5
\$\begingroup\$

I know that opamps can be unstable when driving capacitive loads. The input to the opamp is more complicated than the schematic shows, but an unfiltered full wave bridge rectifier is the proper waveform. The waveform varies between 0Vpp to 15Vpp. Other circuits need the rectifier waveform. My Voltage to Current chip needs a smoother signal, so the current is more consistent. The opamp is an LM358 built as a buffer so the rectifier waveform doesn't get filtered. The LM358 is powered by 24Vdc.

The issue I am having is the rectified waveform increases its voltage linearly at its own slope. while the smoothed opamp output increases its voltage linearly on another slope. The two slopes intersect at some value determined by the resister and capacitor. How can I get the slopes to be parallel or the same slope?

Is there another circuit that will smooth out a rectified waveform into a DC voltage?

schematic link

\$\endgroup\$
8
  • \$\begingroup\$ What is the op-amp powered from? The output of the bridge, or something else? \$\endgroup\$ Mar 29, 2012 at 19:39
  • \$\begingroup\$ The issue I am having is the rectified waveform increases its voltage linearly at its own slope. while the smoothed opamp output increases its voltage linearly on another slope What does it mean? Are you talking about transient or DC curves? \$\endgroup\$
    – clabacchio
    Mar 30, 2012 at 7:08
  • \$\begingroup\$ The opamp is powered from a 24vdc power supply. \$\endgroup\$
    – Z2012
    Apr 5, 2012 at 20:24
  • \$\begingroup\$ Here is a graph of the the quoted text. ![enter image description here][1] [1]: i.stack.imgur.com/CD1yq.png \$\endgroup\$
    – Z2012
    Apr 5, 2012 at 21:00
  • 1
    \$\begingroup\$ Big mistake in my post. The rectified signal is at 350 Hertz not kiloherts sorry. \$\endgroup\$
    – Z2012
    Apr 9, 2012 at 17:51

4 Answers 4

2
\$\begingroup\$

It's a bit unclear, but you want a signal that is the rectified voltage into the IC, you're not trying to power the IC from the opamp, right? If the latter, there are much better ways to do that, like with a voltage regulator chip.

Instead of putting the filter after the opamp, put it before the opamp. At least put most of it there. You can put a little filtering after the opmap if you're worried about slight noise introduced by the voltage follower opamp.

Of course "filtering" is with respect to some other signal, usually ground. You don't show what you consider ground, and what the ground or reference voltage for the IC is. You may also have to attenuate the rectified signal. Unless the IC is running from 15 V or can specifically handle the large input voltage, you can't just feed that directly into it. Mostly likely you have to attenuate the signal into the voltage range of the IC power.

\$\endgroup\$
1
  • \$\begingroup\$ I am using the opamp to tap the rectified signal off the path so the other circuits still get the rectifed signal unfiltered. The opamp has a +24vdc power supply, and the capacitor is filtering to common. \$\endgroup\$
    – Z2012
    Apr 9, 2012 at 15:07
2
\$\begingroup\$

You've got an low-pass RC-filter with cutoff frequency

\$ f_C = \dfrac{1}{2 \pi \cdot R \cdot C} = \dfrac{1}{2 \pi \cdot 100\Omega \cdot 220\mu F} = 7.23 Hz\$.

Note that you can achieve the same value by increasing \$R\$ and decreasing \$C\$. For instance \$R\$ = 10k\$\Omega\$ and \$C\$ = 2.2\$\mu\$F. The opamp wouldn't mind this kind of load.

Like Olin says it's better to place the voltage follower after the filter. This will make the filter characteristic independent of the load.

\$\endgroup\$
0
\$\begingroup\$

I would like to answer my own question. What about using a peak detector that bleeds. The bleeding alows the detector to lock on to a lower voltage after some discharge time. circuit below. I captured the image of an overshoot condition, where the scope on the left is the rectified signal, and the right is the output. the rectified signal operates at 350 hertz between 0 and 10 volts. Everthing is powered by +24vdc. Opamp would be LM358. Main issue is that there is no overshoot.

z2012
enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ What's the impedance of your source? \$\endgroup\$
    – MattyZ
    Apr 10, 2012 at 1:54
  • \$\begingroup\$ @Bitrex The impedance of the source is the output impedance of the output buffer, which is an imverting opamp, from the AD215BY AD215_italic_ bold code \$\endgroup\$
    – Z2012
    Apr 10, 2012 at 15:39
-1
\$\begingroup\$

Is there another circuit that will smooth out a rectified waveform into a DC voltage?

Sure, use a linear voltage regulator such as the 7815: see datasheet for circuit

Or an LM317/LM350.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Not what he needs. A regulator will give you a DC voltage, but there's no relationship with the original signal. Whether the input is 20V or 25V, the output will always be 15V. That's not a signal. \$\endgroup\$
    – stevenvh
    Apr 8, 2012 at 6:04
  • \$\begingroup\$ -1. Worst idea ever? Hard to imagine that someone upvoted this \$\endgroup\$ Apr 8, 2012 at 6:43
  • \$\begingroup\$ Overmore, a regulator isn't meant for/can't smooth out the ripple of a rectified sine. That's what you need capacitors for. If you apply an unsmoothed rectified sine to an LM7815 you won't get a smooth 15V out. \$\endgroup\$
    – stevenvh
    Apr 8, 2012 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.