schottky diode input protection - Page 1
Forum section: electronics / projects

author: brownt
author_rank: Regular Contributor
timestamp: January 17, 2018, 06:20:36 am
content: Hi,I am using a pair of schottky diodes to protect the input lines to a digital potentiometer.The digital potentiometer is powered by the same 5volt line as the microcontroller that drives it via SPI, and the same 5volt line as the schottky diodes.There is a small voltage at the anode/cathode junction about 50mV when I use BAT54S, and about 150mv when I use 1N5817.I wonder why that is so, and how to stop it from happening.thanks

author: georges80
author_rank: Frequent Contributor
timestamp: January 17, 2018, 07:21:39 am
content: Reverse leakage current. Quite a bit worse for the 1N5817 vs the BAT54... Schottky diodes typically have quite high leakage currents (which increase with temperature) versus a standard silicon diode.cheers,george.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 17, 2018, 07:42:55 am
content: That IC more than likely has over voltage diode protection built into it so adding more diodes on the outside is redundant. Maybe a 5 volt crowbar zener if you are worried about it ?

author: brownt
author_rank: Regular Contributor
timestamp: January 17, 2018, 08:05:37 am
content: High leakage current, I see thanks.The IC has ESD protection. When the circuit is connected to an external device at the socket, the external device provides 5 volts and so I wanted to protect against an over voltage there as the dig pot will not function correctly if the voltage across its internal potentiometer is higher than its VCC.Perhaps someone could connect 10 volts by mistake. Would the ICs built in protection be ok for that?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 17, 2018, 09:13:26 am
content: ICs in general have diodes on there input in such a way that 5.6 volts will try to pull up the 5 volt power supply up or - .6 volt will try to put ground to negative. Assuming a solid power supply this will not happen so the IC's internal diodes are protecting the IC within a range of current. The crowbar 5 volt zener protection goes on the power supply, nothing to do with the IC. If for any reason the power supply goes above 5.6 volts the zener will turn on destroying itself with excessive current. What was a zener diode is now a 0 ohm short killing the power supply. The idea is to sacrifice the zener diode to save the IC. It is used more for microprocessor boards where sacrificing a cheap zener to save an expensive microprocessor board makes sense.

author: brownt
author_rank: Regular Contributor
timestamp: January 17, 2018, 11:54:47 pm
content: I think maybe I have not done this correctly. What I wanted to do was prevent greater than 5 volts being applied across the potentiometer, because the digital pot datasheet says that if the voltage across it is more than its VCC, then damage will occur.But this dual diode I have implemented seems to prevent excessive voltage as in ESD, much like the IC probably already has built in. Is what I need then perhaps just some 5v zener diodes?

author: langwadt
author_rank: Super Contributor
timestamp: January 18, 2018, 12:28:23 am
content: Quote from: John Heath on January 17, 2018, 09:13:26 amthe ESD diode will fry long before the zener

author: Yansi
author_rank: Super Contributor
timestamp: January 18, 2018, 12:37:55 am
content: Never use schottky diodes for analog circuit protection. They leak as ... fill in your favorite word.I did once helped a friend to design something, where he fed a signal from a 5V supplied opamp into a 3V supplied ADC. Told him to use a 4148 diode as a protection clamp together with a series resistor. A while after he was back complainig it does not work. After bit of investigation, he used a schottky diode for the high side clamp. But not signal schottky, it was some kind of 0,5A jobbie there. It leaked substantial amount of current, even at 5V!So... use a regular diode instead. BAV99 should be great. Small leakage.If there is a need to have even smaller leakage, there are ways to improve on bootstrapping the diodes, but I think that is really not required here.If using clamps like this, just make sure to provide some small resistance to limit fault current and a use sufficient overvoltage protection at the supply rail the diodes are clamping to.

author: rs20
author_rank: Super Contributor
timestamp: January 18, 2018, 12:41:00 am
content: A somewhat orthogonal point: Those protection diodes (whether built-in to the IC or provided separately) work best in conjunction with a current-limiting resistor. For example, let's imagine using a 100 ohm resistor in a situation where 6V is placed on a pin in a 5V system.If there is no limiting resistor, placing 6V on the nominally 5V-max pin will place a full 1V across the protection diode, leading potentially to a current flow of ~amps and destruction of the protection diode (depending on how strong the 6V source is, this could kill even dedicated diodes.)If there is a 100 ohm limiting resistor, then some current will flow, perhaps with a drop of 0.5V across the diode, and 0.5V across the limiting resistor. This leads to a flow of 5mA, which will typically be totally fine for both standalone diodes as well as the diodes typically built-in to ICs.My point here is that, whereever possible, your first port of call should be adding resistance -- you only fall back to explicit protection diodes if you still need them (even if you obviously needed protection diodes all along, they make little sense without the resistor.) Generally speaking, digital pots should not be carrying a lot of current (for example, they certainly shouldn't be used for directly driving headphones, etc), so adding a little resistance to the pot terminals should have a negligible impact of the performance of the system, while massively improving its robustness.

author: brownt
author_rank: Regular Contributor
timestamp: January 18, 2018, 01:12:24 am
content: I see, seems fair enough thanks. but I still need to prevent greater then 5volts across the pot. Does the dual schottky solution do that?

author: rs20
author_rank: Super Contributor
timestamp: January 18, 2018, 07:01:18 am
content: Your statement "I still need to prevent greater than 5volts across the pot" is probably far too strong of a constraint. Where's the datasheet for your digital pot? Since you haven't provided one, I'll use this one as an example (first one that I found):MCP4017 datasheetIf you look at what theactualrequirements are, in the absolute maximum ratings section:QuoteThat shows you that not only is a voltage of up to VDD+0.3V (i.e., 5.3V) OK, but it also describes the limits of the included clamp diodes. So you don't need to worry at all about the voltage spec; as long as you keep the shunt currents below 20mA, you'll be fine.For instance, the 100 ohm resistor I mentioned earlier will provide protection for voltages up to 7V (20mA * 100 ohm + 5V). If you use a 1k resistor, then suddenly that goes up to protection up to 25V. This is all without using any extra diodes.If you want to use silicon protection diodes (to avoid leaky Schottky diodes), then it's pointless to put them directly in parallel with the dpot pins because the dpot clamp diodes will die before the Silicon diodes even start conducting. If, instead, you place the silicon diodes before the protection resistor, then you can create an extremely robust solution indeed with quite small resistances.There's basically no configuration in which Schottky diodes make sense, because if you put them directly on the dpot pins, you can't guarantee that the diode will actually start conducting before the clamp diodes in the IC, and if you put them before the resistor, there's no point in them being Schottky because even a normal 0.7V drop diode will work perfectly well and will mess with your signal less.The point is, this is all simulatable and calculatable using the datasheet figures; there's nothing magical or different about the job of protecting inputs.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 18, 2018, 10:07:52 am
content: Quote from: langwadt on January 18, 2018, 12:28:23 amHmmm , good point. The zener is only protecting the power supply itself from going over 5 volt. 10 volts getting in through an IC input with ESD would just smoke the IC .

author: brownt
author_rank: Regular Contributor
timestamp: January 19, 2018, 11:10:29 am
content: I see. The MCP4341 is very similar, though both datasheets sayNotice:Stresses above those listed under “MaximumRatings” may cause permanent damage to the device. This isa stress rating only and functional operation of the device atthose or any other conditionsabove those indicated in theoperational listings of thisspecification is not implied.Exposure to maximum rating conditions for extended periodsmay affect device reliability.

author: rs20
author_rank: Super Contributor
timestamp: January 19, 2018, 12:49:59 pm
content: Quote from: brownt on January 19, 2018, 11:10:29 amYah. That's why I'm explaining how to stay below those limits.

author: brownt
author_rank: Regular Contributor
timestamp: January 20, 2018, 05:32:31 am
content: are you saying, that if the current stays below 20mA, then say 10 volts across the pot is not going to cause damage?

author: rs20
author_rank: Super Contributor
timestamp: January 20, 2018, 02:05:10 pm
content: Quote from: brownt on January 20, 2018, 05:32:31 amVoltage and current are dependent. It is physically impossible to put 5 volts in excess of Vdd on a pot pin while having less than 20mA flowing through the protection diode associated with that pin. Just like it's impossible to put 5 volts (the right way) across a Schottky diode while having less than 20 mA flowing. If you're using a resistor to limit the current through the diode, then there's no longer 5 voltsacross the diode.So your question makes no sense, it is physically impossible to set up the hypothetical experiment you are describing. It'snota dumb question though, it helped reveal the slight flaw in your way of thinking. If you want to present a circuit diagram and ask what happens/if it is OK, feel free to post it and of course we can help predict what will happen.So ensure that the current stays below 20mA, and the protection diodes in the dpot will clamp the voltage to less than Vdd + 0.3V (or at least, nothing will be damaged, because the two absolute maximum limits can't quite be simultaneously true.)

author: rs20
author_rank: Super Contributor
timestamp: January 20, 2018, 02:20:06 pm
content: And if you really want to get deep into the philosophy of datasheets, you might observe that it's weird that I am, on some level, disregarding the voltage limits in the datasheet and focussing on the clamp current limits. I mean, whenever you're dealing with absolute maximum ratings, you normally verify that every single one of the absolute maximum ratings is being taken care of. So if you're feeling uncomfortable that I'm disregarding the voltage absolute maximum ratings, that's a good instinct to have.However, the datasheet doesnotprovide a V-I curve or any other characterisation of the protection diodes; so if we're providing a known current, we have no idea what the voltage on that pin is. I mean, I know it's going to be somewhere between Vdd and Vdd + 0.7V probably... but will it be more than Vdd + 0.3V? Who knows. But it'd be pretty dumb if I pumped 10mA into a device with a 20mA max rating clamp current and it failed, and then the IC company told me that my 10mA current had produced a 0.5V drop and thereby exceeded the maximum ratings.So what the datasheet is actually saying is:-- Are you providing a strongly driven voltage level? Then keep it below Vdd + 0.3V and you'll be sweet, don't worry about the clamp current.-- Are you providing a current-limit signal? Then ensure the current doesn't exceed 20mA and you'll be sweet, don't worry about the voltage on the dpot pin.I'm sure this would be explicitly spelled out in an application note somewhere.(BTW, you can't and musn't apply the same logic to MOSFET datasheets, because youdohave all the data there than you need to ensure that the current, power and temperature of the MOSFET is kept within maximum ratings, so you need to do those three checks separately)

author: brownt
author_rank: Regular Contributor
timestamp: January 21, 2018, 01:24:03 am
content: I see. current is being limited below 20mA, so I like what you are saying, because there are other parts to the circuit that seems to cause problems when the dual schottkys are used.It seems that when the dual schottky arrangement is made with the cathode/anode junction floating, there is about 1volt at that junction referenced to ground. This does not happen when there is the dig pot resistance across the junction to ground, and also does not happen with silicon diodes, but does with the BAT54S and 1N5817 schottkys.You can see on the attached circuit, that there is a switch that isolates one connection on the socket, and so when that is done the 1volt is sitting on that connection, which is a problem for the device that attaches to that socket. Its an electronic switch (74HC4053) so it cannot go on the other side of the input protection, because the protection is also for the electronic switch.what is going on there?

author: rs20
author_rank: Super Contributor
timestamp: January 21, 2018, 09:20:12 am
content: What do you mean "there is 1 volt sitting on the connection"? Where are you connecting your probes?

author: brownt
author_rank: Regular Contributor
timestamp: January 21, 2018, 09:48:58 am
content: from ground to the anode/cathode junction, on either pair of diodes.To clarify that. If i get two schottky diodes out of the draw, put them in series and put 5volts across them, there is 1volt at the anode/cathode junction measured from that junction to 0 volts.but with two silicon diodes (1n4001) configured in the same manner, there is only 30mv at the junction.

author: rs20
author_rank: Super Contributor
timestamp: January 21, 2018, 10:41:06 am
content: I think you need to brush up on the basics of electronics. What you're saying is "I put 5V across a pair of Schottky diodes and saw 1V across the pair of diodes." So what's the voltage across the diodes, 5V or 1V?I don't know what you're using as your 5V supply, but I suspect that you are basically shorting out the supply (or causing it to flip into constant current mode), and therefore putting some value of current through the diode which is then causing a 1V drop (2 * 0.5V). There's nothing 5V about it.And your silicon diode measurement is just bogus, your supply is probably giving up due to being shorted out. Try adding a 1 kiloohm resistor in series with the diodes to give your supply a fighting chance.

author: Zero999
author_rank: Super Contributor
timestamp: January 21, 2018, 11:37:06 am
content: Quote from: rs20 on January 21, 2018, 10:41:06 amNo, he's not shorting out the PSU, because the diodes are reverse biased.Quote from: brownt on January 21, 2018, 09:48:58 amYou connected the didoes as per my attachment?V1 = 1VV2 = 30mVIf so, then that's what you should expect to happen.The Schottky diodes have a higher leakage current, than the silicon diodes. In theory the voltage should be 2.5V, assuming the diodes are both the same and the meter doesn't load the circuit at all. In real life, the diodes won't be equal and the meter will load the circuit. Assuming your meter has an input impedance of 10M, the leakage current of the Shottky diodes is 1/106= 1µA and the leakage of the silicon diodes is 0.03/106= 3nA, which seems reasonable, without looking at the data sheets.

author: brownt
author_rank: Regular Contributor
timestamp: January 21, 2018, 12:17:12 pm
content: I see. yes the diodes were connected as in your diagram. Its normal, ok thanks for that. I'll consider leaving them out entirely and instead depend on the IC's built in silicon diode protection, and the low current draw factor as mentioned before.I have seen a similar arrangement in a commercial product. The only difference I can see is that the micro is running on its own supply, whilst the dig pot and schottky protection are running on a separate supply. In this scenario there is no voltage at the anode/cathode junction.what is happening there?

author: Zero999
author_rank: Super Contributor
timestamp: January 21, 2018, 02:30:23 pm
content: Quote from: brownt on January 21, 2018, 12:17:12 pmI don't know. Do you have a schematic?The 0V of the digital pot and MCU will need to be connected together, in order for it to work.

author: brownt
author_rank: Regular Contributor
timestamp: January 22, 2018, 12:25:29 am
content: schematic, no. just tracing it through the PCB. But no worries, I can see that there is no voltage at the anode/cathode junction, because the dig pot always provides a resistance across the junction to ground, if I remove that, the voltage appears the same as in my design.

author: rs20
author_rank: Super Contributor
timestamp: January 22, 2018, 06:07:49 am
content: BTW, a digital pot requires power, ground and some sort of data connection. So they have at least 6 pins (or at least 5/4, in the case of ones that have some of the pot pins inaccessible or tied to ground).You obviously need to connect these in your real circuit, and it's very poor form not to show them as being connected in your schematics. It'd take you 5 minutes tops to create a custom symbol in your schematic editor of choice; we've spent alotmore time than that here being confused due to the lack of clear, unambiguous descriptions of what you're doing nor proper schematics.

author: brownt
author_rank: Regular Contributor
timestamp: January 25, 2018, 11:57:02 pm
content: Assuming that the only requirement was to ensure that the voltage across the pot was less than VCC supplying the IC then I suppose a pair of silicon diodes would be OK. Then the leakage current is not such an issue. Does that sound correct?

author: brownt
author_rank: Regular Contributor
timestamp: January 26, 2018, 12:59:42 am
content: "There's basically no configuration in which Schottky diodes make sense, because if you put them directly on the dpot pins, you can't guarantee that the diode will actually start conducting before the clamp diodes in the IC, and if you put them before the resistor, there's no point in them being Schottky because even a normal 0.7V drop diode will work perfectly well and will mess with your signal less."Is that correct?I thought the diodes inside the device are silicon and so conduct at 0.7 volts, and so the schottky's will conduct first at 0.2volts

author: brownt
author_rank: Regular Contributor
timestamp: January 26, 2018, 02:13:43 am
content: Looking at the datasheet specs and thinking about damage to the dig pot ic when greater than VCC is across the pot ends, what if the dig pot was unpowered, and an external 5 volts was to be put across the pot ends, would that damage the IC?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 26, 2018, 11:18:35 pm
content: A silicon diode can still be beneficial, because the IC's internal diodes usually have a higher voltage drop, especially more series resistance.Current through the internal diodes is almost always undesirable; this might be satisfactory to prevent damage, but may not prevent aberrant behavior (like corrupted internal state, or CMOS latchup) that has to be fixed with a power cycle.Schottky diodes will carry a much larger fraction of fault current than the IC's internal diodes, making it more likely to avoid aberrant behavior as well as damage. (Mind that schottky diodes can have high ESR as well. Choose wisely.) The downside is high leakage current. Nothing for free.Choose the best compromise given your circuit's situation.As for applying a solid 5V to an unpowered pot, yes, that would deliver full supply current (including startup surge current) through the diodes, into the (otherwise unpowered) circuit.Tim

author: brownt
author_rank: Regular Contributor
timestamp: January 27, 2018, 02:53:03 am
content: I see. but as I understand it from datasheet for MCP4341, provided the current delivered through that unpowered pot is limited to 20mA, there should be no damage. Is that correct?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 27, 2018, 03:17:11 am
content: Without having looked at it myself, that is probably correct, yes.Tim

author: max_torque
author_rank: Super Contributor
timestamp: January 27, 2018, 09:54:31 am
content: The trick is to split the input protection into two halves. One half (or a bit more than half depending on your max possible input votlage) immediately after the input pin, followed by your diode protection bridge to gnd and vcc, and then the other half before the microcontroller input. The first half protects the protection diodes, the second half protects the diodes in the micro. Size them inteligently, and you can ensure that the external diodes clamp before the internal ones, so the external diodes actually protect the micro.Regarding leakage, if the input is a high impedance, then yes, the diode leakage matters, because the leakage current has nowhere to go to, but when, in your case, the digipot is connected, it's output impedance is easily low enough to swallow that leakage current. It's also good practice to ensure any inputs to your micro have a known failure state, ie a pull up or pull down resistor, so that should the device connected fail (or be disconnected) the input goes to some known state, meaning your logic can reckoning that state, rather than respond to an erroneous input thinking it's the real input!Often, any input that can cause a system malfunction is set up so that it falls into three known voltage levels:1) Off Scale Low - a low voltage, that the normal device on the input can never get down too.2) Off Scale High - a high voltage, that the normal device on the input can never get up too.3) Normal - a voltage between OSL and OSH, which the device on the input outputs when operating normallytypically, for a 5v input, you'd chooseOSL = <0.5VOSH = > 4.5Vgiving your device a 4v operating span.These voltages are usually furnished by careful choice of pull up, pull down resistors on the input