What the fuck is wrong with this circuit? It seems sound in theory, but when I breadboard it up, I get all sorts of crazy numbers.
Here's the idea: The op-amp is wired up as a precision rectifier a.k.a a "super diode" with an unequal value for the input and feedback resistors such that when the signal is rectified, it is also amplified. When the switch is closed (on), a small signal of ~150mV AC goes in, and a half-wave rectified and filtered DC output of about 2.8V should be the result (when the voltage divider pot on the output is set to 5k ohms). When the switch is open (off), the circuit gets no signal, so there is nothing for the op-amp to pass and rectify, and so there should be no output at all. Normally, with no signal, the voltage at TP1, TP2, TP3, and the output should all be 0V.
When I actually breadboard this thing up and power it, TP1 and TP3 read ~4.5VDC, TP2 reads as ~5.4VDC, and on the output I get ~1.5VDC /with no AC signal on the input/. When I add the signal in, I get no change.
The op-amp I'm using is one half of a TL082 dual op-amp. I've tried each side of two different TL082 chips, and used both a bench power supply and a 9V battery. I've checked and rechecked the connections, even to the point of 2 complete scrap-and-rewires.
The upper diode should be connected to TP2 and not TP3 (see en.wikipedia.org/wiki/Precision_rectifier ) + should be ground but you're setting it to 4.5V thus offseting everything.
Colton Ortiz
OP here. I tried connecting the upper diode to TP2 but got the same result. I'm setting a virtual ground to the non-inverting input such that the circuit can presumably run off of a single +9VDC rail. Perhaps this circuit can only work in a dual rail configuration?
Gabriel Hill
you've discovered new laws of physics present this to the community at once
Angel Torres
>amplifier as a black box
Elijah Gonzalez
the rest of the circuit is simple as shit are you really such a sperg that you want to draw out the entire op amp schematic, especially when it has a bunch of useless auxiliary functions since it probably comes as a discrete package? KYS EE autist
t. EE
Hunter Anderson
>when the signal is rectified, it is also amplified
Have you thought of maybe amplifying it first and then rectifying it?
Christopher Rivera
If you do that, you have to reference 4.5V as your ground in the circuit and pretend the true ground is -4.5V.
>a small signal of ~150mV AC goes in, and a half-wave rectified and filtered DC output of about 2.8V should be the result (when the voltage divider pot on the output is set to 5k ohms).
Too much amplification. If you forget about the rectifying and filtering on the output, the output will be 100 times the "0.150V" input which is "15V" peak (or "21V" peak if that input is in RMS). The upper rail is only "4.5V" so you're way beyond saturated.
Nathan Fisher
>you're way beyond saturated Maybe if he hadn't drawn it as a black box he would have noticed. [spoiler]Just sayin[/spoiler]
Caleb Scott
>opamps >diodes Might as well use some shitty vst. Not only that but those awful solid state circuits fail to grasp simple thing about guitar amps - distorting distorted signal and repeating the process several times is different than raping some opamp once with 1000x gain and plopping bad lp after it.
Isaac James
So you want to input 150 mV AC and have a 2.8 V DC output?
Joseph Turner
OP here
I could do that, but it seems inefficient given that one of them could serve both functions. On top of that, I've tried setting it up as a standard non-amplifying diode where the input and feedback resistors are equivalent and I get the same issue
My understanding is that by setting 4.5 as the virtual ground, it always the output voltage to swing between +9V and 0V with the zero-crossing/bias point as +4.5V. Am I missing something on that front?
Also, yes, the output is supposed to saturate. Once it gets rectified, filtered, and stepped down (at the output), the voltage will be stable, and much lower than the saturation voltage of the op-amp.
Yes is exactly what I want. More specifically, I want to use the presence of signal as a soft "on" switch for another logic device on the output such that: signal present = power to the logic device signal not present = no power to the logic device
The problem is that the whole circuit saturates even when there is no signal on the input. Like, I can totally disconnect the signal source, power the op-amp, and BAM the input is somehow self-saturated. Adding the signal in alters nothing, which defeats my intended use of the circuit. My question is that the math and simulations all say this design is sound. So what gives?
Matthew Ward
super diodes are a meme
James Peterson
>My understanding is that by setting 4.5 as the virtual ground, it always the output voltage to swing between +9V and 0V with the zero-crossing/bias point as +4.5V. Am I missing something on that front? >The problem is that the whole circuit saturates even when there is no signal on the input. Like, I can totally disconnect the signal source, power the op-amp, and BAM the input is somehow self-saturated. >My question is that the math and simulations all say this design is sound. So what gives?
You're forgetting the biasing.
>Normally, with no signal, the voltage at TP1, TP2, TP3, and the output should all be 0V. >When I actually breadboard this thing up and power it, TP1 and TP3 read ~4.5VDC, TP2 reads as ~5.4VDC, and on the output I get ~1.5VDC /with no AC signal on the input
Which is what you should get. The op amp will make the + and - inputs equal so the op amp will output 4.5V + diode drop (0.7~0.9V) to get the 47u cap to float up to 4.5V. Then you have 1/3 voltage divider giving you 4.5 * (5/15) = 1.5V at the output out.
Jonathan Ramirez
/diy/ has a dedicated /ohm/ thread for this very thing:
David Hill
You tied the positive lead to 4.5v
So it's wrong to say there should be no output with the switch open
Luis Collins
Does that mean 150mV audio input and a digital output or do you need a linear relationship between input and output?
Elijah Hill
Yes is exactly what I want. More specifically, I want to use the presence of signal as a soft "on" switch for another logic device on the output such that: signal present = power to the logic device signal not present = no power to the logic device
Just use a voltage regluator
Gavin Gray
Ages ago I had to solve a similar case, I'm drawing from memory. The component values are merely 'order of magnitude' but the circuit is still fully functional. The 1K resistor sets the sensitivity and the 10uF capacitor at the emitter sets the holding time. I tested only the input stage. A 1kHz signal of 120mVpp (about 40mV RMS) yields 3Vp at the collector and clips at about 4V. You can omit the first diode, it only protects the input from 'abuse'.
The idea is to bias the first transistor in such a way that without input it is near saturation (but still in the active region). A positive transition doesn't do much but the negative transition is amplified and charges the capacitor at the emitter of the second transistor. The output stage adapts for TTL or other digital use. Without input signal the output is H and goes L when it appears. It can be inverted if you need active H.
Wyatt Baker
Linear relationship. Like I said, I want to use the signal as a power switch. In the final circuit there will be no physical switch between the signal source and the op amp. The presence of the signal will indicate power. I don't care about the data on the signal. Signal present at op amp = 2.8VDC out Signal not present at op amp = 0VDC out That's it.
Brody Rodriguez
That's brillaint, actually, but 1. Parts count uuughh 2. does it work on lower frequencies as well?
Connor Nguyen
this
Jacob Bennett
>does it work on lower frequencies as well? Yes. The original purpose was to drive a motor from the Doppler frequency output of a 9 GHz motion detector which could go down to a few Hertz on slow motion while the amplitude is a measure of distance. That's why the (relatively) big 10uF capacitor at the input. The speed of the motor followed the 0..3 V output connected to the analog input of a PWM power amp. >Parts count uuughh Not really. Normally you count the pins (solder points), not the parts. Leave off the last stage and compare to your (incomplete) original circuit. There is also a single supply version, LM358 instead of TL082.
Andrew Thompson
Naturally, the problem arises from using diodes to rectify a small signal because of their voltage drop, so by the time it's rectified there's barely anything. One way to solve it is to amplify it first.
Pic related is pretty crude but you get the idea.
Landon Roberts
A gain of 20 may look impressive but doesn't change that much and now you have two diodes in series. The main drawback is that the output is no longer ground-referenced, it floats. This configuration can be used to drive a moving coil indicator, but not a meter. The idea behind the precision rectifier is to use the open-loop gain of an opamp to (almost) fully eliminate the diode voltage as shown in the picture. The current through R2 (and the instrument) precisely follows the input signal, even if it is small.
Luke Barnes
>The main drawback is that the output is no longer ground-referenced
I guess you could do half wave then, and you'd get rid of the two diodes in series problem. The ripples would be all sorts of fucked tho.
Jackson Parker
Here it is.
(4.5^-1 + 4.5^-1 + 9^-1)^-1 = 5.4
I dont know much about EE, but from a basic view your scheme looks like a AC to DC converter. When your switch isn't rectifying why do you assume current isn't flowing? The idea is to control the direction of the current by the loop and diodes, not to cut it off.
Bentley White
(4.5^-1 + 4.5^-1 + 9^-1)^-1 x 3 = 5.4
Mason Ramirez
Yes, that will certainly work. The next step would be to get rid of the inconvenient (+/-) dual supply requirement (which is possible) and pretty soon you realize that a simple transistor circuit can do all you need. The real challenge is a single supply (5V), dc-coupled full wave precision rectifier circuit..
Kayden Rivera
>not using PSIM to simulate
This is your output when the switch is always on
Luis Walker
and when is always off
Adrian Howard
>mfw this entire thread
It's pretty cool recognizing all these words and still having no idea what they mean in this context. It lets me imagine I know what's going on.
Henry Reed
What happens when you remove the diode that is parallel to the resistor?
