+w2aew Nice! The first time I saw a DAC used like that was in a Jim Williams circuit - it was an epiphany for me that DACs had useful applications beyond interfacing microcontrollers with the "real world". :)

@@Analogzoo Late here, but wouldn't the high load capacitance and replacing the 1st OPAMP with a comparator (no compensation capacitor) make the comparator circuit unstable at high frequencies?

Hello, pls could you explain what sort of problem you had met. Im goin to build a circuit like this. So been aware of different issues is a big deal. Thanks a lot

+Adil Malim In dual supply circuits the "pull up" is not needed, since the "pull down" resistor is connected to ground, which is half way between the voltage rails. The "pull up" is added to single supply circuits such as this. See the "Astable Circuit Operation" section of this page (it explains it nicely with schematics and scope traces much better than I can describe it in a youtube comment! :) : www.learnabout-electronics.org/Oscillators/osc42.php

+devttys0 thanks for the link! i see why now! one resistor sets the appropriate schmitt trigger threshold for single supply operation and the other prevents cross over distortion. thanks!

+dwDragon88 Me either. :) Saw it in a Maxim app note and thought it was a pretty clever use of a digital pot.

Interesting, do you mean a second diode from the inverting to non inverting input to keep it within 0.7V? Also maybe decreasing the 100k input resistor could make it a bit faster at high freq?

@@g1ld There are various capacitances in the opamp, and when they're fully charged, and it's going to take a long time to pull all that charge back out. When an opamp goes into saturation, it takes a while to recover. When the next positive half cycle comes in, we're still trying to get from full negative back to zero so we can start tracking it. So the goal is to prevent the output from driving toward negative infinity as it tries to balance the inputs (which can never happen because of the diode). To do that, you can use another diode "inside" the other one in the feedback loop, reversed from the other one, such that it essentially puts the opamp in unity gain on the negative half cycle. Then the output doesn't have to work so hard, and recovery time increases, and frequency response goes up. My main point was that the slew rate is being "abused" a bit with this simple circuit, and it's on the negative (otherwise uninteresting) half cycle. The MCP602 has a unity-gain bandwidth of 2.8MHz. My theory is that the reason the comparator went faster in this test is that the comparator may have less capacitance, but more interestingly, it just gets pulled to ground on the negative half cycle rather than being pulled all the way to a negative rail. Or so it seems to me, an interested amateur! Google "fast peak detector," and look for two diodes.

Thank you for the explanation @Phreadrick !

You don't need the diode since we're not charging a capacitor here. We're only holding the CS line low while the clock sets the digipot output.

+Digger D It's common in RF to just use a Germanium or Schottky diode to rectify the signal and charge up a capacitor for the peak detection. For general input protection I think this would work fine, just remember that there's a 200-300mV drop across the diode, so you'd want to switch in the attenuator if the peak detector output is within ~300mV of your max safe value. I don't know your specific application, but for input protection I don't think you'd need it to be super precise...if the input signal gets anywhere near the max voltage level, you'll want to switch in the attenuator!

devttys0 Thanks for your reply. As I was pondering your video and my problem, I questioned the need for that precision as well. I plan on putting a frequency counter module PLJ-8LED from eBay into a case and I was thinking that adding automatic attenuation of a signal to protect the module would be a nice feature. So your reply and solution should work fine. I will use an LED to indicate attenuation, etc... Thanks again.

+Subhankar De The LM393 has an open collector output, which means that it will pull the output low when it needs to output a 0V state, but doesn't actively drive the output high during a high output state. To ensure a high output voltage when the comparator "goes high", you have to add a pull up resistor on the output pin.

+devttys0 Thanks Craig.

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I have the same question with you