LT1078 precision rectifier - Page 1
Forum section: electronics / projects

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 26, 2017, 04:31:47 pm
content: Hi All,A beginner learning about opamps and this article seem to have caught while i was searching for examples and looked like a perfect fit for my project as well. I wanted something that could rectify the ac signal without any voltage drop and this seems to be ideal.As mentioned in the article herehttp://www.linear.com/solutions/1608the LT1078 can be setup to rectify an AC signal.The working mentioned in the article:For negative inputs, the first section operates as a closed-loop inverter (A=-1) and the second stage is simply a buffer for the positive output. When the input signal is positive, the first opamp output remains saturated near ground and the diode becomes high-impedance, allowing the signal to pass directly to the buffer stage non inverted. The composite effect is a full-wave rectified waveform at the output of the buffer.The design as it is works and it was something that i had picked up quite well but am trying to use (as well learn about the opamp working) it as differential opamp without loosing the rectification unfortunately i loose one half of the cycle. Maybe dave (@EEVblog) could explain it a bit more in his upcoming videos.Would appreciate if somebody could help me understand this.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: September 27, 2017, 02:59:57 pm
content: What do you mean "use it as a differential opamp". Schematics (and more details) of what you are trying to do might help.What signals are you working with and what frequencies?

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 27, 2017, 03:02:37 pm
content: Quote from: TheUnnamedNewbie on September 27, 2017, 02:59:57 pmThe schematic is the same as shown in the picture the input is an AC sine wave at 3vac. The output is fed into an arduino to measure the how much current is put out during a Spot weld.After a little more thought and timeDoes this make any sense. The idea here is to get the same volt as ac input but rectified.Www.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: September 27, 2017, 03:10:27 pm
content: I only now saw your spice schematic. I haven't run any numbers, but perhaps it's interesting to probe the input voltages of your opamp. My first suspicion is that due to the high resistance you are adding for the cable, you might be exceeding the common mode range of your devices.The difference between your application and the appnote as shown, is that in your schematic, both pins have a non-zero voltage. on the negative halfwave, both will be lower than ground and your opamp won't be able to make sense of these voltages.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 27, 2017, 03:38:43 pm
content: Thank you am trying to learn how the opamp works here with respect to the article.QuoteI've attached the capture from pins of Opamp1 (u1). The cable resistance is onlyQuoteIsn't that how the article is ? it is an ac signal in the article and in my schematic as well.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 27, 2017, 03:42:52 pm
content: That circuit works reasonably well with a single supply op. The input will be very close to GND and maybe a few mV (from the OPs offset) negative, but this is OK for a single supply OP. Depending on the OP it might be relatively slow, as the OP needs to come out of saturation. Also some OPs can have a significant input current if the input differential increases (with positive input voltages) - so not every OP will work in this circuit.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: September 27, 2017, 03:48:05 pm
content: anishkgt: In your new circuit, you should find that the output now is a nice rectified signal. The problem is that when you look at the voltages on the input of the amplifiers in the original circuit, they were very large. The differential might be small, but the common mode far exceeded your supply rails.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 27, 2017, 04:47:26 pm
content: It is Ok that the input voltage to rectifier circuit is exceeding the supply (especially going negative). However on a negative input the OP will output a positive voltage and thus way bring the OPs inverting input back to 0 V and thus inside the common mode range.For the rectifier to work the OPs negative supply needs to be at input reference level. So it will not work with a differential signal like shown. The shunt voltage will be rather small - thus an extra amplification stage would be a good idea. However this stage would need a negative supply.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 27, 2017, 04:49:44 pm
content: i had actually got the shunt measurement wrong. The new one attached here should make more sense.

author: alanb
author_rank: Frequent Contributor
timestamp: September 27, 2017, 05:08:58 pm
content: Quote from: anishkgt on September 26, 2017, 04:31:47 pmI seem to remember that Dave has already done a video on this topic. I will try and find it on YT

author: alanb
author_rank: Frequent Contributor
timestamp: September 27, 2017, 05:11:41 pm
content: It's in this episodehttps://youtu.be/jllsqRWhjGMEEVBLOG #490

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 27, 2017, 05:17:57 pm
content: i did come across that video but the heading mislead me to thinking it was not related.Thanks will have a look at it.

author: nugglix
author_rank: Regular Contributor
timestamp: September 27, 2017, 07:28:08 pm
content: While you're at YT you might want to have a look at the following videos.Exactly your topic, I'd say.andVery nice explanation of the workings and some pitfalls.Have fun!