C-multiplier again

amp

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at.

I have used the lighting of the LED to make my "I'm maxed out" indicator out of the very same part that did the limiting.

In a low noise application, you don't want light hitting the LED that does the limiting (at least with a red one) because the LED is photovoltaic.

You don't want light hitting your "low noise reference diode"

I suggested a blue LED under the assumption that the higher forward voltage would make the "SNR" better by making the "signal" bigger. Since then, I have had a reason to rethink that. I think that the impedance of the so called "super bright" RED is lower than that of the BLUE.

BTW: There is no such thing as a superbright LED. LEDs come in 3 brightness classes: (1) Are you sure this is on (2) Way to dim (3) I guess this will do

I use LEDs in full sunlight.

spec

ercap

s measured by Win

Is the leakage current in them very noisy? I have always assumed that the noise current part of the leakage current was equal to the shot noise of a semiconductor device. If it is like a resistor, the noise would be less.

On May 26, 6:27=A0pm, Mike wrote: [....]

I have a slightly improved version:

!\U1 Ref ------[R]--+------------! >-----+- Low noise version ! !/ ! ! ! C1[C] ---[R]---+---[R]-+ ! ! ! ! ! ! /-!-- [R] ---+--< ! ! \+!----------+ U2 ! [C]C2 ! GND

The time constant at C2 can be huge because the leakage of C2 doesn't cause your reference to be low.

The gain of U2 makes C1 look larger than it is. The circuit is second order so it does have a noise peak at the corner but that can be at a very low frequency. The resistors on U2 are low value ones because they add directly to the noise.

Beatcha to it. My green power LED is already the +2 reference for some opamps used as TTL-input level shifters.

John

I have a friend who manages the LED lighting division for a big electrical products company... you buy their stuff at Home Depot. He snuck me a developmental white LED that, at 14 volts and 0.7 amps, looks like a welding arc. It will literally blind you for a minute if you look straight at it. Their initial target market is street lights.

John

I don't know. I tried to measure the noise across a charged one and didn't see anything at microvolt resolution. They don't leak much.

Unlike a wet cap, if you increase the voltage slowly, there's no apparent increase in leakage current right up to the instant it fails shorted, at about 2x rated voltage.

Megohms of leakage into milliohms of ESR will attenuate that leakage noise by roughly a billion.

John

Tkat's R3 all right.

The text says something about surge protection for the opamp input.

John

Well that's rather disappointing.

Admittedly I haven't tried, but I don't think you'll see anything after staring into the average welding arc for a minute. ;-)

Tim

On a sunny day (Thu, 27 May 2010 06:19:51 -0700 (PDT)) it happened MooseFET wrote in :

That is true.

I have never measured that, next time I will have a look with the LEDs I have.

I think there is a fourth class, AAAAUAUWW --- Way too bright. Those are the one I have, even at 10 mA, green, and blue, no way can you look into one.

I try to use clear plastic LEDs, so if you see color it means it is 'on'.

Ultimately you can flash a LED to make it stick out a bit more if it is an important indicator. After that transflective LCDs, with flashing background and beep, I have some very small 5 V beepers that almost make you jump :-)

From the AD797 datasheet:

Maximum Differential Input Voltage : +/- 0.7 V

The AD797's inputs are protected by back-to-back diodes. To achieve low noise, internal current limiting resistors are not incorporated into the design of this amplifier. If the differential input voltage exceeds +/- 0.7 V, the input current should be limited to less than 25 mA by series protection resistors. Note, however, that this will degrade the low noise performance of the device.

The AD587 supplies 10V. There are big electrolytics connected to both inputs of the AD797. In the event of a hard short on either input, the maximum current could be (10 - 0.7) / 49.9 = 0.186 Amp.

This exceeds the rating by 7 times, so the 49.9 ohm resistor is not providing much protection against a hard failure, and it probably can be removed.

Unfortunately, most modern low-noise op amps seem to have similar input restrictions. The best option seems to be to provide as much protection around the circuit as possible to prevent high input currents.

And have plenty of spares on hand:)

Mike

I'm having trouble figuring the values. Can you post that in LTspice?

Mike

Not much protection. The reference is 10V. Both inputs have large electrolytics. A hard short, for example on the low-noise output, will exceed the maximum current and destroy the device. If the resistor is not helping protect the inputs, why keep it?

Mike

That's not what I said.

John

Beats me. Ask the designer.

John

I'm sure it's there precisely because it does protect the diodes.

Depletion fets or SSRs make nice current limiters and active protectors.

John

Against a hard short? How? Since each input has a huge electrolytic to ground, and the voltage difference is close to zero, why not simply add back-to-back power schottkys across the inputs and let them take any surge.

LND150 is nice but around 2k. Got any less than 50 ohms?

Mike

DN3525 6 ohms at 0V, but your fault current is going to be something like an ampere...

Hey, even better! Very low frequency noise will result from temperature fluctuations wiggling Vf, but luckily that's not a problem in our current situation.

John