How can I visualize current flow in this circuit? - Page 1
Forum section: electronics / beginners

author: novicefedora
author_rank: Contributor
timestamp: July 23, 2020, 05:37:44 am
content: This is the circuit.I'm new to electronics so I don't know how the current flows in this circuit. It should start at the negative terminal, from there where would it first go? Then how would the antenna work?

author: tkamiya
author_rank: Super Contributor
timestamp: July 23, 2020, 10:40:11 pm
content: That's a very typical RF probe circuit. Your antenna picks up electro-magnetic waves from air and generate electrical current. Your diode chops it in half and flips under side up (rectifies it) C2 removes RF component and leaves DC component (mostly...) Resister at the end acts as a load so that current turns into voltage across it. You'd measure that voltage to see the level.

author: RJSV
author_rank: Super Contributor
timestamp: July 23, 2020, 11:42:23 pm
content: Ok without lots of RF skill, and building on last comment you could call this a 'half wave' demodulator.The D2 diode will let the positive portion of any received voltage waveform, into charging up the Cap. Then, any negative portions coming in are blocked by D2 but are still allowed by D1: That reversed current thru the first capacitor, for preparation for the next cycle repeat, where another (positive) reversal brings more current. So the second capacitor just sits there, slowly accumulating those positive swings to build up voltage, perhaps several hundred cycles. Meanwhile, that loader resistor is working to cause slow decay to work to drain down, if the RF weakens.You could view those right side components as 'audio' frequency related . The capacitor is slow to the RF but small enough to follow the RF 'peaks' at audio rates, like 300 to 3000 hz.I'm not sure if D1 conducting negative direction affects the coil, certainly it helps the small capacitor to be able to conduct full current on the positive swings. As to the electric fields and magnetic fields, that's some complexity, but just assume an oscillating voltage is 'arriving' to simplify circuit. - RJ

author: asperised
author_rank: Contributor
timestamp: July 29, 2020, 02:07:16 pm
content: Although I'm not the OP but I am a beginner as well. Your explanations make a lot of sense and I can kind off understand how this works. It's a bit like those Bridge rectifiers we were taught in school and the capacitor is to make the voltage stable.My question is, what does the resistor in this scenario do?, and how do you go about selecting the right value of capacitance for such applications?

author: TimFox
author_rank: Super Contributor
timestamp: July 29, 2020, 02:29:55 pm
content: Another way of looking at this circuit is that it is one of the two standard forms of "voltage doubler" rectifier, where the output capacitor charges up to a voltage equal to the peak-to-peak voltage of the input waveform (less overhead for diode drops). This form, which starts out with a capacitor at the input, is preferred for RF probes, since it blocks any DC component at the input.

author: mikerj
author_rank: Super Contributor
timestamp: July 29, 2020, 03:49:11 pm
content: Quote from: asperised on July 29, 2020, 02:07:16 pmWithout the resistor C2 has nothing to discharge into so would stay at whatever the peak voltage it was charged to which is not useful for demodulating amplitude modulated RF. This assumes that whatever you connect to +-volts is too high impedance to provide a useful discharge path. The time constant of C2 and the resistor determines the maximum useful frequency this circuit could demodulate.

author: TimFox
author_rank: Super Contributor
timestamp: July 29, 2020, 03:57:48 pm
content: The 1 nF filter capacitor and a 10 megohm meter gives a time constant of 10 msec, which would be too slow for audio-frequency AM demodulation, but is reasonable for voltage measurement. I don't know how much resistance the germanium 1N34 diodes contribute to this question, but they are probably less than 10 megohms. With a 2000 ohm headphone (showing my age), the time constant would be 20 microsec, which is reasonable for AM (5 kHz audio modulating 1 MHz carrier).