Crystal Set Waveforms are not what I expected - Page 1
Forum section: electronics / beginners

author: rayleask1
author_rank: Newbie
timestamp: September 29, 2015, 12:10:39 pm
content: Hi everyone. After nearly 40 years I thought I would return to radio electronics to see if I could understand it this time round. I can now afford an Oscilloscope and I have been using it to look at some simple crystal sets. Straightaway I have come across something that I don't understand.I have put the scope across the earpiece and I see an AC waveform. Shouldn't it be a DC one? I have also put across it a small ceramic capacitor thinking that should dump any AC but it made no difference. The crystal set works fine and I can both see and hear the waveform. On the scope I have set it to both AC and DC inputs and swapped the input around between the two channels but it made no difference. Looking at the tuned circuit I see what looks like a classic AM signal. Can anyone put me out of my misery?Thanks, Ray

author: janoc
author_rank: Super Contributor
timestamp: September 29, 2015, 12:33:25 pm
content: If you had a DC waveform (= straight line on the scope) through the earpiece, you wouldn't hear anything. Only a click when you connect/disconnect the earpiece.Human ear detects only relative changes in air pressure, it cannot hear how stretched is membrane of the earpiece (which is what you would have in case of DC). And the only way you can get those pressure changes is by running AC through the earpiece.What you should see on the scope when probing the earpiece is low (audio <20kHz) frequency AC, without the high frequency carrier superimposed.The small ceramic cap will "short" high frequencies, depending on its value. If you don't hear any difference is is most likely because the capacitance of the cap is so small that it attenuates only the carrier (radio) frequency, which you wouldn't hear anyway. The purpose of that cap is to get rid of any RF energy that would only load the earpiece without producing useful signal. If you increase the capacitance sufficiently you will start to lose higher audio frequencies too, but then we are talking about cap in tens/hundreds of nanofarads at least.

author: Cliff Matthews
author_rank: Unknown
timestamp: September 29, 2015, 12:39:52 pm
content: Welcome! Can you give the scope mfg and model, someone familiar with your gear may step you through.

author: mikerj
author_rank: Super Contributor
timestamp: September 29, 2015, 01:30:25 pm
content: Quote from: rayleask1 on September 29, 2015, 12:10:39 pmNo, you should be seeing an AC waveform. In fact you are likely seeing the RF carrier waveform (i.e. a sinusoid at the frequency you are tuned to), amplitude modulated by the audio. However even if the carrier frequency is filtered out, the remaining audio waveform will be AC.

author: w2aew
author_rank: Super Contributor
timestamp: September 29, 2015, 01:34:40 pm
content: Quote from: rayleask1 on September 29, 2015, 12:10:39 pmIt depends on the capacitor value and the frequency involved. The capacitor's impedance is Xc = 1/2*pi*f*C, so small capacitor values will have little effect on low frequency signals.QuoteThe AC/DC coupling on the scope doesn't dictate what type of signals you can measure, it affects how they're processed by the scope. There are many instances where you might want to us AC coupling on a DC signal, and vice-versa. Here's a video tutorial on that...

author: rayleask1
author_rank: Newbie
timestamp: September 29, 2015, 06:11:03 pm
content: Thanks to everyone for their fast replies. The scope is an old Kenwood CS-4125. What I was expecting was this kind of thing, Waveform C taken from Wikipeadias Crystal Set site.https://en.wikipedia.org/wiki/File:Amplitude_modulation_detection.pngSurely the Diode would prevent all negative voltage? The capacitor I used was taken from a commercial crystal set I brought and its notes state that its there to allow residual RF to bypass the earpiece. But I think this is beside the point. mikerj states "even if the carrier frequency is filtered out, the remaining audio waveform will be AC". How can this be if the diode is only allowing voltage at the earpiece to be present in one direction? I can compare both the RF and the Audio using the scope and yes it looks like the Audio is the RF minus the carrier albeit at a lower voltage.Its looking as though my level of understanding is even less than it was a couple of hours ago.Cheers, RayThanks to w2aew for the Youtube link, I will take a look at that now.

author: w2aew
author_rank: Super Contributor
timestamp: September 29, 2015, 09:02:58 pm
content: Quote from: rayleask1 on September 29, 2015, 06:11:03 pmYou'll note from the picture that you provided that the detected audio is essentially following the envelope of the RF. More specifically, it is a varying DC voltage (because it doesn't change polarity). However, this is usually passed through a series capacitor (AC coupled) to the headphones, thus removing the DC component (DC offset), thus leaving the AC audio component being sent to the headphone.

author: rayleask1
author_rank: Newbie
timestamp: September 29, 2015, 10:12:55 pm
content: Thanks w2aew for your email. I'm sorry but I still having brain problems. My circuit diagram is almost identical to this -http://www.usefulcomponents.com/main_contents/projects/choccy_block_crystal_radio/choccy_xtal_radio_scem.jpgThere is no series capacitor just that diode that prevents a polarity change. Theres no denying its AC flowing through the headphones I am just flumoxed how it gets there.Thanks, Ray

author: Zero999
author_rank: Super Contributor
timestamp: September 29, 2015, 11:00:30 pm
content: The transformer will remove the DC component.In any case, even if there is a DC component, the crystal earpiece should work fine with some DC bias.

author: mikerj
author_rank: Super Contributor
timestamp: September 30, 2015, 10:08:17 am
content: Quote from: Hero999 on September 29, 2015, 11:00:30 pmYep, a crystal earpiece is a fundamentally capacitive load, so there's no need for a separate coupling cap.Quote from: rayleask1 on September 29, 2015, 10:12:55 pmAt the cathode of the detector diode, the voltage will be varying but will always be >= 0v. The transformer will remove this DC bias and produce a purely AC voltage. However, you don't need a transformer if you use a crystal earpiece. In this case you have a varying DC signal applied to a capacitor (i.e. the earpiece itself), which will naturally give an AC current.

author: rayleask1
author_rank: Newbie
timestamp: September 30, 2015, 12:46:52 pm
content: Thanks for your email. "In this case you have a varying DC signal applied to a capacitor (i.e. the earpiece itself), which will naturally give an AC current."This sounds like the bit I haven't got my head around. I put my scope in series between the diode and the earpiece and I see an audio signal DC. If I put the scope in parallel with the earpiece I see an AC audio signal. So it sounds like it. "varying DC signal applied to a capacitor will give an AC current"? Is this the case? I would have thought a varying DC current would come out of it albeit with a phase shift?Thanks, Ray

author: rayleask1
author_rank: Newbie
timestamp: October 02, 2015, 11:43:39 am
content: Yep, got it. Thanks to everyone.Ray