hybridfree/eevblog-posts huggingface dataset:
Question about measuring the current in a circuit . - Page 1
Forum section: electronics / beginners

author: SH@RK
author_rank: Contributor
timestamp: July 21, 2012, 02:31:35 am
content: HiI have a question about measuring the current in a circuit"" Why there is a big difference when I change from mA to A range in the same circuit ? ""I have a 10watt LED and tried to find the current it derive and faced by this problemI used tow DMM and the same result.The voltage is the same in each try (8.11 V DC) .The circuit should look like thisSee the photosFirst :- MastechAt the mA range it reads 65.5 mAAt the A range it reads 0.14 A = 140 mASo it is far away reading (140-65.5 = 74.5 mA )The right side is the "mA range"Second :- Fluke 87VAt the mA range it reads 108.2 mAAt the A range it reads 0.148 A = 148 mASo it is far away reading (148-108.2 = 39.8 mA )The right side is the "mA range"I am thinking that the resistance of the DMM changes between the mA & A ranges .So is my thinking is true ??And which reading is the accurate to find the power delivered to the LED ??Thanks

author: Rerouter
author_rank: Super Contributor
timestamp: July 21, 2012, 02:35:32 am
content: your facing the problem of burden voltage, your milliamp range will use a rather large resistance to give a bigger signal, so it will drop more voltage across itself and cause your discrepency, if you want to confirm this, have 1 meter read the current on the 2 different ranges, and have the other read the voltage across the first meter,

author: Psi
author_rank: Super Contributor
timestamp: July 21, 2012, 02:39:18 am
content: Quote from: SH@RK on July 21, 2012, 02:31:35 amYep, the meter measure current by reading the voltage drop across an internal resistor, so that voltage is lost.The result is your load (LED) gets less volts and so draws less current than it would without the meter.This current sense resistor is a different value for the low and high current ranges, so the current detected will be different.The 10A range will be more accurate but still not 100% correct

author: IanB
author_rank: Super Contributor
timestamp: July 21, 2012, 04:12:18 am
content: Quote from: SH@RK on July 21, 2012, 02:31:35 amYes, it surely does. Use a second meter to measure the resistance of the first meter on various resistance ranges and be illuminated.One way to avoid this problem is to use Dave's uCurrent device.

author: SH@RK
author_rank: Contributor
timestamp: July 21, 2012, 05:42:52 pm
content: Thank you allQuote from: Rerouter on July 21, 2012, 02:35:32 amAs you said the resistance at mA is larger than A rangeThe voltage in both cases was about 8 V deliverd by the Power supplyThe voltage accross the DMM was not stable (raising by the time LED became warmer)First :- MastechAt the mA range it reads 0.965 V and the resistance is 6.45 owmAt the A range it reads 0.332 V and the resistance is 0.23 owmSecond :- FlukeAt the mA range it reads 0.185 V and the resistance is 1.9 owmAt the A range it reads 0.098 V and the resistance is 0.0~0.1 owm (was not stable btween 0.0-0.2 but mor at 0.0-0.1)will read the other replies and post again laterthanks
Question about motors. - Page 1
Forum section: electronics / beginners

author: GeorgeTsak
author_rank: Contributor
timestamp: June 07, 2016, 12:06:57 am
content: I have some old floppy drives and I am really interested on the "big" motor (spindle) that spins the disk. It has four wires. As far as I know it is brush less, DC, 3 phase motor. Can a floppy's drive motor be used as power generator?

author: Dave
author_rank: Super Contributor
timestamp: June 07, 2016, 07:11:32 am
content: Yes, it can. The coils inside are probably going to wired in a wye arrangement. Using a multimeter, find the terminal which has the same resistance between it and any of the three other terminals. That is the center of the wye and you won't be needing it. Connect the other three terminals to a 3-phase bridge rectifier and test your circuit.

author: GeorgeTsak
author_rank: Contributor
timestamp: June 07, 2016, 07:33:51 am
content: Thank you very much! It work perfectly!
Question about my circuit board - Page 1
Forum section: electronics / beginners

author: Hextejas
author_rank: Frequent Contributor
timestamp: December 15, 2018, 06:02:35 pm
content: Looking at the back of this board, is what I marked as "A" part of the ground plane and is the part where B-----------> points to, part of the same plane?Thanks

author: rstofer
author_rank: Super Contributor
timestamp: December 15, 2018, 07:10:32 pm
content: It certainly looks like one plane. You can do a continuity check with a DMM.I have no idea why there is a slot but it probably serves as some kind of fault limit scheme.

author: Brumby
author_rank: Unknown
timestamp: December 16, 2018, 12:14:45 am
content: Quote from: rstofer on December 15, 2018, 07:10:32 pmThe slot will force all current flowing through that arm to pass through a single point. The symmetric top section will be the same. This can help with decoupling - and I'll bet the solder joints you can see near the neck of each is for a capacitor to do just that.
Question about npn transistor gain and voltage output - Page 1
Forum section: electronics / beginners

author: little_carlos
author_rank: Regular Contributor
timestamp: December 10, 2014, 02:51:52 pm
content: Hey guys, i have a question with this thing thats been in my head for over a weekim using a pic 16f628A and im conecting a output to a bc547 npn transistor, the thing is that the pic output gives about 4V, and i need to amplify so the transistor swtich a relay, wich is for 5v, but i only get 3.5V of output from the transistoris there any other npn transistor that would give me the 5V of output with only 4V in the base? or any configuration to increase voltage gain in the bc547? thanks for your time

author: TKt48
author_rank: Newbie
timestamp: December 10, 2014, 03:02:49 pm
content: TTL outputs can sink relatively high current in low state, but output current in high state is minimal. You can use pnp transistir connected by emitter to power rail and base via resistor to pic output. Use then negative logic to drive relay. Or connect relay directly to the output and it will switch on when state is low.

author: Paul Price
author_rank: Super Contributor
timestamp: December 10, 2014, 03:05:57 pm
content: You can use a NPN transistor, just activate the relay coil with the relay in the collector circuit to +5V, do not drive the relay using an emitter-follower, drive the base of the NPN relay driver with (for instance) a 470 ohm to 10k in-series base resistor, emitter is grounded. Put a 1N4001 diode or equiv. (cathode to +5V) across the relay coil.

author: Zero999
author_rank: Super Contributor
timestamp: December 10, 2014, 03:13:53 pm
content: It sounds like you're not providing enough base drive current.How much current does the relay need? If you don't know that, what's the coil resistance (measure with a meter) and calculate the current using Ohm's law.It's actually impossible to get 5V out with a 5V power supply because there will always be some loss in the transistor. It's possible to get very close but the voltage across the relay will always be slightly less than the power supply voltage.

author: kaindub
author_rank: Regular Contributor
timestamp: December 10, 2014, 03:55:13 pm
content: I think you have the transistor connected as a voltage follower. The clue is that the emitter voltage is 0.7v less than the input.You need to connect the output of the pic via a resistor to the base of the transistor.Groung the emitter.Connect the relay between 5v and the collector
Question about op amp errors - Page 1
Forum section: electronics / beginners

author: king.oslo
author_rank: Frequent Contributor
timestamp: November 06, 2012, 06:30:44 pm
content: Hello there,I have a couple of questions I found hard to get answered whilst searching the web:1. If I model offset voltage as a voltage source in series with one of the inputs, how to I learn from the datasheet: in series with which input should it be modelled,and of which polarity is it?2. I model offset current as a current source from one of the inputs. How do I determine which input? Is the polarity of the input current affected by the type of input transistor? Or does the polarity of the input current tell the engineer which input the bias should be modelled at, or both?Thanks.M

author: Kremmen
author_rank: Super Contributor
timestamp: November 06, 2012, 06:51:54 pm
content: Offset voltages and curents are by nature error or stray signals and there is no universal rule regarding their polarity. Often they can be ignored provided the input circuit provides a ground path preventing a capacitive integration of the offset magnitude.When the offsets cannot be ignored a way must be provided to compensate or zero the offset signal. In such a case you usually cannot calculate it away due to its somewhat unpredictable nature.

author: Bambur
author_rank: Regular Contributor
timestamp: November 08, 2012, 10:06:31 am
content: Quote from: king.oslo on November 06, 2012, 06:30:44 pmTry looking at the op amp model in Fig 7-122 on page 144 of chapter 7 in Op Amp Applications Handbookhttp://www.analog.com/library/analogDialogue/archives/39-05/op_amp_applications_handbook.htmlfor some hints on modeling.

author: king.oslo
author_rank: Frequent Contributor
timestamp: November 08, 2012, 03:01:49 pm
content: Thanks guys. I will check out the opamp handbook this evening. Where can I learn how to minimise op amp errors? For example, I imagine low resistance at the input minimise input current errors.Thanks.M

author: Kremmen
author_rank: Super Contributor
timestamp: November 08, 2012, 04:24:33 pm
content: This classic from Carter and Mancini has probably been linked a number of times on this forum but one more time for luck:http://www.ti.com/lit/an/slod006b/slod006b.pdfI can't recall if there is a chapter dedicated to errors specifically, but after reading&understanding the test, you won't really need one.

author: king.oslo
author_rank: Frequent Contributor
timestamp: November 08, 2012, 06:20:47 pm
content: Thanks. I will read it tonight.M
Question about Op Amp's negative non-inverting input on single supply. - Page 1
Forum section: electronics / beginners

author: alexg
author_rank: Regular Contributor
timestamp: October 10, 2016, 04:25:00 pm
content: Hi All, this seems to be very simple question and I already did some research about it but this one question that I still can't find very definitive answer.I have MCP601, it is powered by single supply and wired into voltage follower/buffer configuration. On non-inverting input I have a piezo disk and when I tap piezo disk it sends waveform into Op Amp's input and apparently even though waveform has both, negative and positive components, on Op Amp's output I can only see the positive half of waveform, which is good and what I need. What are the op-amps input voltage limits? For example if Op-Amp is powered by 3.3v, can input voltage be 5v, 10v? Can it be -5 or -10v, is it a simple matter of Op-amp just not being able to make output voltage high/low enough, it I can actually damage it and need to protect inputs from overvoltage?

author: bktemp
author_rank: Super Contributor
timestamp: October 10, 2016, 04:34:57 pm
content: Most datasheets have a section called "Absolute Maximum Ratings":For the MCP601 it gives a maximum input voltage range of Vss-0.3V to Vdd+0.3V. So for single supply 3.3V this gives -0.3V to +3.6V.The input pins tolerate a current of up to 2mA, so it has build in protection diodes, limiting the voltage if the current is not higher than 2mA.If you exceed one of both values, the chip may get damaged.

author: alexg
author_rank: Regular Contributor
timestamp: October 10, 2016, 04:49:31 pm
content: Thanks for reply, how would you recommend to ensure that opamp does not get fried then? I'm thinking about this course of action:1. Smack piezo bender as hard as possible without damaging it and see how much voltage it outputs, then use divider to make sure that this voltage stays below OP AMP's max input voltage? But since there are built in protection diodes inside of an Op Amp, then probably no need in lowering input voltage in case if piezo does not output more than 2mA of current.2. Measure how much current piezo bender outputs and if that current less than 2mA then it's safe. How can I measure its output current with oscilloscope? I am googling it right now but if someone does not mind posting suggestion directly here, I'd sincerely appreciate that.Thanks again.

author: bktemp
author_rank: Super Contributor
timestamp: October 10, 2016, 04:57:06 pm
content: Dividing the voltage probably doesn't work, because piezo discs can produce hundreds of volts if you hit them really hard.Since the MCP601 has a high input impedance, the easiest solution is adding a series resistor: Let's say you use 10k, the input voltage can go up to +23.6V or down to -20.3V before reaching 2mA. You can probably go even higher, but if you go to high it may attenuate the signal.Since 20V is not enough, you can add additional diodes capable of much more than 2mA directly to the piezo disc, limiting the voltage to Vcc+0.7V and Vcc-0.7. After that only a small resistor like 1k is necessary to protect the opamp.
Question about OpAmps... - Page 1
Forum section: electronics / beginners

author: Raav
author_rank: Newbie
timestamp: April 21, 2014, 03:57:46 pm
content: http://www.reddit.com/r/AskElectronics/comments/23i5xs/real_quick_question_about_opamps_setup_as/(Please note: my level is novice)Posted there, but in essence my question kinda boils down...it was stated in #600 Fundamental Fridays on OpAmps that the feedback resistor always has a higher resistance value than the r1 resistor going to ground.I was playing with the equation for non-inverting amplifiers and again it is Av(gain)= R(feedback)/R(to:ground) +1But I was trying to figure out how to get a ratio of 2:1 and the only way I can tell how to do that is by using the same resistor value.Av=(1000ohms/1000ohms)+1-or- 1+1=2So my question is, can the negative feedback resistor be the same value or even lower (to get maybe 1.5:1?)

author: wiss
author_rank: Frequent Contributor
timestamp: April 21, 2014, 04:00:07 pm
content: Sure! Zero for unit gain.

author: c4757p
author_rank: Super Contributor
timestamp: April 21, 2014, 04:02:38 pm
content: Damn, wiss beat me to it!Yeah, "always" is a bit silly. You shouldn't usually go below unity gain (Av = 1) - some op amps aren't stable - but there's no reason you can't gotounity gain, or below it on a decent op amp (only possible in the inverting configuration).

author: wiss
author_rank: Frequent Contributor
timestamp: April 21, 2014, 04:08:08 pm
content: A unit-gain stable opamp (most of them) will be stable with output tied to -in, gain 1 in non-inverting conf and gain 0 in inverting. You should not put another amplifiet in the loop! Unless you know what you are doing...

author: Simon
author_rank: Unknown
timestamp: April 21, 2014, 04:18:43 pm
content: err yes for non inverting same value resistors gives 2 gain and no problem with that.

author: Fank1
author_rank: Regular Contributor
timestamp: April 21, 2014, 09:21:42 pm
content: For a gain of 1 you don't need any resistors.Just hook the output back to the inverting input.

author: Raav
author_rank: Newbie
timestamp: April 21, 2014, 10:23:36 pm
content: Ok excellent, thank you for the replies. I thought Davey said something about feedback always being higher than ground resistor but maybe he said that in a certain context.Thanks again for the replies.Funny enough I was talking to my Step Father In Law (quite a title) but he is a friggin electronics genius, and I told him I was working out OpAmps math and theory and he gave me an old oscilloscope so I am going to try to rustle up a 741 or something with an OpAmp in it and actually put in a few different sets of resistors just to see what it is doing IRL
question about oscillation when testing a battery - Page 1
Forum section: electronics / beginners

author: No.15
author_rank: Contributor
timestamp: October 02, 2020, 02:01:14 pm
content: I have an Amrel DC load tester that I have been using to test batteries. I have successfully used it to test many 12v and 24v batteries with no problem.I am now trying to test some 3.65v 280ah cells and I am having issues. Soon after I start the test the load goes into protection wit an oscillation to high error.I can disable this feature but I don't want to harm the load. I can measure a small amount of ac voltage on the battery in the mv range and was wondering if this is causing the issue.questions areCan someone explain to me why this is happening?If I disable this protection will I harm my load? Assuming yes because why else would the load go into protectionWhy is there a measurable ac voltage on a dc battery?These are eve 3.65 280ah lifepo4 batteriesThanks

author: SmokedComponent
author_rank: Regular Contributor
timestamp: October 02, 2020, 02:51:39 pm
content: I had this happen while I was testing power supply with a constant current DC load, happened at low voltage, high current. I assume supply's and load's control loops started fighting each other and went berserk.Not sure why this happens on a battery... perhaps because of some current limiting circuitry?I stopped the oscillation in my case by connecting a few uF capacitor across load terminals.
Question about oscilloscope hacks - Page 1
Forum section: electronics / beginners

author: mehdi
author_rank: Regular Contributor
timestamp: December 08, 2015, 09:19:46 pm
content: HiThere are some oscilloscope hacks for some models that add functionality to the oscope (like, increasing the bandwidth)Now I've got a question:Are manufacturers OK with publishing these hacks? (I mean, is it like a software crack which is illegal or not?)I've got a R&S HMO1002 and there are some software upgrades for it (bandwidth upgrade, bus decoding, ...)Everyone can mess around with their oscopes (at their own risk/responsibility), but what about sharing it?(I need more info about the whole thing; have there been any legal actions in the past?)Thanks

author: Fat
author_rank: Regular Contributor
timestamp: December 08, 2015, 09:53:45 pm
content: Well, this should generate a lot of opinions. Here's how I look at it, if you buy a product and it has additional functionality that you can open then it's yours and you can do with it as you like. They are the ones that set it up like that. I'm not talking about finding an open door in a store and helping yourself, just generating the most from the product they produced and I own.I feel the same about a smart phone. The network buys them, adds bloatware and sells it to me. If I want to crack it and take that crap off I own it, so why not?I don't think the manufactures are too happy about the hacks, but they keep producing products with the very same firmware that has been exploited for years. Rigol is a prime example. They don't appear to have done anything to prevent hacking to open bandwidth. While I say this, I bought a Rigol scope this year and debated paying less and hacking it, but I bought the official 100 meg version from an authorized distributor.Fat
Question about power - Page 1
Forum section: electronics / beginners

author: florian_1034
author_rank: Newbie
timestamp: June 22, 2016, 07:49:27 am
content: Let's say I have a motor connected to a load. Motor efficiency is 100%. And we have two formula: P=U*I and P=F*v. So my question is: If I measure the power consumed from electric grid by the motor and then I measure the power output by the motor, this two powers will be equal? Is electric powerthe same as mechanical power?

author: Kremmen
author_rank: Super Contributor
timestamp: June 22, 2016, 08:35:09 am
content: Yes to the first question, no to the second.An electric motor is a converter of electric to mechanical power. If the motor is 100% efficient then the conversion is perfect and by definition all electrical power is converted to mechanical. That takes care of Q1.Only no motor really is 100% efficient so there will always be various losses dependent on the nature of the particular motor. At least there will always be thermal losses in the windings and magnetic circuits. Sometimes noise is loud enough to figure in the efficiency and there might be other loss mechanisms that don't immediately come to mind.Electric and mechanical power are 2 distinct and separate things so they are not "the same". There might be a near 1:1 correspondence between one and the other as when converted by a motor but the same they are not. Definitely not in nature but in practice neither in magnitude.A minor comment regarding the formulas. They are correct as stated but the mechanical one applies to linear motion whereas motors are usually (not always!) rotary in nature. So a more appropriate form would be:P = T*w, whereT = torque,w (omega) = angular velocity.In the SI system (the only correct one and those who disagree can exit stage left) torque is given in newtonmeters  and angular velocity in radians per second [1/s] and then you get power directly in watts .

author: danadak
author_rank: Super Contributor
timestamp: June 22, 2016, 09:44:40 am
content: Electric power is the rate at which electrical energy is transferred by an electric circuit.Mechanical power. In physics, power is the rate of doing work. It is the amount of energy consumed per unit time.They are essentially the same, rate of energy transfer/consumption per unit time. Both nominally in Joules/sec,but can be expressed in various ways.Regards, Dana.

author: florian_1034
author_rank: Newbie
timestamp: June 22, 2016, 09:51:51 am
content: Thanks a lot Kremmen ! Your response is well explained and very helpful to me.

author: Galenbo
author_rank: Super Contributor
timestamp: June 22, 2016, 10:01:23 am
content: Quote from: florian_1034 on June 22, 2016, 07:49:27 amWith your 100% assumption: yes. P is universal, but always define the "box"If you then add an ideal gearbox and an ideal generator, you have an 100% efficient frequency converter.A not-100% version existed, to convert 60Hz to 50Hz on boats going from EU to USA.
Question about principles of electronics and station solder  - Page 1
Forum section: electronics / beginners

author: javiercj
author_rank: Newbie
timestamp: March 11, 2013, 12:24:47 pm
content: HELLO FRIENDSmy name is javier, I am from SpainI am begineners in electronicsI need help me.Hopefully would have known before this man david eevblog.I am Spanish and beginner electronics.Work with drive units of cars, computers, soldering and desoldering components micros motorolla eeprom-etci do have same knowledge of electronics, I need to learn the knowledge of the components and operation of the circuits.I have read manual of electronica-youtube videosetc but I need to know many things, they can helpyou can advise manual notes or books.I have seen in the Forum that they advise:Practical electronics for inventors and arts of electronics (2 Edition) and "Electronic Principles" by Malvino (3 diccion)?Y other question:than soldering station advise me to be good and have to weld penAnd allow soldering and desoldering with hot air gun for desoldering and remove Tinthank you. hakco model fx-888d only has for welding with penthanks

author: paulsev
author_rank: Contributor
timestamp: March 11, 2013, 12:38:04 pm
content: The Art of Electronics, by Paul Horowitz and Winfield Hill, is a popular textbook dealing with analog and digital electronics. Dave has done a book review of this in one of his very early videos.http://en.wikipedia.org/wiki/The_Art_of_Electronics

author: RjSa
author_rank: Contributor
timestamp: March 11, 2013, 03:03:55 pm
content: Hi Javier,As mentioned by paulsev, The Art of Electronics is like the bible of Electronics.I know you are looking for some basic electronics books but if you want some information on microcontrollers there are two excellent boks writen in spanish.The first one is "Microcontroladores PIC16F84. Desarrollo Basico y Aplicaciones" by Jose M. Angulo. (Basic and Assembler languages)The other book is "Microcontrolador PIC16F84. Desarrollo de proyectos" from Editorial RaMa. I find this to be an excellent entry level microcontroller ebook since it comes with all the simulations on Isis.I hope you find these suggestions useful in the future.

author: RjSa
author_rank: Contributor
timestamp: March 11, 2013, 03:16:28 pm
content: Also, you can check this thread:https://www.eevblog.com/forum/beginners/electronics-primers-course-material-and-books/

author: javiercj
author_rank: Newbie
timestamp: March 12, 2013, 09:14:56 am
content: Many thanks to all Pauselv and rjsa'll start with books and when in doubt ask youthank you.Aquestion where I can buy hakco soldering station, since I'm from Spain, and through amazon makes me buy from Spain.Know that website you can buy this season of dave, or other similarthank you

author: nukie
author_rank: Frequent Contributor
timestamp: March 12, 2013, 11:34:10 pm
content: In Spain you buy a JBC if wallet permits.
Question About Probe Leads - Page 1
Forum section: electronics / beginners

author: eev_carl
author_rank: Regular Contributor
timestamp: January 02, 2018, 09:09:37 pm
content: Hi,When I use a 1.2m BNC-BNC connection between my scope and function generator, I can get a square wave without any overshoot. However, when I replace this with a 1.2m scope probe and a 0.6m lead from the function generator, I get an overshoot of several hundred millivolts.Is this from the extra inductance in the cable length or is something being lost in all the alligator clips and breadboard contacts?Thanks,Carl

author: w2aew
author_rank: Super Contributor
timestamp: January 02, 2018, 09:14:22 pm
content: Most likely ground lead inductance from the probe ground connection (and anything else, like a breadboard, etc. in the ground path), plus the lack of proper termination.This video might help you better see the ground lead inductance issue:

author: eev_carl
author_rank: Regular Contributor
timestamp: January 02, 2018, 09:44:09 pm
content: Thanks. That video is a great help and right to the point."...assorted ground connections, but most people don't pay too much attention to them"Is my BNC-BNC connector an even shorter ground connection than the wrapped wire shown in the video?

author: eev_carl
author_rank: Regular Contributor
timestamp: January 02, 2018, 09:59:15 pm
content: There are a pair of wrapped wire lead attachments that come with the Rigol 1054Z similar to what's shown in the above video. Doing some A/B testing on my test circuit, I was able to reduce the overshoot by 100mV on the +-2.5V square wave.

author: w2aew
author_rank: Super Contributor
timestamp: January 02, 2018, 10:08:02 pm
content: You might also like this video, which improves on the spring-type ground contact:

author: Electro Detective
author_rank: Super Contributor
timestamp: January 02, 2018, 10:36:18 pm
content: Awesome videos as usualA dab of contact cleaner or isopropyl alcohol on the probe and soldered springy thingies before connecting should get the business going 100%especially if not used for a while, you'd be surprised how much fine dust, tarnish and air born muck can hit on the connectors that appear clean and shiny
question about resistor across power pins - Page 1
Forum section: electronics / beginners

author: sanman
author_rank: Contributor
timestamp: June 14, 2019, 01:58:56 pm
content: hello!I'm working on a board that contains a TI SN74VC8T245. This device uses a VCC-A and VCC-B at pins 1 and 24 respectively.The manufacturer has bodged a resistor between the two pins, connecting at the bypass capacitors for each pin.I was just curious why this was being done. Any insight?Thanks for the help!

author: cur8xgo
author_rank: Regular Contributor
timestamp: June 14, 2019, 03:51:34 pm
content: Quote from: sanman on June 14, 2019, 01:58:56 pmJust guessing here:-That resistor net effect is to keep the two Vcc's near each other-So if one goes down, the other goes down, if one comes up, the other comes up-Sort of synchronizing thingsOtherwise, one Vcc might be up a long time (relatively speaking) before the other, with undesirable results.Also notice the "keep ~OE high until vcca and vccb are up"...similar reasoning
Question about RFI in AC circuits - Page 1
Forum section: electronics / beginners

author: Jay112
author_rank: Regular Contributor
timestamp: May 16, 2016, 02:40:25 am
content: Can anyone please tell me what the following paragraph means? It's from a digikey description of a zero voltage switch:"Zero-Crossing Detector - Synchronizes the output pulses of the circuit at the time when the AC cycle is at zero voltage point; thereby eliminating radio-frequency interference (RFI) when used with resistive loads."How does synchronizing the output at the time when AC is at zero voltage eliminate RFI? I just want to learn.

author: Audioguru
author_rank: Super Contributor
timestamp: May 16, 2016, 03:08:19 am
content: The voltage of AC swings to a positive peak then drops through 0V then goes to a negative peak then rises through 0V to go positive again, over and over. If a switch is activated at a voltage peak then the high voltage causes interference. But if the switch is activated when the voltage is 0V then there is no interference. The AC 0V is detected by a zero-crossing circuit.

author: Tomorokoshi
author_rank: Super Contributor
timestamp: May 16, 2016, 03:20:01 am
content: Another way to think about it:Let's say my Triac gets turned on at the peak of the AC cycle. We are then switching the load circuit from 0V to 170V in the turn-on time of the Triac. Although it's a resistive load, there is some inductance and capacitance in the line. The "step" function of the Triac turning on looks like a square wave edge, which will look like a blast of energy at a wide frequency range, particularly across the AM band.By starting at the zero-cross, there is no meaningful sudden voltage switching - it just happens to have a higher load on that next cycle.Turning off at the zero-cross is also important.
Question about SMPS - Page 1
Forum section: electronics / beginners

author: zaoka
author_rank: Frequent Contributor
timestamp: September 25, 2013, 10:25:52 pm
content: Why some SMPS are using upconverter circuit that makes around 400V and than its going to SMPS power supply?I see that a lot in these new power supplies for TVs.Why is that needed?

author: c4757p
author_rank: Super Contributor
timestamp: September 25, 2013, 10:32:34 pm
content: Power factor correction. That circuit regulates the current waveform of the power supply so it doesn't draw power in the harsh spikes characteristic of a simple diode-capacitor rectifier.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: September 25, 2013, 10:35:57 pm
content: Hi,The up converter is normally called a boost circuit in smps language.The purpose of the boost stage is obtain a sinusoidal input current, power factor correction. Without the PFC the power factor will be about 0.65 for a capacitor input filter. with the PFC power factors of 0.95 or better are possible. In some areas PFC are mandatory for power consumptions over a certain limit.I suggest a search for 'PFC'Jay_Diddy_B
Question about sourcing image sensor - Page 1
Forum section: electronics / beginners

author: giacomo
author_rank: Regular Contributor
timestamp: March 18, 2021, 02:03:25 pm
content: From the image the camera is an OV5640, from a quick google search it is made from OmniVision and WaveShare has use it for building the module pictured.You may try to ask to OmniVision directly there is a "contact sales" in the home page as wells as a list of official distributors

author: PeterJ
author_rank: Newbie
timestamp: March 21, 2021, 03:23:53 am
content: Lots matches Aliexpress .https://www.aliexpress.com/premium/OV5640.html?d=y&origin=y&catId=0&initiative_id=SB_20210320192034&SearchText=OV5640However in my opinion would be quite risky buying from such suppliers and large qty u want is doubtful capacity of these suppliers.

author: Ian.M
author_rank: Super Contributor
timestamp: March 21, 2021, 07:33:47 am
content: There's not much about the OV5640 on the Omnivision website. Its over 10 years old and appears to be discontinued. The only remaining OV564x series sensor is end of life:https://www.ovt.com/sensors/OV5645Therefore you are vanishingly unlikely to find a reputable supplier for assembled OV564x camera modules with enough stock available.1000 to 5000 units is very small volume as most of these camera sensors end up in cellphones with typically millions of units sold. Also, pre-assembled camera breakout boards are a niche market, catering to the maker community, who typically don't care much about *what* module they get as long as the specs vaguely match their needs and there is an Arduino library or Linux kernal module to support it.Can you use aSony IMX219sensor, as used in the current Raspberry Pi official camera modules? Its an 8 megapixel sensor with a four lane MIPI interface. However the Pi camera module only breaks out two MIPI lanes. Its also fixed focus.If so, Farnell/Element14 have over 40K units available:https://au.element14.com/w/c/embedded-computers-education-maker-boards/raspberry-pi/embedded-daughter-boards-modules-raspberry-pi?product-range=camera-modules&range=inc-in-stock
Question about ssd's and possibility to add nands. - Page 1
Forum section: electronics / beginners

author: darknezz19
author_rank: Newbie
timestamp: August 21, 2015, 04:05:38 pm
content: I have a Corsair Nova2 60GB which has a ps3105-s5 for the controller and 8 nand chips on one side and 8 blank tsop foot print on the other side.I see other model ssd's using the same controller but with different nand chip arrays. What I'm wondering is if these controller chips can auto detect what the nands sizes are or if it's a hardware thing where the nand has certain pins getting power or ground or resistor values for example to tell the controller what size they are.I know that the ps3105-s5 firmware is a universal one and the same across all devices as they can all be flashed with the firmware located in the link below. Thank you for your time.http://www.mydigitaldiscount.com/mydigitalssd-msata-half-slim-and-2.5-inch-sata-ssd-firmware-update-p2.html

author: rvalente
author_rank: Frequent Contributor
timestamp: August 21, 2015, 06:33:11 pm
content: I believe yes, the should be smart enough to detect the correct size of the memories after formatting, does not make much sense to have many versions of a very similar product.But the question is, does it worth changing the NANDs? Do you have them in hand for swap (like a dead SSD as donor?)
Question about starting to use Altium to create the microcontroller - Page 1
Forum section: electronics / beginners

author: willielin
author_rank: Newbie
timestamp: August 27, 2016, 05:45:03 am
content: Hi all,This is my first time to use Altium and microcontroller.....When i start to define the pin of the microcontroller (LPC824, TSSOP20), I get confused that there are more than one type in each pins, like pin 19 got IO and A, so what should i put in the 'electrical type' .Datasheet of the LPC82x :http://www.nxp.com/documents/data_sheet/LPC82X.pdf?fasp=1&WT_TYPE=Data%20Sheets&WT_VENDOR=FREESCALE&WT_FILE_FORMAT=pdf&WT_ASSET=Documentation&fileExt=.pdf

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 27, 2016, 03:28:26 pm
content: First of all, do you know what "electrical type" means, and how it's used?Tim

author: willielin
author_rank: Newbie
timestamp: August 28, 2016, 06:00:07 am
content: Quote from: T3sl4co1l on August 27, 2016, 03:28:26 pmNot sure actually... Do you know where can i get all those base knowledge of microcontroller? At least I can understand the data base...

author: Mechatrommer
author_rank: Super Contributor
timestamp: August 28, 2016, 06:18:44 am
content: Quote from: willielin on August 27, 2016, 05:45:03 ammy way is you put the name "IO/A/etc" in "Component Description" or "Designator" or "Display Name", if you like me, i want "Display Name" and "Designator" to be simple such as "IO" or "PB1" or whatever. and the for multipurpose pin like this you just leave the electrical type as default or "passive". this is during component making. when you put the component in schematic (after compilation) for real application, you double click the component, uncheck "lock pins", click ok, and double click the pin, you may choose the suitable type for your particular application, such as "input" if you want set the pin as adc, or digital input. or set the pin as "output" if its an output. ymmv.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2016, 04:23:50 pm
content: Quote from: blueskull on August 28, 2016, 06:04:56 amAll port pins. Might as well set the power (VCC/GND) pins to power. Or passive everything, that's the "ignore this pin" default.Tim
question about step down buck converters  - Page 1
Forum section: electronics / beginners

author: little_carlos
author_rank: Regular Contributor
timestamp: March 20, 2016, 02:07:17 pm
content: Ive seen that from older to newer, they work with higher frecuencies, whats the main reason to do that? the only thing i know is that, the size of the low pass filter gets reduced, but, is that the only reason? does it makes them more efficient? could that cause bigger interferienece?Thanks for your time

author: SeanB
author_rank: Super Contributor
timestamp: March 20, 2016, 02:12:40 pm
content: Smaller size of the magnetic components, smaller value capacitors needed on input and output, smaller switching transistors and most importantly the loop response is so much faster leading to better regulation on step changes. The higher frequency also means less copper needed in inductors making them cheaper to wind for the same output power, and you can use cheap ceramic SMD capacitors instead of large electrolytics in many cases.Higher frequency also means that if you want a smooth DC output with reduced ripple you need a smaller post switcher LC filter yes, but the inductor in the converter is doing much more than low pass filtering.

author: danadak
author_rank: Super Contributor
timestamp: March 20, 2016, 02:49:03 pm
content: The downsides are the generation of noise into the RF area, theneed for Cs that act like a C at high frequency, eg. esr curvesstill predominantly capacitive at the higher frequencies. Need forbetter bypassing. Switches that incur higher Pdiss because ofswitch drive dissipation. Increased coupling on PCB trace to tracein sensitive signal paths......To name a few.Regards, Dana.
Question about strange trace around PCB - Page 1
Forum section: electronics / beginners

author: Jay112
author_rank: Regular Contributor
timestamp: May 02, 2016, 11:28:14 pm
content: Can anyone please explain why this solderable breadboard looks like it has a copper trace that goes all the way around it, but which doesn't seem connected to anything?http://www.ebay.com/itm/10Pcs-5x10cm-Single-Side-Copper-Prototype-Paper-PCB-Breadboard-2-3-5-joint-hole-/401051491158

author: alsetalokin4017
author_rank: Super Contributor
timestamp: May 02, 2016, 11:42:27 pm
content: Can be used for common ground and/or power. Cut trace in convenient place to make 2 traces, one for + and one for -. Connect by bridging from a pad with hole over to the common trace(s). Or drill your own hole(s) where convenient.

author: Jay112
author_rank: Regular Contributor
timestamp: May 02, 2016, 11:49:52 pm
content: Quote from: alsetalokin4017 on May 02, 2016, 11:42:27 pmThat's good to know! Thanks!
Question about T4148 diode - Page 1
Forum section: electronics / beginners

author: bshi02
author_rank: Regular Contributor
timestamp: May 21, 2024, 04:33:19 am
content: I have 29 inch CRT TV which only display black screen.After I switched on it and I heard inrush of current,I measured voltage of pin #9 of STR-S6707(6.2-6.4 DCV). and I found that DCV(placing red probe to cathode terminal of diode) of ZD808(it is written B75 on its surface) was 6.8-7V and D805 and positive pin of C811 was 6.4-6.6V , D807 and positive pin of C812 was 13-13.2V, But D806(it is written T4148 on its surface) was 0.5-0.7V like below schematic which I drawed STR-S6707 and its adjacent components in circuit . It is doubt whether D806 is faulty diode or not.

author: m k
author_rank: Super Contributor
timestamp: May 31, 2024, 12:11:09 pm
content: STR-S6707 datasheet covers STR-S6709.Maybe it's demo circuit can give some thoughts.

author: kevin.gibbs
author_rank: Regular Contributor
timestamp: August 22, 2024, 02:10:18 am
content: In general, 0,5-0,7V should drop on the diodes. Therefore, it is likely in good working order. You can unsolder one of the leads for complete confidence and then check the diode with a multimeter.
Question about the IEC C7/C8 "Figure 8" connector - Page 1
Forum section: electronics / beginners

author: TheNerd389
author_rank: Contributor
timestamp: March 23, 2021, 06:56:51 pm
content: I am currently working on a school project (a high efficiency, low performance benchtop power supply) that requires a C7/C8 connector to be routed inside an enclosure. I know this is odd, but I don't really want to disassemble the power brick I'm using in order to bypass the connector.The idea is to use a panel-mount C8 jack on the enclosure and wire that to the power brick on the inside. My problem is that the strain relief on commercial C7 cables is huge, and I'd like to use something more compact. I decided to try and find a C7 connector that I could assemble into a cable myself, but I can't find any.Can someone explain why these connectors either don't exist or are so hard to find?P.S. The internal cable I am planning to use has one other noteworthy feature. It's also a Y-adapter so that several of these devices can be daisy-chained together. I've already double checked that I'm within the current limits of the connector for any reasonable array of these devices. Each one will draw around 0.2 amps from the grid at most, and the design is based on chaining a maximum of 5 units together.

author: Jeroen3
author_rank: Super Contributor
timestamp: March 23, 2021, 07:06:23 pm
content: Isangledan option?Btw: they also sell diy.https://www.allekabels.nl/stekker/1242/1583768/c7-apparaat-stekker.html

author: TheNerd389
author_rank: Contributor
timestamp: March 23, 2021, 07:25:08 pm
content: Quote from: Jeroen3 on March 23, 2021, 07:06:23 pmI considered angled, but it would be a pretty tight fit as well. The enclosure is 180mm wide, 220mm deep, and 50mm tall. The brick itself is 133mm x 59mm x 30mm, and the connections are on the 30mm x 59mm sides. They add to the dimension that's already the most problematic.The strain relief on the built-in cable takes up another 30mm on the longest side. As a result, the long dimension is around 160-165mm before accounting for the C7 connector.Here are some images of the design for some context. (still very much a work in progress)

author: TheNerd389
author_rank: Contributor
timestamp: March 23, 2021, 07:29:14 pm
content: Quote from: Jeroen3 on March 23, 2021, 07:06:23 pmI have no idea how you found this link, by the way.I'll see if they ship to me in the US.Just checked. They do not ship to the US.
Question about the WS2813B addressable LED. - Page 1
Forum section: electronics / beginners

author: MarkS
author_rank: Unknown
timestamp: March 13, 2023, 04:45:03 pm
content: I need to use this in an upcoming project and I'm lost as to the need for a current limiting resistor. The previous model, WS2812, needed one and showed as much in the datasheet. This model not only doesn't mention it, but doesn't provide the current draw to calculate the resistor if needed. All I can gather from the datasheet is this blurb: "...and voltage programmable constant current control part, which achieves highly consistent color effect."The best I can tell is that I just slap a 5V rail to it and done. However, I need 280 of these for this project and that's not cheap. I'd hate to blow them all because of something I missed.Datasheet is attached. Nevermind!Quite clearly posted on page 1:"NO extra components needed, even the capacitor."
Question about the Wye-power splitter - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 04, 2022, 11:37:19 pm
content: I'm looking at a Wye(Y)-power splitter configuration for a 1-to-4 splitter. I'm looking at these two sources below, but they seem to show contradictory material in terms of what resistor value to use. Let's assume a characteristic impedance of 50-ohms.In thisfirst link, with N=4, it states that our resistors need to be Z0*(3/5), or 30-ohms. However, in thissecond link, they show two different resistor values (with R2 = 2* R1) where the value depends on Z0. In this case, R2 would be 33.3 {100/3} and R1 would be 16.67 {50/3}. Why would the first configuration use different resistors than the second one? Plus, do they contradict one another in terms of the number of resistors to use? The second link had me at 11 resistors total. I'm not sure regarding the first link.

author: TimFox
author_rank: Super Contributor
timestamp: March 04, 2022, 11:58:31 pm
content: The second link has three different configurations: the configuration that uses two different resistor values is for a branched network "2: Branched Wye resistive power splitter"Their first configuration, "1: Direct N Way Wye Resistive Power Splitter" also gives 50 x (3/5) = 30 ohms for a single-input, four output (N+4) case, with equal resistors from the input and the four outputs to an isolated central node.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 05, 2022, 05:40:12 am
content: So, I was looking at the wrong one it seems? They both were Wye networks, but the second one seems to be option #1 compared to the first link. If that's the case, then what good is option #2, the Wye network of the second link? I suppose when N=2, both option 1 and option 2 are the same.They're all resistive networks, but they're just configured differently but achieve the same result. Which one do you prefer and why? (As an aside, the delta network uses 9 resistors when N=4?)

author: TimFox
author_rank: Super Contributor
timestamp: March 05, 2022, 02:46:27 pm
content: The second link discusses practical complications about fabrication for extremely high frequencies.The "regular" wye with the minimum number of resistors is what is usually used at "normal" frequencies.
question about thevenin and norton calculations - Page 1
Forum section: electronics / beginners

author: kars
author_rank: Contributor
timestamp: September 17, 2017, 04:25:17 pm
content: hi, for school i have to make some assignments but im a bit stuck on one of them.in the picture, Uv= 8v and i had some resistor values witch i already managed to assign to the right resistor: R1=12ohm R2=47ohm R3=68ohm and R4=33ohm (Uxy had to be 3.5v). calculating I results in 0.2A.next they ask how you can calculate Uthevenin, RtheveninInortonand Rnorton. the only problem is that i just dont understand how, my textbook makes calculating steps i dont get and yt videos use completely different schemas so i dont really get any help from them either.could anyone help me with this?

author: Assafl
author_rank: Frequent Contributor
timestamp: September 17, 2017, 05:01:11 pm
content: What is the load (or if it is there - which resistor is it?). For these you first remove the load, calculate the voltage, and then remove the source and calculate the thevenin equivalent resistance.

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 05:16:01 pm
content: I assume the load is to be connected between X and Y, so the question is what are the source characteristics looking into the X-Y terminals?

author: MattHollands
author_rank: Frequent Contributor
timestamp: September 17, 2017, 05:38:45 pm
content: Always a bit worried that I am providing too much when it is a school question, but I have attached my partial working for this problem.Basically, to find the Norton version, you need to find the short circuit current, and the open circuit voltage. From these you can easily create a norton-equivalent circuit.NB. (Rx||Ry) symbolises the resistance of resistances Rx and Ry in parallel.