Why is second analog multiplier(marked with red color) used to detect the amplitude of the input signal in this circuit instead of diode? The point of this circuit is to maintain the output level of the first multiplier constant all the time.
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Looks more like a true detector with no offset and some gain – Tony Stewart EE75 Aug 22 '20 at 23:08
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It makes sense. But why is it used here? – Sundark12 Aug 22 '20 at 23:15
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I dont know why is it beign used here. I dont understand its concpet of squaring the voltage, it seems like it is giving him some DC offset when I square it and that is it. – Sundark12 Aug 22 '20 at 23:26
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1It seems to be AGC output, but why is squaring used to detect amplitude? – Sundark12 Aug 22 '20 at 23:31
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true RMS detector? is this a test to guess where you got this? – Tony Stewart EE75 Aug 22 '20 at 23:36
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1Do some more reading https://www.analog.com/media/en/training-seminars/design-handbooks/ADI_Multiplier_Applications_Guide.pdf?doc=AD633.pdf – Tony Stewart EE75 Aug 22 '20 at 23:40
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I read the book you sent me, I am sorry but I still dont understand the point of second multiplier in order to detect the amplitude. In the book you sent me squaring input voltage is usally part of something bigger, but in my circuit its only prupose is to detect the amplitude and I seem to fail to understand why is it beign used. – Sundark12 Aug 24 '20 at 21:47
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Where did you get this RMs detector cct? – Tony Stewart EE75 Aug 24 '20 at 21:48
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I am sorry but I do not remember that, that is the reason why I am posting it here. If I knew where I got it I think it would have come with an explanation. – Sundark12 Aug 24 '20 at 21:54
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Thats pretty vague then if you have no idea what its for. i suggest as V^2 is proportional to power it measures power into a fixed load. – Tony Stewart EE75 Aug 24 '20 at 21:56
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As I said previously it seems to be AGC but I am confused why use multiplier instead of some diode type peak detector. – Sundark12 Aug 24 '20 at 21:59
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I’ve never seen this before yet it shud be obvious with the equation it is solving when you equate the op amp inputs, no? – Tony Stewart EE75 Aug 24 '20 at 22:12
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I guess that is just how circuit works. Equations fit and that is all is needed. – Sundark12 Aug 24 '20 at 22:19
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When you square a sine wave, you get a DC term and a term at twice the frequency. The RC filter is used to eliminate the double frequency term. The DC term is proportional to the square of the input level and is fed back to the input to keep the signal level constant at the output of the first multiplier. The multiplier eliminates the need for a precision rectifier and has no offset voltage to worry about.
Barry
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So DC terms contains information about the peak then that peak is compared to reference? – Sundark12 Aug 22 '20 at 23:46