Bogged Down With Diode Detection Theory - UK Vintage Radio Repair and Restoration Discussion Forum

MajorWest · 26th Sep 2014, 5:17 pm

Feel free not to read this if you dislike over-complicating. I have been a bit baffled with respect to simple diode detector stages and much of it has to do with my usual fixation with current flow.
So far, what I have gathered is when an antenna picks up RF signals and they are passed through tuning to a valve diode, only the positive cycles can pass through the anode of the diode. Negative cycles of A.C. are blocked. The result is what is called a D.C. pulse at the cathode of the diode. This is called demodulation. Also, as we know any remaining R.F. is filtered out.
Where I get a bit bogged down is there is current flow going on around the valve all the time. This is partly D.C., as well as A.F. and even a bit of R.F. as well.

Now the small point that really gets me is that DC current flow in valves is an electron flow from cathode to anode. Meantime, AC current doesn't work quite the same way so I'm not really sure how exactly to class an audio frequency "ripple". Do you think the current flows the same as in pure DC?

In the past, books I found never went into too much depth over this but the latest one I have goes as deep as I ever managed to dig so far. However, sometimes I think there are some little points that are left unexplained.

Tyso_Bl · 26th Sep 2014, 6:20 pm

Quote:

Originally Posted by MajorWest

Where I get a bit bogged down is there is current flow going on around the valve all the time. This is partly D.C., as well as A.F. and even a bit of R.F. as well.

I'm not sure what you are trying to ask here, if we're talking about 'detection' what is happening can be visualised as half wave rectification of the applied radio frequency energy. The peak voltage of the half wave pulses will be proportional and follow the voltage applied to the rectifier. The train of pulses when then applied to a 'smoothing' capacitor will then follow the level of the applied radio frequency, a high value resistor connected to the capacitor will allow the capacitor to discharge when the level of the applied radio frequency drops instead of increasing. The effect is then that the varying voltage across the capacitor follows the level of the RF signal fed in to the rectifier/detector.

The best explanation I have ever had is in Scroggies Foundations of Wireless, an early 4th edition.

Tyso

kellys_eye · 26th Sep 2014, 11:23 pm

You may be confusing AC (as the transition across a 0V boundary) with the 'ac' of a waveform that is always above the zero level - effectively a DC voltage but with alternation in ampitude rather than in polarity.

G8HQP Dave · 28th Sep 2014, 7:41 pm

Think of a diode AM detector as a low voltage low power half-wave rectified power supply. If you understand one then you can understand the other. In both cases the diode only conducts on AC peaks, to top up the capacitor. Between these times, the load takes some charge out of the capacitor so the voltage drops a little.

There are two AC signals: the carrier and the modulation. It may be best to think of the modulation as being slowly varying DC. Note that the distinction between AC and DC is only a myth taught to learners to get them started.

G6Tanuki · 28th Sep 2014, 7:55 pm

The way I always think of it is to consider a half-wave-rectified power supply with inadequate smoothing.

What in a power-supply we view as the annoying hum/ripple component is in a detector the audio we want to listen to.

What in a power-supply is the smoothed DC-component becomes in a detector the averaged signal-level which we use to drive the receiver's AGC.

kalee20 · 28th Sep 2014, 9:32 pm

Quote:

Originally Posted by G6Tanuki

The way I always think of it is to consider a half-wave-rectified power supply with inadequate smoothing.
What in a power-supply we view as the annoying hum/ripple component is in a detector the audio we want to listen to.

No, it's not!

The annoying hum/ripple component in a power supply is the residual RF in a detector after filtering - which is usually very small.

The DC (or rather the very slow variation in DC, as the mains drifts up or down a bit) in a power supply is in a detector the AF we want to listen to.

PJL · 29th Sep 2014, 11:32 am

A typical radio valve detector is unusual in that the cathode of the diode is normally to earth. This makes it a little more difficult to visualise but to understand its behaviour you need to be able to do this.

The circuit normally comprises an AC source from the last IF transformer, connected to the anode of the diode with the other end connected to a parallel C & R to earth.

It is easy to see that the first positive AC cycle will cause the diode to conduct heavily and negatively charge the capacitor but this is a start up situation and the next cycle will charge the capacitor a little more. In visualising the behaviour you must look for the steady state condition.

Imagine the waveform on the diode anode, a sinewave with a DC offset from the voltage across C. In the steady state, the voltage on the anode must never go positive and for this to be the case, the DC voltage on C (Vc) must have negative half the peak to peak AC voltage (Vpp).

It is now easy to calculate the DC (or average) current through the diode as it is in series with the parallel R & C so it is the same as the current flowing through R which will be given by I=V/R or Vpp/2R.

Up till now we have been talking about a constant IF signal with no audio signal (the carrier). If we were to halve the amplitude of our IF signal then from our work above the DC voltage Vc would change from Vpp/2 to Vpp/4 so it is proportional to the amplitude change.

If we were to change the amplitude in direct proportion to our audio signal then the DC voltage Vc should follow our audio signal. This just leaves one thing to discuss which is the value of C. If it is too small then it will not smooth the IF signal and will leave a high IF AC component in Vc. If it is too large it will smooth the audio signal and cut-off the higher pitch notes.

We have not discussed the AC current waveform which has complex dependencies as it is not relevant in gaining an understanding of the detector operating principles.

MajorWest · 29th Sep 2014, 5:33 pm

Quote:

Originally Posted by kellys_eye

You may be confusing AC (as the transition across a 0V boundary) with the 'ac' of a waveform that is always above the zero level - effectively a DC voltage but with alternation in ampitude rather than in polarity.

That's about right. I assume the DC ripple that passes to the cathode via the anode still has amplitude. I do know also the resulting DC voltage from the smoothing capacitor was used in simple AVC systems and could be tapped.
Yes, it is very similar to a rectifier. The simplest system I came across was the one where the DC ripple is smoothed by a capacitor and smoothing choke built into the speaker (electromagnet).
I think maybe where I was curious is the effect of voltage ripple on current. I can only guess the DC current here must rise and drop in sympathy as it were with the ripple voltage.
It's not a huge deal except not long ago I found out AC current doesn't really flow and the ripple or pulsed DC post rectification does sort of resemble A.C. Certainly it can very easily be converted to A.C.

26th Sep 2014, 5:17 pm	#1
MajorWest Retired Dormant Member Join Date: Jul 2014 Location: Stoke On Trent, Staffordshire, UK. Posts: 247	Bogged Down With Diode Detection Theory Feel free not to read this if you dislike over-complicating. I have been a bit baffled with respect to simple diode detector stages and much of it has to do with my usual fixation with current flow. So far, what I have gathered is when an antenna picks up RF signals and they are passed through tuning to a valve diode, only the positive cycles can pass through the anode of the diode. Negative cycles of A.C. are blocked. The result is what is called a D.C. pulse at the cathode of the diode. This is called demodulation. Also, as we know any remaining R.F. is filtered out. Where I get a bit bogged down is there is current flow going on around the valve all the time. This is partly D.C., as well as A.F. and even a bit of R.F. as well. Now the small point that really gets me is that DC current flow in valves is an electron flow from cathode to anode. Meantime, AC current doesn't work quite the same way so I'm not really sure how exactly to class an audio frequency "ripple". Do you think the current flows the same as in pure DC? In the past, books I found never went into too much depth over this but the latest one I have goes as deep as I ever managed to dig so far. However, sometimes I think there are some little points that are left unexplained.

26th Sep 2014, 11:23 pm	#3
kellys_eye Octode Join Date: Jan 2014 Location: Oban, Scotland, UK. Posts: 1,129	Re: Bogged Down With Diode Detection Theory You may be confusing AC (as the transition across a 0V boundary) with the 'ac' of a waveform that is always above the zero level - effectively a DC voltage but with alternation in ampitude rather than in polarity.

28th Sep 2014, 7:41 pm	#4
G8HQP Dave Rest in Peace Join Date: Sep 2008 Location: Solihull, West Midlands, UK. Posts: 4,872	Re: Bogged Down With Diode Detection Theory Think of a diode AM detector as a low voltage low power half-wave rectified power supply. If you understand one then you can understand the other. In both cases the diode only conducts on AC peaks, to top up the capacitor. Between these times, the load takes some charge out of the capacitor so the voltage drops a little. There are two AC signals: the carrier and the modulation. It may be best to think of the modulation as being slowly varying DC. Note that the distinction between AC and DC is only a myth taught to learners to get them started.

28th Sep 2014, 7:55 pm	#5
G6Tanuki Dekatron Join Date: Apr 2012 Location: Wiltshire, UK. Posts: 14,007	Re: Bogged Down With Diode Detection Theory The way I always think of it is to consider a half-wave-rectified power supply with inadequate smoothing. What in a power-supply we view as the annoying hum/ripple component is in a detector the audio we want to listen to. What in a power-supply is the smoothed DC-component becomes in a detector the averaged signal-level which we use to drive the receiver's AGC.

29th Sep 2014, 11:32 am	#7
PJL Dekatron Join Date: Sep 2005 Location: Seaford, East Sussex, UK. Posts: 5,997	Re: Bogged Down With Diode Detection Theory A typical radio valve detector is unusual in that the cathode of the diode is normally to earth. This makes it a little more difficult to visualise but to understand its behaviour you need to be able to do this. The circuit normally comprises an AC source from the last IF transformer, connected to the anode of the diode with the other end connected to a parallel C & R to earth. It is easy to see that the first positive AC cycle will cause the diode to conduct heavily and negatively charge the capacitor but this is a start up situation and the next cycle will charge the capacitor a little more. In visualising the behaviour you must look for the steady state condition. Imagine the waveform on the diode anode, a sinewave with a DC offset from the voltage across C. In the steady state, the voltage on the anode must never go positive and for this to be the case, the DC voltage on C (Vc) must have negative half the peak to peak AC voltage (Vpp). It is now easy to calculate the DC (or average) current through the diode as it is in series with the parallel R & C so it is the same as the current flowing through R which will be given by I=V/R or Vpp/2R. Up till now we have been talking about a constant IF signal with no audio signal (the carrier). If we were to halve the amplitude of our IF signal then from our work above the DC voltage Vc would change from Vpp/2 to Vpp/4 so it is proportional to the amplitude change. If we were to change the amplitude in direct proportion to our audio signal then the DC voltage Vc should follow our audio signal. This just leaves one thing to discuss which is the value of C. If it is too small then it will not smooth the IF signal and will leave a high IF AC component in Vc. If it is too large it will smooth the audio signal and cut-off the higher pitch notes. We have not discussed the AC current waveform which has complex dependencies as it is not relevant in gaining an understanding of the detector operating principles.