Help me understand diodes, please - Page 1
Forum section: electronics / beginners

author: hirada
author_rank: Regular Contributor
timestamp: November 06, 2017, 04:40:45 pm
content: On my quest to come up with a kind of an all passives comparator to eventually ignite a thyristor, I've tested the circuits below. I've been already confident, that they would not do what I wanted, but for completely different reasons and whilte testing, I figured, I did not remotly understand the results I've been measuring.At the upper one the voltage stopped raising at 2.8V, which I believed to be the leaking current of the Zener being the reason it never even remotely reached 3.3V. By itself, the lower circuit would have been also explainable, too, because the current is too low to make the LED conducting, so asides a minimal leakage current all voltage drops there.But this explanation does not make any sense any more when compared to the results of the upper circuit. And what happened in the middle, after I've removed the LED, is beyond me.It may be notable, that the differential voltage to 5V is always dropping over the 820K resistor, there is no further voltagedrop (that I could measure, that is) behind the node.Can anybody shed some light, please? Or maybe some reading, that helps me getting a clue?

author: mjkuwp
author_rank: Unknown
timestamp: November 06, 2017, 04:48:02 pm
content: what are you using as your supply? Do you mean you are physically testing these parts? it's not a simulation?

author: hirada
author_rank: Regular Contributor
timestamp: November 06, 2017, 04:52:36 pm
content: Supply is a 7805 (from a 12V batterypack) and this is not a simulation, but build on a breadbord.Edit:Of course the middle circuit does not change when the small 820R resistor is removed.And here is the link to the Zener, just in case:http://www.produktinfo.conrad.com/datenblaetter/175000-199999/180041-da-01-en-BZX55_ZPD_3_3_V.pdfI do not have any data on the LED. But with an 820R and 3.3V it is light. Dim, but visible, and has a forward voltage of around 2V.

author: jm_araujo
author_rank: Regular Contributor
timestamp: November 06, 2017, 05:09:51 pm
content: Any chance you connected the zenner backwards?

author: kosine
author_rank: Regular Contributor
timestamp: November 06, 2017, 05:14:22 pm
content: You appear have a 20k resistor after the supply. The current to the circuit will be tiny as a result, 100uA sort of tiny.Zeners need at least 1mA to start regulating, preferably a bit more. Try using a 1k resistor or even a bit less. You should see voltages closer to what you expect.

author: hirada
author_rank: Regular Contributor
timestamp: November 06, 2017, 05:41:15 pm
content: Thanks for your replies. Well, the 820k resistor after the power supply is needed to get the capacitor charged slowly. That is in fact the problem I am facing, that the charging current is way lower than the current needed after the capacitor to do the job.The Zener is correctly (reverse) biased and works on its own. But, great hint, as in series with the LED it always drops only 2.8V, as 1.9V are dropped over the LED. Brings us a bit closer to understanding.The Idea, which I am aware that it is prone to fail, is to load the capacitor slowly, once it has reached the zener voltage, it discharges and therefore can shortly can deliver more current than the power supply though the 820k resistor.I was hoping for a kind of minimal hysteresis within the zener, so the current burst until the capcitor voltage droped below the zenervoltage would have been long enough to then ignite a thyristor. Or flash a the LED.Again, this idea was a bit to idealistic, I ignored the knee of the z-diode, but that does not stop me from wanting to understand the odd voltages I have been reading. Despite its otherwise probably useless design.Edit: Please note the large (820k vs. 820R) Resistor in the second attachement. The fist diagram makes the LED actually dim.

author: jm_araujo
author_rank: Regular Contributor
timestamp: November 06, 2017, 06:00:53 pm
content: Quote from: hirada on November 06, 2017, 05:41:15 pmIt seems to me that you described a relaxation oscillator. You need a device with a negative resistance zone like an unijunction transistor or a tunnel diode, both unusual components (don't know if you still can find them). Found a nice writeup about UJTs here:http://www.electronics-tutorials.ws/power/unijunction-transistor.html

author: hirada
author_rank: Regular Contributor
timestamp: November 06, 2017, 07:20:06 pm
content: Thanks very much, I'll look into those to maybe solve my initial problem. But for now I would just like to understand what I've been measuring. Or rather why.If I have a nearly full voltage drop over the LED, as in the lower schematic, why do I have lesser drop, when an additional zener is in series?And why do I have only 1.4V drop if I just use the zener, as in the middle diagram? The voltage over the zener always equals that over the capacitor. Same is true for the led in the last diagram.But what is going on here?

author: MK14
author_rank: Super Contributor
timestamp: November 06, 2017, 07:54:34 pm
content: Quote from: hirada on November 06, 2017, 07:20:06 pmThe lowest schematic (in your original 1st post), seems to be in error. Maybe the Led circuit bit, was open circuit or something ?Because once the Led starts significantly conducting (at say 2 V), the voltage remains fairly constant, even if the current rises.I.e. it should be about 2V NOT 4.7V, if I understand what you are saying, correctly. 4.7V sounds more like what one would expect to see if a capacitor charges up from a 5V supply, via a 820K resistor. Because electrolytics, can have a bit of leakage current.EDIT: Correction, due to post, immediately after this.The meters internal resistance (probably 10 M), is likely to be why the 5V drops to 4.7V .820K is quite a high value, so it will be influenced, by quite small currents, such as the meters internal resistance.

author: Damianos
author_rank: Frequent Contributor
timestamp: November 06, 2017, 08:40:26 pm
content: Just some practice to lead you to find the answers by yourself; for each step keep notes:- Measure the voltage of the source, without any load.- Connect the 820kOhm resistor only and measure the output.- Calculate the voltage ratio of the divider made by the resistor and the input resistance of your meter. Does this agree with the above measurement?- Add the capacitor and measure again. Is there any change?- Leave the capacitor to charge for a while and then connect the voltmeter and notice if the voltage changes and in what direction.- For the continuance check for the leakage current of the diode in the datasheet.- ...Regards,Damianos

author: hirada
author_rank: Regular Contributor
timestamp: November 06, 2017, 10:10:30 pm
content: Quote from: MK14You are absolutely right. I must have reverse biased the diode there. Attached are the correct measures, this time without the capacitor, that I still do not comprehend. But it begins to make some sense now.So we have current in uA (calculated by the voltagedrop over R, cannot measure this low):R: 6R+Z: 4.5R+L: 4.2R+L+Z: 2.6The reverse current at 1V (and room temperature) is 2uA according to the datasheet. But what does this practically mean? When just using the zener (with roughly 1.33V drop) we still have a current of 4.5uA.And further, a LED, that is not turned on, does still seem to conduct some current? As once it is led, the voltagedrop is around 2V. Or does it need to be able to drop 2V to actually turn on?I suppose, the current dictates the voltagedrop over the diods? But if we have that drop over the zener way below its working voltage, what sense do they make? I was under the impression, they should block completely until the zener voltage? Or is mine defective? Currently I do not have any other to test.I am still missing something.Edit: Will do the same tomorrow (without the cap) with an 82k resistor instead of 820k. No idea, what will change, but the show must go on. Thanks so far

author: hirada
author_rank: Regular Contributor
timestamp: November 06, 2017, 10:42:10 pm
content: I think I have it for the zener. Kosine had the clue a while back already, I was too much fixated on the voltage and thought, if there is not enough either current or voltage, it won't do anything at all and behave more or less like an open switch. Same for the led that is not lit or a standard diode below ~0.7V. As diodes are no resistors, I did not expect leaking current in the uA range to drop relevant voltage.This is obviously wrong. Using the 820R the Zener drops 3V and using 12V supply instead of 5, it is limiting @3.3V, as one would expect. Bit disappointed about its off-behaviour, but there we are.That still leaves for tomorrow, why then the voltage with the cap is ~0.6V higher than without, but will redo this to avoid any mistakes I may have made today.

author: MK14
author_rank: Super Contributor
timestamp: November 06, 2017, 10:49:50 pm
content: Quote from: hirada on November 06, 2017, 10:42:10 pmAssuming the capacitor is fairly big, it will reach the peak voltage (assuming the multimeter is disconnected while the capacitor charges up). So the internal resistance of the multimeter, won't drop the voltage any longer, because the capacitor has already reached the peak available voltage.Without the capacitor, the 820K has to supply all the current, so the voltage drop across it will occur again.Even so, 0.6V seems a bit high (I would have expected around 0.3V), so maybe there is more to the explanation.If that theory is right, if you leave the multimeter connected, you should see the voltage eventually drop down to the low voltage point, in time.

author: tpowell1830
author_rank: Frequent Contributor
timestamp: November 07, 2017, 01:34:16 am
content: Quote from: hirada on November 06, 2017, 04:40:45 pmAlthough you don't have a value written down for the capacitance, it doesn't matter, the capacitor will never charge up to 5 volts because of the 820k resistor on the front. The leakage from the devices will not let this happen.You need to replace the 820k resistor with a much lower value and put an active device in the circuit (such as JFET or BJT darlington) in order to get oscillations, if that is your goal...Edit: I understand now, sorry for the negative sounding comment.

author: hirada
author_rank: Regular Contributor
timestamp: November 07, 2017, 08:53:46 pm
content: After rewatching daves video on zener in addition to your input and the testing I think I got my mistake and some slightly deeper understanding what is going on here. Thanks big time.I have redone the measurement with the capacitor, values attached, and I do not think the multimeter is hindering the charge up of the capacitor. Otherwise I would need to see a voltagedrop after connecting the probes, which I do not - the voltage is swaying around 2.83-2.84V rather constantly and not dropping at all. I also can witness the raise of the charge.Also, after removing the voltage probes, the current remains pretty stable. In the configuration posted on top.However when I moved the amperemeter around, I got surprisingly different values, up to the point, where the capacitor would not charge at all, unless the amperemeter had been removed and replaced by jumperwires. Lower figure. Tried different wires, no chance. Although the burden voltage should only be around 300-600uV in this range. I currently do not get this, but will retry on thursday, too tired today and probably made a stupid mistake. I was able to borrow this additional multimeter and probably messed something up.Quote from: tpowell1830Thanks. Good idea. I've tried to measure the leakage current of the capacitor, but was unsuccessful. See aboveEdit:The 2.8u in the lower diagramm is an error. Forgot to delete it. I was not able to measure any current in this configuration, as (or because) the capacitor would not charge unless the a-meter would be bridged.

author: MK14
author_rank: Super Contributor
timestamp: November 07, 2017, 10:00:48 pm
content: Quote from: hirada on November 07, 2017, 08:53:46 pmIf an ammeter (mode in multimeters), suddenly does funny things, especially appearing to be open circuit. You (or it may have been broken BEFORE you used it) may have blown its fuse. If not applicable in this case, worth remembering in the future.Assuming the multimeter has/had a fuse(s), which the vast bulk of do, these days (except some really, really cheap ones, or some old ones). It could have been around 500mA (it depends on what multimeter, you used. One of my older ones has 400mA fuses, but it can be less, or more) on a mA range, so would easily blow if you connected it, directly across your 5V supply (an easy mistake to make). Unless the 5V supply was rather weak.

author: Damianos
author_rank: Frequent Contributor
timestamp: November 08, 2017, 01:52:45 pm
content: Quote from: hirada on November 07, 2017, 08:53:46 pmThere were some steps proposed to help you understand what happens, but you are trying to conclude things in a wrong way. You have made nodes and branches in the circuit so your conclusions are irrelevant, because there are interactions.Having the diodes connected, these are the dominant load in the circuit...What are the specifications of the two instruments?...

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: November 08, 2017, 02:38:17 pm
content: You cannot trigger an SCR like this. The current delivered by the capacitor will be too small to turn it on. They typically take a few tens of mA gate current, whereas your capacitor charging through 820k will only deliver a microamp or two.There are various approaches. A 555 IC will do quite well. Its output can deliver up to 100mA, so should trigger most SCRs through a 47R resistor and a capacitor (say 100n) from 5v. Or, use a 'unijunction equivalent' (NPN and PNP with the collector of each feeding the base of the other)You could also use a mosfet to buffer the capacitor voltage, only issue there is that the timing will not be very accurate due to device threshold voltage spreads.When you mention hysteresis I think you mean a diac. These work at more like 30v though. I don't know of any that will trigger at under 5v.

author: dmills
author_rank: Super Contributor
timestamp: November 08, 2017, 02:45:24 pm
content: Your initial circuits can actually flash an indicator...Just replace the LED with a NE2 neon and crank the supply up to a 150 volts or so! The neon will flash every time the cap voltage reaches 90V or so (They ignite at 90V or so and go out once the voltage drops to ~60V), a favourite trick for adding warning indicators to a high voltage cap bank.Regards, Dan.

author: hirada
author_rank: Regular Contributor
timestamp: November 08, 2017, 07:50:34 pm
content: Quote from: MK14 on November 07, 2017, 10:00:48 pmRight again you are. My mighty 3uA has indeed blown the hoplessly underdimensioned 400mA fuse. What shocks me though is that they cost an arm and a leg. I can get a tenpack of multimeters at aliexpress for what mouser charges for a single fuse...Anyway, all further testing has to be delayed until the postman rings twice.Quote from: Damianos on November 08, 2017, 01:52:45 pmAnd I was under the impression, I have done that?The suggestions you kindly made stoped where things started to get interestingly. So I went on to the best of my knowledge. Which isn't much, otherwise I would not be posting here.Specification I only have for one DMM, Inputimpedance is 10M, Burdenvoltage is 100uV/uA, what exactly are you looking for? For the other I haven't found any data.Quote from: Damianos on November 08, 2017, 01:52:45 pmThat is exactly what I am fighting with to get in my head. I simply do not comprehend why, especially the LED is way more dominant than the zener, probably because its revers biased, with keeping a steady 1.6V drop. Well, hence the topic.Quote from: IanMacdonald on November 08, 2017, 02:38:17 pmI am aware that this idea does not work, but what you say is quite interesting. As I was under the impression, that once the capacitor is loaded, its output current is soley dependend on the resistance in the unload path, not on the current it has been loaded with and since the 820k is connected directly to the 7805 output, it will not discharge in this direction.So if there was only a 200 Ohm resistor between the capcitor and the SCR, current would be 5/200=25mA. Now the voltagedrop with high current is too fast to keep the zener alive, but at least I got my general misconception of zeners by now, so the whole idea was nonsense. Part of learning.I would like to avoid using an IC, as this eventually will become a delay for a power supply and I do not have any stable supply voltage while the filter capcitors get charged thru an current limiting resistor. Otherwise a simple comparator would do the job just fine. Building a passive comparator is how this all started. In a wrong direction, but again, that's learning.Quote from: IanMacdonald on November 08, 2017, 02:38:17 pmThat sounds interesting. Do you have link for that? I currently cannot imagine this. Timing is irrelevant, we are talking 10+ seconds anyway, one more or less does not matter at all.Quote from: IanMacdonald on November 08, 2017, 02:38:17 pmYep. I was trying to emulate an diac working below 5V. Or rather treat the zener as one. Which would basically be the same idea as a comparator in this case.Quote from: dmills on November 08, 2017, 02:45:24 pmGreat, but the LED is just a placeholder for a relay. But at least the NE2 seems to have a hystersis. We need more of that...Again thanks all for input

author: David Hess
author_rank: Super Contributor
timestamp: November 08, 2017, 08:21:01 pm
content: Motorola/On application note AN964 covers the various options. Diacs, sidacs, and PUTs are still produced and there is always the neon bulb. I might use an NPN and PNP transistor configured as an SCR with a zener diode to make a custom controlled breakdown device for triggering if necessary.

author: hirada
author_rank: Regular Contributor
timestamp: November 08, 2017, 10:37:33 pm
content: Quote from: David Hess on November 08, 2017, 08:21:01 pmAs posted by jm_araujo, the UJT or PUT seem indeed to be the solution to my initial problem. Muito obrigado, in case you should be still reading here. Whatever practical problems may occur here once they arrive.However, in case you do not mind, I would like to see a starting point to your idea to recreate this from more basic components? Any link with a diagramm as a starting point?That does not mean, my quest to understand the diode behaviour a little more is finished, but I need to wait until I can continue testing. Until then I can lookup sidacs, which I never have heard about before

author: MK14
author_rank: Super Contributor
timestamp: November 09, 2017, 12:08:46 am
content: Possible solution. (Your title, originally confused me, as to what you were trying to achieve. I thought from the title, you wanted to learn about diodes. I did not realise it was a delay for PSU circuit, you wanted).Top of page 18 (Delay Timer), seems to have a circuit that may do what you want (with suitable modifications):http://www.ti.com/lit/ds/symlink/lm431.pdfAlthough it IS an integrated circuit. It really is designed to be a much improved, zener diode, replacement. The IC only needs a few micro amps, to power it (plus any external resistors, current consumption, depending on configuration).I.e. It does not need supply rails as such, as it can be built into a circuit (like a zener or transistor) can be, WITHOUT power rails, if you understand what I am getting at.

author: David Hess
author_rank: Super Contributor
timestamp: November 09, 2017, 06:26:12 am
content: Quote from: hirada on November 08, 2017, 10:37:33 pmAny books and application notes on thyristors will cover SCRs and other devices. The advantage of using an NPN/PNP pair to make an SCR is that you may add resistors to control its characteristics and high voltage bipolar transistors are readily available.While I would not normally consider it for a trigger, thelambda diodemade from an N and P channel JFET operates like a higher voltage tunnel diode and might be suitable for what you want.It is too bad that no p-channel depletion mode MOSFETs are available. They would be handy for making a high voltage version of the lambda diode.

author: xani
author_rank: Frequent Contributor
timestamp: November 09, 2017, 09:18:32 am
content: If all you want to do is to charge cap to desired zener voltage (minus one diode drop) slowly, see attached picture. R1 ensures Zener diode gets enough current to be at "rated" voltage so you dont have to worry about calculating where exactly on the "knee" your circuit lands. You'd still need something to trigger the charge dump, diac would probably be simplest way to do it (see 2nd image)I'd still not want to use it to trigger SCR if you want to do "trigger it few seconds after device power-on". Just drop a 555 there and be done with it. There is a nice writeup here about triggering methods:http://www.littelfuse.com/~/media/electronics_technical/application_notes/switching_thyristor/littelfuse_triggering_and_gate_characteristics_of_thyristors_application_note.pdf

author: hirada
author_rank: Regular Contributor
timestamp: November 09, 2017, 09:19:17 pm
content: Quote from: MK14 on November 09, 2017, 12:08:46 amThe title is still correct, just the thread has become double laned by now. Probably because I talk to much in trying to explain what and why I did things. Nevertheless I am really, really thankful for the help for the root cause of my diode issues, be it the ujt or the lm431. Or the more grass roots approach of David, which I have to look more into it. At the end it will be a delay, but I would like to understand what I am doing and so far I've failed at a rather basic stage, the diodes. So this is not really over.To your suggestion: According to the first look I would not mind the the lm431, as it does not seem to need a power supply. I am not against ICs, in this case I just have no stable supply to reliably power them for time they are needed. The supplyvoltage would slowly rise as the smoothing capacitors are being charged.For a PUT as for the lm431, as long as the reference is raising faster than the voltage over the capacitor, I should be safe. In my theory.Anyway, sorry for the confusion. I've replaced the lm7805 with a lm317 and will do more testing once the fuse arrives. It is still about diodes and why they do what they do and hopefilly I will then be able to ask more specific questions about their behaviour.Quote from: xani on November 09, 2017, 09:20:24 amAll I originally wanted was to discharge the capacitor partly over the zener, once the zener voltage has been reached. Stupid idea by wanting a zener to behave like a low voltage diac. And the current required to make the zener work would require a too big of a capacitor to get the Tau of ~10sec. Something like 0.5F or there about.But big thanks for the link.

author: MK14
author_rank: Super Contributor
timestamp: November 10, 2017, 12:26:30 pm
content: Quote from: hirada on November 09, 2017, 09:19:17 pmI understand. Threads, do sometimes go in odd directions and can be hard to accurately name, initially.The LM431, because it only seems to need a few microamps (2uA typical, 4uA guaranteed) to run, plus any current the rest of the circuitry needs, may allow your longer time constants, such as 10 seconds, and yet have a capacitor, which is not ridiculously big. Anyway, they are fairly cheap (probably a lot less than a dollar, each), so could make an interesting experiment, for more messing/learning about diodes.The LM431 seems to have been designed, to eliminate or reduce, some of the issues, people can have with normal zener diodes. Such as they (zeners) are nonadjustable, somewhat/partly inaccurate (i.e. about +/- 5% or so) and the voltage is not that precisely regulated, especially at lower currents, etc.In theory, you could get similar results, by using multi transistor circuits, and conventional diodes or zeners, but these days it is often not worth the effort, when there are small, low cost ICs, which will do the job.Other solutions in this thread should also work. But some of them would have limitations E.g. Neons need high voltages (around 90V plus) and some of the alternative transistor types mentioned, can be harder to obtain and/or a bit pricey.But I suspect it could be done, with a pile of discrete components (including transistors). Doing it that way, could be more educational, than using ICs. But these days, using ICs is the more common way, fairly often.tl;drLM431 is one way which might work. But there are probably millions of other ways of doing it. Just like most things in electronics.

author: hirada
author_rank: Regular Contributor
timestamp: November 15, 2017, 11:28:18 am
content: Quote from: MK14 on November 10, 2017, 12:26:30 pmOr can simply delay shipping, if a shop does not have it's inventory under control