Understanding this signal conditioning circuit - Page 1
Forum section: electronics / projects

author: Red_Micro
author_rank: Regular Contributor
timestamp: February 22, 2020, 02:55:44 am
content: Hi,The circuit attached is used to condition a signal coming from a CT that detects very low currents. I don't fully understand this circuit. What's the purpose of C4? It is obviously reducing the original amplitude of the signal. I'm trying to simplify this circuit where possible using a simpler precision rectifier. Any comments on the overall operation of this circuit will be appreciated.

author: bob91343
author_rank: Super Contributor
timestamp: February 22, 2020, 06:41:41 am
content: My guess is that C4 is used to stabilize the amplifier. In other words, without it we have the gain set by the feedback resistors. With it, the low frequency gain is reduced to unity.

author: mk_
author_rank: Regular Contributor
timestamp: February 22, 2020, 07:23:15 am
content: try a square at your signal source CT instead a sine and see what happens

author: awallin
author_rank: Frequent Contributor
timestamp: February 22, 2020, 07:33:44 am
content: C4 and R6 make a high-pass with a time-constant of C4*R6=44 ms.if you are measuring 60Hz AC it has a period of 17ms.so C4 is there just to AC-couple the whole measurement, if you are not interested in signals that are slower than ~44ms.

author: Red_Micro
author_rank: Regular Contributor
timestamp: February 22, 2020, 03:39:13 pm
content: Another thing is the circuit is using dual power supply. I think that using a precision rectifier will allow me to use just single power supply, right? I want to use an ADC and scale the signal between 0-3.3V.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 22, 2020, 04:30:53 pm
content: The circuit is rather odd. The first OP part is still simple, an inverting AC coupled amplifier with quite a bit of gain. C4 is for AC coupling, not to amplify the OPs offset.The odd part is the kind of rectifier with the 2nd OP. It is only working above some threshold, so far from linear and not really suitable for a small signal.If the purpose is measuring mains AC with a µC, I would consider only doing AC amplification and the rectification in software. The usual µC internal ADCs are fast enough to sample some 50/60 Hz and the math is simpler than a nonlinear circuit. One could get away with OPs supplied with the same supply as the µC - so nothing to worry about higher voltage coming to the ADC. AC does not need precision OPs - just simple CMOS is sufficient.

author: Red_Micro
author_rank: Regular Contributor
timestamp: February 22, 2020, 06:18:40 pm
content: Quote from: awallin on February 22, 2020, 07:33:44 amI was just thinking C4 = 22uF is quite high, maybe it can be reduced.

author: laxch123
author_rank: Newbie
timestamp: February 23, 2020, 08:52:57 am
content: C4 is usually used as either high pass filter or capacitive coupling. Here, input impedanced of the op-amp is high enough. So C4 is just high pass. You can simplify the circuit by removing the second op-amp.

author: RoGeorge
author_rank: Super Contributor
timestamp: February 23, 2020, 09:02:48 am
content: C4 is there to stop any DC current going into the transformer.If you put DC into a transformer, that will magnetize the magnetic core, thus throwing the transformer out of specs.

author: David Hess
author_rank: Super Contributor
timestamp: February 26, 2020, 07:19:03 pm
content: C4 could be moved to the output and its value decreased to match the output load resistance if the 3.6 Hz high pass cutoff frequency is to be kept. This risks saturating the operational amplifier if the input offset voltage is too high.

author: graybeard
author_rank: Frequent Contributor
timestamp: February 26, 2020, 07:48:55 pm
content: Quote from: Red_Micro on February 22, 2020, 02:55:44 amC4 is just a DC blocking capacitor. I creates input high pass roll-off corner at about 3.6 Hz. It should have no significant effect on the 60 Hz input signal.It is shown as a polarized capacitor being used in a AC circuit. This is not good practice since it will see reverse bias.By inspection:C1 & C5 make the amplifiers act as ~670 Hz low pass filters. Becasue of the diodes C1 appears to only a low pass for negative inputs, it does nothing for postive inputs. Since the circuit is non-liner there may be subtle effects that a simulation would reval that a quick look does not.D4 & D5 are input protection diodesR5 || R6 || C3 create a single pole low pass at ~339 HzThe final stage in absolute value circuit. The 500K resistors seeem a bit large to me.