Author Topic: clipping a fast negative pulse  (Read 5042 times)

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Offline brumbarchrisTopic starter

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clipping a fast negative pulse
« on: May 08, 2019, 09:43:17 am »
So I have to deal with a big and fast negative pulse, as in the attached picture. The requirement is to reliably and as accurately as possible (nice spec, right?) clip or clamp this pulse to a fixed voltage level (does not need to be adjustable) between -0.8V and -1.4V across temperature (-20C to 60C) and across the parameter variations that the components employed would have.

What are your suggestions?

Best regards,
Cristian
 

Offline OwO

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Re: clipping a fast negative pulse
« Reply #1 on: May 08, 2019, 09:58:25 am »
Source impedance? is adding another series resistor acceptable?
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Offline brumbarchrisTopic starter

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Re: clipping a fast negative pulse
« Reply #2 on: May 08, 2019, 10:05:37 am »
This pulse is coming off of a ~2mH coil which has some 10Ohm DC resistance, when supply current is suddenly cut. We also have a 1k damping resistor in parallel with the coil.
 

Offline dzseki

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Re: clipping a fast negative pulse
« Reply #3 on: May 08, 2019, 10:17:29 am »
Usualy a diode is used for such application.
HP 1720A scope with HP 1120A probe, EMG 12563 pulse generator, EMG 1257 function generator, EMG 1172B signal generator, MEV TR-1660C bench multimeter
 

Offline AndyC_772

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Re: clipping a fast negative pulse
« Reply #4 on: May 08, 2019, 10:21:15 am »
Do you mean it needs to be clamped to a fixed, accurately controlled, known level (which happens to be between those limits)?

Or, is the requirement simply to keep the voltage within those limits in order to not damage something?

If it's the latter, then a fast silicon diode to GND may be all you need. If it's the former, you'll need something much more involved and complex.

Offline brumbarchrisTopic starter

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Re: clipping a fast negative pulse
« Reply #5 on: May 08, 2019, 11:06:18 am »
This is what I need, sorry for not making it more clear to begin with:
Quote
Do you mean it needs to be clamped to a fixed, accurately controlled, known level (which happens to be between those limits)?
Yes!

A simple diode will not do, as it is by far not accurate enough. To put a figure on it, let's say the clamping voltage needs to have a tolerance of +/-20mV, and of course, it has to be within the already stated limits.

Best regards,
Cristian
 

Offline AndyC_772

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Re: clipping a fast negative pulse
« Reply #6 on: May 08, 2019, 12:32:47 pm »
In that case, I'd probably look toward using a fast comparator to control an SPDT analogue switch.

When the input is above threshold, the switch output is connected to the input. When it's below threshold, the comparator output flips state and switches the threshold voltage directly through to the output.

You'll almost certainly want some input scaling, an output buffer, and an actual protection diode too. Maybe use two diodes in series to give you a worst case (ie. most negative possible) input of ~-1.5V.

Offline OM222O

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Re: clipping a fast negative pulse
« Reply #7 on: May 08, 2019, 01:59:57 pm »
simple solutions include: a schottky diode going from output to ground (with the positive lead (anode) connected to ground) and a snubber RC circuit.
Edit: I'm not sure how fast each method is, but they are the go to for this application, you should be able to find suitable parts if you go through a lot of datasheets  :-DD
« Last Edit: May 08, 2019, 02:01:37 pm by OM222O »
 

Offline brumbarchrisTopic starter

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Re: clipping a fast negative pulse
« Reply #8 on: May 08, 2019, 02:59:54 pm »
The variable forward drop on the schottky is far to great for this application. Definitely larger than 20mV, even for the same device, across temperature.

Comparator + SPDT analogue switch is a possible option, albeit a bit more IC heavy than the single opamp solution I had in mind, which is depicted in the attached picture. Of course, this needs to use a fast opamp, but also the comparator would have to be fairly fast.
 

Offline OM222O

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Re: clipping a fast negative pulse
« Reply #9 on: May 08, 2019, 03:24:16 pm »
The variable forward drop on the schottky is far to great for this application. Definitely larger than 20mV, even for the same device, across temperature.

Comparator + SPDT analogue switch is a possible option, albeit a bit more IC heavy than the single opamp solution I had in mind, which is depicted in the attached picture. Of course, this needs to use a fast opamp, but also the comparator would have to be fairly fast.

that solution is also using a diode  ??? how can that be any more accurate? It adds accuracy to the triggering point if that's what you're after, but the output voltage will still vary significantly based on the diode temperature, current, etc

The main questions is: why do you want such a precise negative voltage for? if it's meant to be used for something, then just use a charge pump to create the negative voltage. if it's meant for protection, then you don't need 20mV accuracy?  :-//
 

Offline dzseki

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Re: clipping a fast negative pulse
« Reply #10 on: May 08, 2019, 04:09:28 pm »
Also a question: do you need to actually clamp that pulse or you just need an output voltage as per specified during the pulse?

What about a "comparator like" feedback loop using an opamp that drives a shunt transistor, if the voltage is above say 0V the transistor is closed, if the pulse gets negative the opamp and the transistor would clamp the output at the threshold voltage.
HP 1720A scope with HP 1120A probe, EMG 12563 pulse generator, EMG 1257 function generator, EMG 1172B signal generator, MEV TR-1660C bench multimeter
 

Offline StillTrying

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Re: clipping a fast negative pulse
« Reply #11 on: May 08, 2019, 04:47:06 pm »
There's not enough clear information for sensible suggestions. Such as whether it's the 64V pulse that needs to be clamped, or it just needs to be prevented from getting to the next stage. And if the next stage is only ever going to see a very stable -0.8V then that's easy. :)
.  That took much longer than I thought it would.
 

Offline OM222O

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Re: clipping a fast negative pulse
« Reply #12 on: May 08, 2019, 05:01:45 pm »
Also a question: do you need to actually clamp that pulse or you just need an output voltage as per specified during the pulse?

What about a "comparator like" feedback loop using an opamp that drives a shunt transistor, if the voltage is above say 0V the transistor is closed, if the pulse gets negative the opamp and the transistor would clamp the output at the threshold voltage.

the main question still remains: WHY WOULD YOU WANT TO CLIP A PULSE TO A VERY SPECIFIC VOLTAGE WITH LESS THAN 20mv ACCURACY!
if a negative rail is needed, this a really bad way of generating it, and if it's done for protection purposes, then the accuracy should not matter  :-/O
 

Offline capt bullshot

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Re: clipping a fast negative pulse
« Reply #13 on: May 08, 2019, 05:01:56 pm »
So the circuit doesn't have to absorb the pulse energy but just to limit the negative voltage to a stable level, and you can apply a high source impedance like that 2k2 resistor shown in your circuit?

So one might consider a fast enough Rail-to-Rail I/O OpAmp to buffer the signal. Apply some coarse input protection using diodes, set its negative supply voltage to the desired clamping level and have your output clamped to the negative voltage.

Or do you need a signal path without active devices in the non-clamping state, like your circuit? One can speed up the OpAmp by doing some biasing tricks to prevent output saturation in the non-clamping state. This will increase the clamping action at the falling edge of your signal.
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Online SiliconWizard

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Re: clipping a fast negative pulse
« Reply #14 on: May 08, 2019, 06:42:37 pm »
the main question still remains: WHY WOULD YOU WANT TO CLIP A PULSE TO A VERY SPECIFIC VOLTAGE WITH LESS THAN 20mv ACCURACY!

I'm not sure why either. Especially since the OP mentioned at first that it should clamp between -0.8V and -1.4V. So if 20mV accuracy is needed, does that imply that this clamping voltage is going to be adjustable? Because this is the only way I see that those two requirements would match?

 

Offline OM222O

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Re: clipping a fast negative pulse
« Reply #15 on: May 08, 2019, 07:10:19 pm »
the main question still remains: WHY WOULD YOU WANT TO CLIP A PULSE TO A VERY SPECIFIC VOLTAGE WITH LESS THAN 20mv ACCURACY!

I'm not sure why either. Especially since the OP mentioned at first that it should clamp between -0.8V and -1.4V. So if 20mV accuracy is needed, does that imply that this clamping voltage is going to be adjustable? Because this is the only way I see that those two requirements would match?



He mentioned it's NOT ADJUSTABLE and the value doesn't matter. it just has to be accurate for some reason  ??? :wtf:
 

Offline David Hess

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Re: clipping a fast negative pulse
« Reply #16 on: May 08, 2019, 09:15:39 pm »
A simple diode will not do, as it is by far not accurate enough. To put a figure on it, let's say the clamping voltage needs to have a tolerance of +/-20mV, and of course, it has to be within the already stated limits.

The variable forward drop on the schottky is far to great for this application. Definitely larger than 20mV, even for the same device, across temperature.

When diodes are used as precision clamps, then usually two are used with one compensating for the voltage drop and change in voltage drop with temperature of the other.  This leaves the logarithmic change in voltage with linear change in current (60 millivolts per decade) but you gave no specification for current.

If more precision is required and the speed is not too great, then a shunt regulator can be used as a clamp.
 

Offline brumbarchrisTopic starter

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Re: clipping a fast negative pulse
« Reply #17 on: May 09, 2019, 09:33:19 am »
Hello everybody, thank you for your replies and patience.
Here are some clarifications:

Quote
WHY WOULD YOU WANT TO CLIP A PULSE TO A VERY SPECIFIC VOLTAGE WITH LESS THAN 20mv ACCURACY!
The signal is going to be then "signal processed" and eventually fed to an ADC input. As there will be a large number of products that will be manufactured, we will need to account for component-to-component variation between various samples. I guess one of the concerns, something I should have probably mentioned in the first post, is that once the input signal goes above -0.8V it must be not altered or distorted in any way. That is where the actual information that we are interested in is.

Quote
Especially since the OP mentioned at first that it should clamp between -0.8V and -1.4V. So if 20mV accuracy is needed, does that imply that this clamping voltage is going to be adjustable?
As mentioned in the OP, the level does not need to be adjustable, but it has to be precise. So for instance if you suggest a solution that always clamps at -1.2V+/-20mV that is good. Equally good to a solution that clamps at -0.9V+/-20mV. Of course, the nominal clamping level needs to be known (if it is 1.2V or if it is 0.9V) and reproduce able across samples.

Quote
If more precision is required and the speed is not too great, then a shunt regulator can be used as a clamp.
Unfortunately speed is quite high, and I do not think a basic shunt regulator would provide this accuracy, particularly if it is BJT based.

Quote
that solution is also using a diode  ??? how can that be any more accurate?
It is definitely more accurate as the opamp will drive its output so as to compensate any variations in the diode voltage drop until the VOUT node (which in this case is connected directly to the inverting input of the opamp) reaches the same value. AFAIK, the opamp drives its output so as to always keep its inputs at an equal level.

Quote
So one might consider a fast enough Rail-to-Rail I/O OpAmp to buffer the signal. Apply some coarse input protection using diodes, set its negative supply voltage to the desired clamping level and have your output clamped to the negative voltage.
Could you elaborate a bit more on that please? It's "sounds" relatively similar with the opamp based variant that I was considering, and the schematic of which I indicated in one of the above posts.

Best regards,
Cristian

 

Offline capt bullshot

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Re: clipping a fast negative pulse
« Reply #18 on: May 09, 2019, 01:07:29 pm »
See that quick and dirty schematic, hope you get the point. Component choice is up to your requirements, choose R2 to limit the current into the OpAmp input to a level typically lower than 5mA, better 1mA (check the datasheet and appnotes).



Anyway, your approach appears rather unusual, for what you describe (accurate and fast measure the input signal above the clamping level) I would consider coarse clamping by diodes and wouldn't rely on a stable clamping level for the rest of the circuitry. Maybe you should re-think the rest of the design once more.

Edit: AFAIR, some special OpAmps exist that allow for signal clamping to a defined level (set by a voltage applied to a pin), independent of the supply voltage. I believe, they were mainly used in video applications.
« Last Edit: May 09, 2019, 01:10:19 pm by capt bullshot »
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Offline OM222O

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Re: clipping a fast negative pulse
« Reply #19 on: May 09, 2019, 01:11:34 pm »
A basic simulation here:
http://tinyurl.com/y4eeqnjd
it shows that the negative edge is clipped well, but so is the positive edge due to the supply rails of the op amp (+-15v) so that can be a limitation.
Also it has so sink / source considerable current in order to do that (about 30mA which is on the high side for op amps, they're not meant for driving loads)
I'm not sure if this will actually work in real life when you build it.
 

Offline David Hess

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Re: clipping a fast negative pulse
« Reply #20 on: May 09, 2019, 05:09:53 pm »
There is another way to make a precision clamp with diodes and it is very fast.

Use the signal, now low impedance, to drive the input of a diode bridge.  Set the top and bottom currents through the bridge to produce the correct clamp levels into the load impedance.  Some oscilloscope input protection circuits used this method as shown below in the Tektronix 1GHz 7A29 vertical amplifier.
 

Offline Marco

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Re: clipping a fast negative pulse
« Reply #21 on: May 09, 2019, 06:31:30 pm »
That's more a clipped voltage follower, the input does not provide the power and is not itself clipped.

The only reason I can think of for the requested circuit is that the flyback voltage on the primary of a transformer needs to be limited to a precise voltage to also generate a precise voltage on the secondary.
 

Offline T3sl4co1l

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Re: clipping a fast negative pulse
« Reply #22 on: May 09, 2019, 07:06:26 pm »
Wait.  Why are you clipping the pulse you fought so hard to perfect?

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Offline brumbarchrisTopic starter

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Re: clipping a fast negative pulse
« Reply #23 on: May 09, 2019, 08:24:29 pm »
Quote
Wait.  Why are you clipping the pulse you fought so hard to perfect?

In order to further signal process it and feed it to an ADC eventually.


Quote
A basic simulation here:
http://tinyurl.com/y4eeqnjd
it shows that the negative edge is clipped well, but so is the positive edge due to the supply rails of the op amp (+-15v) so that can be a limitation.
Also it has so sink / source considerable current in order to do that (about 30mA which is on the high side for op amps, they're not meant for driving loads)
Hey, nice simulation. But it actually suggests the circuit would work. Clipping the positive side of the signal is not a problem, as I do not need to clip a sinewave, but a mostly negative waveform, as indicated in the OP. The opamp will have to be appropriately selected, of course, speed and current output are critical, offset and other usually important parameters not so much, in this case.

Quote
See that quick and dirty schematic, hope you get the point. Component choice is up to your requirements, choose R2 to limit the current into the OpAmp input to a level typically lower than 5mA, better 1mA (check the datasheet and appnotes).
Thank you for the suggested schematic, kapitän, but would that circuit be any more accurate than the clamping diodes? Their forward voltage varies quite a lot!


Best regards,
Cristian
 

Offline Marco

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Re: clipping a fast negative pulse
« Reply #24 on: May 09, 2019, 09:48:12 pm »
What's there to ADC if your clipping circuit works?
 


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