op-amp input protection - Page 1
Forum section: electronics / projects

author: Harvs
author_rank: Super Contributor
timestamp: June 12, 2013, 04:08:29 am
content: I've got a high impedance input that I need to add over-voltage protection for a buffer configured op-amp for inputs up to about 100V through a 1M series resistance (above that will have MOV protection on the input.) The attenuation network is simply resistors switching in through reed relays. Voltage <0V will be clamped by the op-amps ESD diodes, but the op-amp has no ESD diodes to the positive rail (which seems to be pretty normal for very low bias current op-amps.)I'd like to achieve mV accuracy of the input voltage (with the attenuation network off), so the protection scheme needs low leakage currents. The op-amp input bias current is typically 7pA (MCP6V26.)So the approach in most of the application notes I can find is a diode connected JFET from the input to the positive rail. But these app notes are from the 90's and recommend parts that are now obsolete. My googling skills have failed to unearth anything from within the last 10yrs on this subject.Is this still the typical way it's done, or is there another part used these days? Either way, any recommendations for a part? Digikey's search can't sort by reverse leakage current...Thanks

author: Paul Price
author_rank: Super Contributor
timestamp: June 12, 2013, 09:41:02 am
content: :'(Google it!Connect this diode either to the positive rail, or to a voltage less than that which is above any expected positive level you will encounter to minimize the already small reverse-bias leakage.In other words, create a positive voltage clamp, if necessary using a diode like that shown below.

author: Harvs
author_rank: Super Contributor
timestamp: June 12, 2013, 11:11:26 am
content: Quote from: Paul Price on June 12, 2013, 09:41:02 amYes, I have.Thanks for the suggestion, but that diode has a leakage at least an order of magnitude greater than I was after. To compare to the device listed in this app note:http://www.ti.com/lit/an/sboa058/sboa058.pdfThat JFET has in the order 1pA - 10pA leakage compare to > 1nA at useful temps for that diode. 1nA of leakage will straight away eat all the 1mV error budget across a 1M resistor.However, I found another app notehttp://www.analog.com/static/imported-files/tutorials/MT-069.pdf,where they suggest using a C-B connected BJT (garden variety 2n3904).I know this is bread and butter stuff for a few of the characters that lurk on here, that's why I've asked...

author: Harvs
author_rank: Super Contributor
timestamp: June 12, 2013, 11:19:02 am
content: Obviously my part searches were some what crap earlier, I've found that JFET in other packages.http://www.digikey.com.au/product-detail/en/MMBF4117/MMBF4117CT-ND/3042811So yeah, <10pA.

author: Harvs
author_rank: Super Contributor
timestamp: June 12, 2013, 12:14:16 pm
content: No I hadn't seen that before. Very cool, I'll use that!Obviously I still need the JFET voltage clamp on the opamp input, but this is a very useful little circuit configuration.Thanks!

author: Christe4nM
author_rank: Unknown
timestamp: June 12, 2013, 12:33:56 pm
content: This is how I did it in a previous project, basically the default way as I understand it. To the left of this circuit was a large voltage divider to select an input range (1mV up to 100V) and at the bottom a selectable gain network. I don't know about the diode leakage of these diodes though. So you might indeed be better of with the other options mentioned.[edit: nah this is not gonna work for you, sorry. Leakage starting at 1 nA according to datasheet. Leaving post for future readers looking for just a standard circuit]

author: Paul Price
author_rank: Super Contributor
timestamp: June 12, 2013, 12:51:45 pm
content: What is the highest positive voltage allowable that would damage this input circuit? NE-2 neon bulbs have essentially infinite resistance until their striking voltage is reached.

author: Harvs
author_rank: Super Contributor
timestamp: June 12, 2013, 01:19:57 pm
content: Quote from: Paul Price on June 12, 2013, 12:51:45 pmInteresting. On the input side, it needs to keep the current across 1M resistor to under 2mA under fault conditions. So the limiting factor will be breakdown voltage of resistor(s) or PCB isolation distance. NE-2 bulb could be a bit low though as I want to handle inputs up to 100VDC. I just looked at a few and they were in the range of 65-75V for strike voltage. Maybe there's others though?

author: Paul Price
author_rank: Super Contributor
timestamp: June 12, 2013, 01:24:59 pm
content: Yes there are many voltages choices for neon discharge lamps for striking voltage. This technique of protection was even used in older Tektronix scopes on the inputs

author: Harvs
author_rank: Super Contributor
timestamp: June 12, 2013, 01:38:01 pm
content: Quote from: AcHmed99 on June 12, 2013, 01:25:33 pmNot quite, if you read the fine print they've got diodes connected to the negative rail that have a breakdown voltage that will absorb ESD spikes. They go on to say it wont protect against slow overvoltage conditions so external over voltage clamping must be used.They go on to sayQuote

author: ignator
author_rank: Regular Contributor
timestamp: June 12, 2013, 03:25:01 pm
content: I'm really confused by why you need any protection. If your schematic is correct, the 1Meg resistor provides all the current limit needed. The spec for MCP6V26 indicates all pins are ESD protected, and this is by the substrate back to back diodes inherent in the device.Typically a 10Kohm on up provides current limit for lightning protections that can exceed 750 volts in my domain area of avionics. The ESD built in diodes then can dissipate the lightning joules without damage. It's low impedance circuits like power input that a PITA to protect.Or I'm missing something.ESD protection on all pins (HBM, CDM, MM) ? 4 kV,1.5 kV, 300V, this is from section 1.1 of<http://ww1.microchip.com/downloads/en/DeviceDoc/25007B.pdf >

author: Harvs
author_rank: Super Contributor
timestamp: June 13, 2013, 01:37:57 am
content: You are correct that the 1M resistor does provide all the current limiting needed.But, as I said above, the same document you referenced specifically says that the input pins are ESD protected from positive voltages by the breakdown voltage of the ESD diodes connected to negative rail only, and this does NOT provide protection against slow over voltage conditions (like would be experience by applying a high input voltage while the attenuator is switched out.)Check out sections:4.2.1.2 Input Voltage Limitsattached is graphics snippets out of the document

author: ejeffrey
author_rank: Super Contributor
timestamp: June 13, 2013, 02:02:23 am
content: You can use an LED as an extremely low leakage protection diode as long as you do a good job of encapsulating it in black epoxy so that it doesn't act as a photodiode. Of course the clamp voltage will be higher.

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 02:24:26 am
content: Ejeffery: Surely you jest. You meant to type NEON LAMP but you wrote LED instead. A LED will kick in at a few volts, he wants protection perhaps above 100V.

author: P_Doped
author_rank: Contributor
timestamp: June 13, 2013, 02:35:47 am
content: I believe the issue is just how to place a protection element for positive overvoltage conditions where the protection element will not disturb the input current.The voltage which triggers the protection element should be (1) lower than the positive destructive voltage of the opamp and (2) ideally higher than the largest input voltage that the opamp can amplify so it retains the usable input swing of the opamp.I'm assuming that ejeffrey is suggesting connecting the LED between the input pin and the positive supply (in the same configuration a normal ESD diode would be). It would appear to me that this LED meets (2) and quite likely (1) as it will conduct at 2-3 volts above the positive rail which, if it's too much, can be brought into the process's voltage range by lowering the positive supply a little bit (sacrificing the top end of the common mode range).Assuming the leakage (or lack thereof) is correct, it sounds pretty viable to me.

author: Rufus
author_rank: Super Contributor
timestamp: June 13, 2013, 02:39:55 am
content: Quote from: Harvs on June 13, 2013, 01:37:57 amI don't really know how to read that or between the lines. Makes no sense having a 2mA polarity less current limit and a +ve voltage limit which couldn't possibly cause 2mA to flow into the pin and then insisting that both limits apply.They often do that for parts with normal diode clamps but it is read between the lines as that voltage (often Vdd + 0.3) won't cause any current flow and exceeding that voltage (which you have to to get any current) will not cause damage if the current is limited to 'x'.As you have discovered providing +ve protection has serious impact on input currents so you would think they would be a bit more helpful with the chip and/or advice.BAV199 dual diodes have a typical 3pA leakage at 25C and 75v, sadly max is a lot worse.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 13, 2013, 02:47:57 am
content: Yes, I am suggesting using an LED in the same way you use a normal diode for input protection: from the opamp input to the positive rail. LEDs have a much, much lower leakage current than silicon diodes.

author: Rufus
author_rank: Super Contributor
timestamp: June 13, 2013, 02:54:33 am
content: Quote from: ejeffrey on June 13, 2013, 02:47:57 amBut can you buy LEDs with specified leakage?You would be winging it or testing it same a using BAV199s and hoping you get near typical parts.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 13, 2013, 03:04:12 am
content: No, it isn't specified in the data sheet. Still, any off the shelf red LED will negligible reverse leakage -- femtoamps. This is just intrinsic to the semiconductors used. The higher bandgap means there is zero thermal population of the conduction band. The problem, and the reason it isn't specified on the data sheet, is that you have to do a really good job of making it light-tight. Any stray light will cause an error current. This is not something you want to do for a normal design, but it sounds like the OP is really trying to push the limit -- in which case it might be worth it.I don't know if anyone makes encapsulated diodes with a large bandgap semiconductor designed for this type of application. In that case you could probably find a data sheet specification.

author: Harvs
author_rank: Super Contributor
timestamp: June 13, 2013, 04:40:53 am
content: Rufus, I agree it doesn't make a lot of sense how they've written it, and when I originally decided on the part I was assuming I'd be fine with the +/-2mA limit like you said.I also find it a bit frustrating that all the major semi manufactures discuss this issue in app notes, but give no help as to what actual part could be used for protection (i.e. they just say to use a JFET, despite most of them having ~1nA leakage).The BAV199 looks like it might be a good choice, the BAS116 looks like the identical single diode version. That's a huge spread from typical to max though.An Analog Devices app note suggest using a C-B junction of a BJT (2N3094), or the B-E junction if the voltage is low enough (even lower leakage). That leakage isn't spec'ed anywhere either, but I found a passage in Bob Peace's Troubleshooting Analog Circuits where he said that a 2N3904 used in this configuration will have <10pA leakage at room temp. So that would certainly be an easy answer.I really just need less than about 100-200pA at normal temps, so up to about 50 C.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 13, 2013, 06:21:37 pm
content: There may be a better way. Since your opamp is a unity gain follower, you can connect the protection diodes to the output. As long as the input is in-range there will be nearly zero voltage across the diodes, so nearly zero current. In essence, this works like a guard ring. When the output clips at the rails, the diodes will start conducting. The opamp output can easily sink the necessary current.This is not a panacea. The offset voltage isn't zero, so there will still be some leakage current. You still want to use low-leakage diodes. Also, just like a guard ring, the capacitance and resistance of the diodes will affect AC performance. Still, for DC applications this can help quite a bit.