
| S Aug 22, 2019 at 6:36 | history | edited | Chetan Bhargava | CC BY-SA 4.0 |
More clarification/insight?
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| S Aug 22, 2019 at 6:36 | history | suggested | dhrovner | CC BY-SA 4.0 |
More clarification/insight?
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| Aug 22, 2019 at 3:39 | review | Suggested edits | |||
| S Aug 22, 2019 at 6:36 | |||||
| Aug 22, 2019 at 1:41 | comment | added | dhrovner | Doesn't class C require the transistor to go into and out of cut-off? I don't see that happening. Unless going in and out of saturation will also accomplish class C operation? But in any case I don't yet see how this circuit is amplifying the transient noise to the point of Q1 cutoff nor getting out of saturation (if indeed it is in saturation) and into the active region for a short time. Thanks for your patience with me! | |
| Aug 22, 2019 at 0:45 | comment | added | dhrovner | Thank you Tim. I added to my question above as I could not fit it here! | |
| Aug 21, 2019 at 23:59 | history | answered | TimWescott | CC BY-SA 4.0 |