hybridfree/eevblog-posts huggingface dataset:
Undercurrent Charging - Li-Ion Battery Pack - Page 1
Forum section: electronics / beginners

author: ed_reardon
author_rank: Regular Contributor
timestamp: May 04, 2017, 09:34:51 am
content: Hello all,Working on a project using batteries, which is a first! I have on hand some 3.7V Li-ion battery packs and a charging IC, the battery spec' is a charge current (not specified as maximum, only 'charge current') of 1500mA but the IC I have on hand tops out at 500mA.This is only for prototyping so I'm not concerned about charge time etc (it can run in to tens of hours, it's only for proof of concept) but not knowing an awful lot about the vagaries of Li-Ion batteries I just wanted to check that a) this IC can charge the battery, b) this doesn't present any danger.Probably a numpty question but I want to tread very carefully with Li-ions as they're not something I've used before and have read many a dire warning about charging/discharging them!You can call me an idiot for asking, I'll take itCheers,Ed

author: Zbig
author_rank: Frequent Contributor
timestamp: May 04, 2017, 10:45:38 am
content: You should be fine. There's nothing wrong with charging Li-Ion/Li-Po below its rated/maximum current.

author: rob77
author_rank: Super Contributor
timestamp: May 04, 2017, 10:54:47 am
content: the lower the charging current is, the longer the battery life will be in terms of charge/discharge cycles. (high charging currents reduce battery life)important is to charge with constant current followed by constant voltage - but that is handled by your charger IC.

author: zoidiano0
author_rank: Newbie
timestamp: May 04, 2017, 10:57:45 am
content: I normally charge my almost death batteries with a very small current.make sure they don't get fried, but once u get the right voltage u can turn the current up to specI do normally charge them at the beginning with Half rated current and then turn it yo a most higher current. 0.8 to 1 amps it's what I useOn most lower capacity (lower than 1500mah) and older ones I use 0.2 at the beginning and then. 0.4 or 0.6Using a lower current charger make sure that ur battery last longerSent from my Moto G (4) using Tapatalk

author: ed_reardon
author_rank: Regular Contributor
timestamp: May 04, 2017, 11:12:36 am
content: Many thanks all.These are onlyfabricobbledprototype units, so cell life isn't that important in the grand-scheme but it's good to know that lower charge currents increase battery life-span.

author: mrratburn
author_rank: Newbie
timestamp: May 04, 2017, 02:23:19 pm
content: 500ma is my standard "safe" charging current for lipos. Pretty much any battery can charge safely (albeit slowly) at this current. If you want to charge at a higher current, you have to be careful, especially since the USB standard typically does not give you more than 500ma unless it's a dedicated charger.

author: FlyingSquirrel
author_rank: Contributor
timestamp: May 04, 2017, 07:20:22 pm
content: also a good tip to prolong the life of a LiPo is to try not discharging it far bellow the storage voltage (3.7-3.8V per cell)plus it also spares you from having to put it to storage voltage after use
Understand low power gate timer TPL5110, any alternates? - Page 1
Forum section: electronics / beginners

author: sairfan1
author_rank: Frequent Contributor
timestamp: July 21, 2020, 05:35:18 pm
content: I was reading datasheet of TPL5110 it says "the TPL5110 has a unique One-shot feature where the timer will only power the MOSFET for one cycle."I do not understand what does it mean by "One-shot feature" and "turning on MOSFET for one cycle" here what does it mean by one cycle?I want to turn on ESP8266 by this timer i want to know which MOSFET would be suitable in the circuitIs there any better alternate approach for same application?
Understand operation of buck-boost XL_6019 - Page 1
Forum section: electronics / beginners

author: Vindhyachal.takniki
author_rank: Frequent Contributor
timestamp: April 14, 2020, 05:32:12 am
content: 1. Need to design a buck-boost. Inout 10V-30V, output 12.36V/1.55A. Saw this circuit from XLSEMI pic attached using XL6019.2. It has weird implentation of L1,CDC,L2 & D1 as in pic.But wiki shows different implentation, no where I saw two indcutors used, D1 is also in reverse diection,?3. Is it ok to use as given by XLSEMI? Cannot understand its operation

author: Chalcogenide
author_rank: Regular Contributor
timestamp: April 14, 2020, 07:39:03 am
content: Although the circuit in the picture obtains the same effect as of a Buck/Boost converter, it is actually a SEPIC converter.https://en.wikipedia.org/wiki/Single-ended_primary-inductor_converterThe circuit shown in the image is used by countless cheap "CC/CV Buck/Boost" converter modules on Aliexpress and eBay.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 14, 2020, 11:47:54 am
content: Notice the inductors can be coupled, as Fig.17http://www.xlsemi.com/datasheet/XL6019%20datasheet-English.pdfThis improves performance.Note that C2 and COUT need to be placed very close to D1, and CDC, and SW and GND pins of the reg.Note also that the current limiting add-on is likely to oscillate; I think this may depend on load, YMMV.Tim
Understand PSU spec - Page 1
Forum section: electronics / beginners

author: WombatWong
author_rank: Newbie
timestamp: October 12, 2018, 08:46:57 am
content: Hi All,I like to understand the PSU specification. Reference to a Meanwell power supply HRP-600-24. The AC input is 100-240VAC 8.5A 50/60Hz. DC output is 24V, 27A.This is a 600W PSU, based on DC 24V*27A, I got around 600W. However, I do not understand 100-240VAC 8.5A can get 600W.Can anyone explain to me?

author: Kleinstein
author_rank: Super Contributor
timestamp: October 12, 2018, 08:57:58 am
content: The 8.5 A rating is the maximum input current, mainly intended for choosing fuses and lines and switches. It is kind of worst case and thus applies to the 100 V. With a 230 line voltage the actual input current would be lower. There is also a limited efficiency and power factor, so that the current times voltage at the input side would be higher than the output side power. So of to 850 VA in for some 648 W (24 V*27 A) out sounds plausible.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 12, 2018, 09:50:32 am
content: Yeah, 850VA input is quite reasonable. It has PFC, so it's not drawing, say, 1.2kVA for 600W, which is quite nice!It may be that it's rated for 600W, but goes up to ~800W (plus losses) when operated near or beyond current limiting conditions, at certain input voltages. It is probably not 2kVA (at 240V), but if it is, that would be important just to be sure that, in case it needs it, you can deliver it safely.Input current rating is only a maximum figure, for wiring and fusing purposes. It is not indicative of actual operating current or power draw.Tim

author: grifftech
author_rank: Frequent Contributor
timestamp: October 12, 2018, 02:03:29 pm
content: the 8.5 amp max input could be when first plugged in

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 13, 2018, 02:34:52 am
content: Likely inrush is much more, compliant with whatever standard (if any) applies. Probably under 100A peak.Tim
understand the negative signal in case of differential signalling - Page 1
Forum section: electronics / beginners

author: m4rtin
author_rank: Regular Contributor
timestamp: December 27, 2014, 04:42:57 am
content: In case of single-ended signals the signal is referenced against common ground. In addition, as I have understood, ground and signal path form the closed circuit. Now differential signals do not have a ground. Does this mean that electrical circuit is formed between the signal+ and signal-, e.g. between Tx+ and Tx- in case of100BASE-TX?

author: Richard Crowley
author_rank: Super Contributor
timestamp: December 27, 2014, 04:58:13 am
content: Yes, exactly. That kind of balanced/differential signaling has been used for everything from old-school analog telephone lines to MIDI and most recently as LVDS (low-voltage differential signaling) used in HDMI and SATA and the internal connection to flat-panel displays, etc. etc.http://en.wikipedia.org/wiki/Low-voltage_differential_signaling#Applications

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 27, 2014, 05:43:32 am
content: Note that there is still an implicit ground, whether you provide one or not: common mode current, interference and so on. Differential traces on a PCB almost always exhibit more "lone trace" character than "differential" character, as does shielded twisted pair.On the rare occasions when a twisted pair is specifically designed to behave as such (like Ethernet), it's almost always with some horrible catch, like how Ethernet is transformer (AC) coupled only. So your encoding scheme has to take that into account.You can't simply shove, say, RS-422 down CAT-5 and expect it to work anywhere near as well as Ethernet does!Tim
Understanding a circuit - Page 1
Forum section: electronics / beginners

author: Moriambar
author_rank: Unknown
timestamp: September 05, 2023, 02:32:17 pm
content: Hi there,you can find a circuit in the attachment,which I found onlinethat I'm really struggling to understand, especially the left partSo here's my analysis:At the start C2 looks like a short so there's no voltage drop across it,the same across C3 and R4.There's basically where I'm stuck: the capacitor will have to act as... capacitor, but where does the current flow, since there's no voltage drop anywhere? I mean: being the trannys bjt don't they need a current to be activated? I mean everything would be different if they were n-enh-fets...I have more questions though:why putting C3 in parallel with R4? I don't understand what they're trying to achieve and how...last question: is the purpose of C7 to make, with R8//R9 a low pass filter, smoothing the transitions?Cheers!

author: DavidAlfa
author_rank: Super Contributor
timestamp: September 05, 2023, 02:44:29 pm
content: Did you read the site you took the circuit from?QuoteSo C3 (Discharged) will initially behave like a short, so it energizes the coil at max. power.Later C3 gets charged and the current can only flow through R4, reducing the relay current.
Understanding a concept about power mos - Page 1
Forum section: electronics / beginners

author: molotov
author_rank: Newbie
timestamp: September 11, 2015, 03:19:09 pm
content: Hello,First post here. Great site, I have learned a lot from your youtube videos.Anyway, I have a basic conceptual question about some power mosfets I bought for a project.I have some STP75NF20 which are rated for 200v and 75amp in the TO-220 package.I am just wondering how something so small and with such small leads can handle that type of power.Where as wire rated for that current and voltage would typically have a diameter larger than the entire mosfet.And what is the proper way to make the high power connections to the device with large diameter wire?

author: Jony130
author_rank: Newbie
timestamp: September 11, 2015, 03:39:06 pm
content: Please read thishttp://www.mcmanis.com/chuck/robotics/projects/esc2/FET-power.html
Understanding a drawing for making a foot print - Page 1
Forum section: electronics / beginners

author: horse_ebookmarklet
author_rank: Contributor
timestamp: October 13, 2022, 03:14:11 am
content: Trying to use aTYPE-C-31-M-12USB C Connector for a USB2.0 application, power and data.In the datasheet, there is this dimension, labeled "4-1.01". Based on scale and other clues, I take this to mean the tab is 1.01mm tall.But what is the "4-" about? What does "4-1.01" mean?

author: ledtester
author_rank: Super Contributor
timestamp: October 13, 2022, 03:25:17 am
content: It's a multiplicity -- the number of things having that dimension."2-1.10" -- there are 2 tabs having width 1.10"4-1.01" -- there are 4 tabs having height 1.01"16-0.20+/-0.05" -- there are 16 pins having width 0.20 +/- 0.05.

author: horse_ebookmarklet
author_rank: Contributor
timestamp: October 13, 2022, 03:31:40 am
content: Ok, that makes sense. Thank you for the explanation!
Understanding a floating input pin on an IC - Page 1
Forum section: electronics / beginners

author: tiofilo75
author_rank: Contributor
timestamp: May 02, 2013, 10:08:45 pm
content: Hi I am trying to use the IC HT9200B and need some clarification on what it is meant by "floating pin". I am trying to figure out what this statement means:QuoteNow I want to use the serial mode so I am confused by the floating portion of the statement. I thought that having a floating input pin was a not a good thing so can someone shed some light please.

author: digsys
author_rank: Unknown
timestamp: May 02, 2013, 11:07:19 pm
content: Firstly, a floating Input is really only "dangerous" if the environment is noisy AND it is tied to an IRQ or similar (in most cases).If it isn't tied to internal hardware / IRQs, then it can't do much "damage". It's just NOT good "practice" to leave Inputs floating.In this case, they rely on the External Clock being dynamic, and therefore the Input is really never floating. Same as if you driveit from logic ICs etc No need for pull-up / down resistors. IF you check the specs of all I/O, you'll usually find details of how theyare set up, and what you MAY need to be careful of.

author: jmole
author_rank: Regular Contributor
timestamp: May 02, 2013, 11:18:11 pm
content: It's basically saying that in parallel mode, the pin has an internal pullup resistor. In serial mode, the internal pullup is disconnected, leaving the pin "floating", or in a High-Z state, disconnected from any other internal circuitry (aside from the gate(s) of whatever transistor(s) it's connected to).

author: JuKu
author_rank: Frequent Contributor
timestamp: May 03, 2013, 05:32:33 am
content: Quote from: digsys on May 02, 2013, 11:07:19 pmIt is not that simple. Floating input is an ESD hazard. Also, the noise can cause rapid switching of input states, wasting power (bad, if you running on battieries). This will also make the power rails inside the chip very noisy, causing all kinds of random behavior. Also, it can easily float to in-between voltages, which causes both input transistors to conduct. I haven't seen a chip actually burn because of that, but would you think a chip works as it should if i/o section power is shorting to ground every now and then?Open input is worse than just bad practice, it is bad engineering.

author: CarlG
author_rank: Regular Contributor
timestamp: May 04, 2013, 02:35:28 pm
content: I fully agree with JuKuIn your particular case, the CLK pin will obviously be driven by whatever is running the serial interface to the HT9200B, so the node will not be floating in your circuit.The manufacturer have provided each pin with a PU where needed, and enabling/disabling them by the serial/parallel selection pin, so that you don't have to add discrete PU's to the unused pins.
Understanding address decoders of a RAM IC - Page 1
Forum section: electronics / beginners

author: gkmaia
author_rank: Frequent Contributor
timestamp: April 15, 2019, 03:27:55 am
content: I am trying to understand how the decoders of this 256k RAM would look like.As I have 16 address inputs, that made me think about on 8x256 decoder for rows and another 8x256 decoder for columns. But that would only give me 65,536 registers.So to achieve 262,144 registers with 16 address inputs, how the decoders should look like?

author: jmelson
author_rank: Super Contributor
timestamp: April 15, 2019, 03:40:48 am
content: I count 18 address bits. That's 256 K words of 8 bits each.Jon

author: Nerull
author_rank: Frequent Contributor
timestamp: April 15, 2019, 03:42:21 am
content: 0-17 isn't 16 inputs...

author: gkmaia
author_rank: Frequent Contributor
timestamp: April 15, 2019, 08:01:06 am
content: That is right. I did count it wrong.In a case of 18 lines how does the decoders would be configured?

author: Berni
author_rank: Super Contributor
timestamp: April 15, 2019, 09:17:40 am
content: Well its just a decoder.18 signals go in in and 262144 signals come out. Depending on what address is passed in the decoder selects the appropriate output and signals the appropriate memory row to be selected.Tho in the actual chip design its probably not one big decoder for practical reasons. You would likely have the memory array split up into many parts (could be only 4 could be over 1024). Then you could have a decoder with only as many output as there are memory sections, this would then activate that section where a separate smaller address decoder would find the right byte in that section. Because the sections of memory are now smaller means that the decoder in each section can also be smaller. But this is just an implementation detail, functions just the same as one big decoder so the datasheet will show it like that for simplicity.
Understanding Arduino Laser Sensor Module Componets  - Page 1
Forum section: electronics / beginners

author: Andrew74
author_rank: Newbie
timestamp: October 23, 2017, 06:45:30 pm
content: Hi my name is Andrew and I'm new to the forum but long time EEVBlog YouTube watcher. I recently have been trying to learn how to create simple circuits by following the Arduino starter examples which has been going well. I have moved on to trying to create a laser sensor project using the KY-008 laser module. This works fine but I want to understand what components they are using to build the module. I linked a picture of the Module below. It looks like fairly simple components but I don't understand what all the jelly bean components are doing. My understanding was that the laser sensors would output a variable voltage depending on how much light is being shined on to it. According to the oscilloscope it's output for a digital output of 0v or 3v. This digital output is exactly what I wanted for my project. Although, this project is pointless if I don't understand what the Module is doing or how it works to get a digital output.https://m.banggood.com/Laser-Receiver-Module-Non-modulator-Tube-Laser-Sensor-Module-p-915633.htmlThanks in advanced,Andrew

author: rstofer
author_rank: Super Contributor
timestamp: October 23, 2017, 07:15:37 pm
content: You can chase the traces on the back side of the PCB to come up with a schematic.One 10k resistor is in series with the LED and I don't see where there is any on-off signal for the LED so it seems like it is on all the time EXCEPT that a 10k resistor is much too high for the LED to actually turn on. You'll have to chase this...The other 10k resistor is pulling the output up to 5V. The sensor pulls the voltage down.In the end, it is an on-off type of sensor. The output is high when the sensor detects the laser and low when it doesn't detect the laser.The capacitor just provides a little transient reduction and is connected across Vcc and Gnd.Not much to it...

author: Andrew74
author_rank: Newbie
timestamp: October 24, 2017, 01:46:55 am
content: Is this a special sensor that has a digital output? Is it safe to assume the sensor is a phototransitor because it has three legs? Based on my readings there are three lights sensors the photodiode, photoresistor, and phototransitor. Are you able to tell me what type of sensor is used on the module? I would like buy some at a cheaper rate than $5ea. I just don't know what to search on digikey.Thanks,Andrew
understanding bit shifting  - Page 1
Forum section: electronics / beginners

author: khatus
author_rank: Regular Contributor
timestamp: April 07, 2020, 08:27:23 pm
content: https://www.best-microcontroller-projects.com/pwm-pic.htmlControlling the PIC PWM outputThe PIC microcontrollers are very easy to use with PWM as they have built in PWM generators - all you do is set up the relevant control registersCode:// Timer 2 PIC PWMPR2=199; // 4 MHz clock -> 5kHz PWM frequencyT2CON = (1<<TMR2ON);// Initialize Control PIC PWMCCPR1L = 30; // Initial DutyCCP1CON = 0x0f; // PWM mode set and 5,4 duty = 0...and then control the duty cycle:CCPR1L = j;Note that the above control only controls the upper 8 bits of the 10 bit PWM module (the other two bits are in a CCP1CON).Here how doesT2CON = (1<<TMR2ON);works??

author: Siwastaja
author_rank: Super Contributor
timestamp: April 07, 2020, 08:49:50 pm
content: TMR2ON is #defined elsewhere (likely in some .h file provided by Microchip), and hopefully equals to 2. This means, the compiler preprocessor replaces TMR2ON with 2.(1<<2) means that value 1 is shifted left by two places:0000 0001->0000 0010->0000 0100

author: admiralk
author_rank: Regular Contributor
timestamp: April 07, 2020, 09:23:43 pm
content: It is not a very good idea to use equality (=) to do that as it could also change all the other bits, if they are not already 0. If you are only wanting to change one bit, you should or (|=) the value. SoCode:T2CON |= (1 << TMR2ON);is the proper way to do it. That way you only set the TMR2ON bit and leave the rest unchanged.
Understanding BOURNS 3680 series digital potentiometer - Page 1
Forum section: electronics / beginners

author: Houseman
author_rank: Regular Contributor
timestamp: May 11, 2021, 05:09:19 am
content: Hi forum.I have searched through the net without effort.I want to use a Model 3680 Digital Display Precision Potentiometer but something goes beyond my understanding.Attached is the photoThere are three decades with + and - on the front panelIf for example I have a 20k value how can I with 999 values make all the scale? is it a 20ohm jump between values?And how can I relate let's say a 8.5k ohm value with the number shown on the display?Do I have to do the math every time? (8500/20 = so enter 425?)Is there no 1-to-1 direct relation between the values and the digits on the panel?Thank You all as usual.Steve

author: ataradov
author_rank: Super Contributor
timestamp: May 11, 2021, 05:55:50 am
content: They are available in different number of sections. And the number of sections defines resolution. The resistance range is mapped to the whole range of switches (it would be 0-999 for the 3 section one shown in the picture).You can see specification for resolution. For a single section (3681), the resolution is 10%, which makes sense, since you have 10 numbers to pick form.Also note that only 1K, 5K and 10K values are readily available. The rest are made to order.1K in 3 decade version and 10K in 4 decade version would read the resistance directly. The values 0-999 and 0-9999 would be directly mapped to Ohms reading.

author: Houseman
author_rank: Regular Contributor
timestamp: May 11, 2021, 10:21:37 am
content: Aha, now I understand.So no 20k value is available.Thank You so muchRegards

author: Terry Bites
author_rank: Super Contributor
timestamp: May 11, 2021, 12:10:39 pm
content: Its just like a rotary or slider pot (arrgh) but with discrete steps. To connect seehttps://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiEgcv3y8HwAhUE-hQKHSl0A9MQFjACegQIBRAD&url=http%3A%2F%2Fwww.bourns.com%2Fdata%2Fglobal%2Fpdfs%2F3680.pdf&usg=AOvVaw3SgwdpwVXo0S7WbFGusjwCYou get 1000 steps. Each step is R/1000 so if you get a 10k version, each step is 10 ohms.
Understanding Circuit 3.106C of tAoE - Page 1
Forum section: electronics / beginners

author: Moriambar
author_rank: Unknown
timestamp: May 13, 2021, 11:48:35 am
content: Hello,I'm trying to optimize the time I have at work by studying a bit of tAoE.I'm in chapter 3, page 204 and I'm looking at the circuit C in fig. 3.106, you can find it in the attachment.Without the pnp tranny Q3, the circuit (essentially fig. B) is designed to drive a high current load via 3.3V logic.I understand that Q1+R1 form a current sink creating a drop of ~8V on R2, without Q3 this drives Q2 and switches on the load just fine (as long as switching speed is not critical etc).Unfortunately I don't understand why Q3 should limit current.My reasoning is this. First of all, when the 3.3V pulse is not applied to Q1, Vce|Q3 = 0, so the transistor cannot turn on. When the pulse is present we have that the tranny can turn on.When it's turned on... well I think that Vbe|Q3=-0.6V (one diode drop) and this means that ~1A goes through R5 since there's a 0.6V of drop at its 2 terminals and its value is 0.5 ohms.I don't think this is correct. Most of all, even if it is, how does this limit the current/protect a short (in the circuit after Q2 I presume)?Thanks

author: Ian.M
author_rank: Super Contributor
timestamp: May 13, 2021, 12:05:07 pm
content: R1 and Q1 limit the gate pulldown current to about 180uA. Not coincidentally, that's enough to develop 9V across R2, which is more or less the most that it could be as Q1 collector voltage will always be higher than its emitter voltage.When Q2 is on, a voltage dependent on the load current is developed across R3. When it reaches about 0.6V Q3 starts turning on significantly, reducing the gate drive to Q2, increasing its on resistance and limiting the current. Under 1A load current, and Q3 has negligible effect. N.B. Q2 must be rated for linier operation and over 12W dissipation.If you can run LTspice at work, some poking around in a sim of it should convince you fairly rapidly, otherwise I guess you'll have to make do with Falstad!

author: Moriambar
author_rank: Unknown
timestamp: May 13, 2021, 12:08:19 pm
content: Quote from: Ian.M on May 13, 2021, 12:05:07 pmah I see... so basically it's the Voltage dropped by the 0R5 resistor that turns on Q3, raising its collector voltage towards the emitter, thus shutting off Q2!Unfortunately I can barely look at this forum while at work (and I have my copy of tAoE handy since working from home). But your explanation is excellent, thanks!
Understanding comparator output current - Page 1
Forum section: electronics / beginners

author: kicken
author_rank: Contributor
timestamp: November 20, 2016, 08:38:56 am
content: Having finally gotten the hang of 555-timer circuits (i think) I thought I'd expand my knowledge a bit by trying to incorporate some binary counter ICs and a comparator. I create a circuit that I think should roughly count seconds using two 4-bit counters. I'm also using two 4-bit comparators to determine when the count has reached 59. At that point I want to turn on a transistor.I got confused however when trying to figure out a resistor value for the transistor base. I went tothe datasheetfor my comparator to see what the current limit is on it's output and it listed a bunch of negative values for the output high and also no maximum value. I was planning on using a 5V source for my circuit, and if I am understanding the datasheet correctly that means the output pin in a high state has a typical current of -1mA.I did a bunch of reading regarding negative current, sink vs source current, active high vs active low, etc and so far things are not much clearer to me than before. If my understanding is correct the the datasheet is saying that roughly 1mA would be provided by the comparator to drive the transistor. Does this also mean I don't actually need a current limiting resistor on the transistor base as the current is already limited?Would appreciate any insight into this datasheet. I'm having a hard time understanding what it's saying and what it means for my circuit planning.

author: danadak
author_rank: Super Contributor
timestamp: November 20, 2016, 02:23:44 pm
content: For bipolar transistor as switch -If you are trying to use the transistor s a switch then preferably you want itsaturated. To do that rule of thumb is to "force beta" to ~10, eg. Ibsat =Icsat / 10That will minimize Pdiss in the transistor switch and realize Vcesat. Not theirare some transistor if you interchange collector and base youc an get evenlower Vcesat values.Regards, Dana.

author: Zero999
author_rank: Super Contributor
timestamp: November 20, 2016, 11:25:51 pm
content: How much current do you need the transistor to switch?Perhaps you should consider a logic level MOSFET, rather than a BJT? MOSFETs don't draw any current from the control circuit, apart for a short length of time when it's switched on/off.

author: kicken
author_rank: Contributor
timestamp: November 21, 2016, 03:37:20 am
content: I'm basically just using the transistor as a switch to pull the load pin on my counter ICs to ground so they will reset to 0 on the next clock cycle. To do this I just connected the collector to my +5v along with a 1k resistor and tied the reset line in just after the resistor. If my understanding is correct then when the transistor turns on the line will be pulled low and about 5mA will be flowing through the transistor. I mostly just stuck in a 1k because it's what I have on hand. I'm guessing I could use a higher value and reduce the current even more yet still have it work ok. I've not yet bought any of the counter or comparator parts yet so I could add some bigger resistors to the list.
Understanding DC offset in AC mains - Page 1
Forum section: electronics / beginners

author: adauphin
author_rank: Unknown
timestamp: March 15, 2018, 12:13:32 pm
content: Being a beginner for the most part, going to dive into the reaches of the 1054Z, I would like to find a good article or video on DC voltage in the AC mains.This has likely been discussed in some depth multiple times, I'm just trying to get a grasp on why there is, or can be DC voltage on the mains. I've watched many videos on AC and DC triggering and coupling, I have a slight understanding on how a small amount of DC can raise the amplitude of the signal, just looking to know how and why DC can be bad or good.One reason is I have an occasional hum in my torridial transformer in one of my amps. I would like to narrow down the source and one culprit is my microwave. This happens when the oven is idle, also happens when the plasma tv is on.This hobby is becoming increasingly fascinating and all the videos have helped tremendously. If I'm going to try to track down the cause, I'm very curious as to how and why DC is present in any form on the mains.I'm not going to blindly go probing the mains..so no worries there. I have other projects involving an older preamp and possible leaking caps I need to investigate as well.Thank you for all the great info and help here...

author: coppice
author_rank: Super Contributor
timestamp: March 15, 2018, 01:06:58 pm
content: Mains is universally supplied by a transformer, which is incapable of transferring a DC voltage to the load. However, small local DC components can occur in the voltage waveforms to individual loads, due to the combination of half wave (or other asymmetric) loads creating a big DC component in the current waveform, and the impedance of the wiring.Power supplied from a UPS is different. A UPS should not produce a DC component in its output, but what happens in practice varies somewhat.

author: adauphin
author_rank: Unknown
timestamp: March 15, 2018, 05:02:01 pm
content: Ok...I see what you're saying.I guess I just need to track down the culprit within my house. I also want check a few older components for any bad caps that may introduce some DC, but for the most part I want to speculate the microwave is one.I can hear the hum from the torodial 5 feet away when the microwave is running, the torodial goes quiet when it shuts off. Occasionally the torodial will hum without the microwave being on, then just opening the breaker then closing it again will quiet the transformer.I thought maybe the Fios entering the house could be a problem also. I checked the inside wiring and outside ground, but still doesn't rule it out.

author: tecman
author_rank: Frequent Contributor
timestamp: March 15, 2018, 06:09:50 pm
content: DC on the mains can create headaches for the utility. When I was in University many years ago, one of the seniors made a big inductor. He used a couple of car batteries to pump DC into the power line, eventually causing a pole transformer to fail spectacularly with oil spewing out. Seems, by design, he got the transformer to saturate on each half cycle causing it to thermally fail in a relatively short time.paul
Understanding decoding unknown protocols from 433mhz receivers. - Page 1
Forum section: electronics / beginners

author: optoisolated
author_rank: Unknown
timestamp: July 27, 2019, 07:54:44 am
content: Hi all,I'm trying to understand how best to go about trying to decode and understand an relatively unknown serial bitstream I'm receiving on a 433MHz receiver, extracted from Fine Offset WH3018 Weather Station. I've seen similar projects for the WH1080 series, but this model seems very different. It's not helped that I've not really tried to decode unknown bitstreams before, or even how to capture it using something like an Arduino to prototype a replacement display for it.The receiver module has the 3 obligatory pins; Data, VCC and GND. I've powered it, and connected data to an Arduino module and to a Logic Analyser. I'm struggling to understand the packet structure though. They seem to come in big of 6 second blocks. Or at least there's a blanking period of about ~140ms every 6 seconds. I seriously doubt the packets are that big, but i can't see any logical delineation for the ending of the bitstream so I don't really know where to start decoding this thing. I've attached some screenshots of the logic analyser output for example.I am expecting to see something along the lines of this, with perhaps one or two extra nibbles for the Solar reading, which the 1080 lacks. It makes sense it would be similar to the 1080 as it's just an evolution of the same unit.I have a rough idea of the sort of data that will be coming in (basic weather info from the transmitter outside, including wind speed, wind direction (in the form of one of 8 directions NNE, NE etc), outside temp, humidity, solar output, battery check, and I believe (if it's anything like the 1080, starting with a preamble, an ID, and then ending with some form of checksum).Any guidance or advice on how to approach this problem would be greatly appreciated.Thanks

author: optoisolated
author_rank: Unknown
timestamp: September 03, 2019, 09:28:46 am
content: I've done some more digging and I think I may be approaching this the wrong way. I am going to deconstruct the transmitter and see what I can glean from that end; i think these transmitters are going to be hard to sort out actual data from general noise as I don't see any way for it to do any carrier discrimination when the unit isn't being transmitted to. Once I get my head around how the transmitter is being triggered and what the data packet actually is (Especially the header) I will be in a far better position to figure out the receiver.

author: electricMN
author_rank: Regular Contributor
timestamp: September 03, 2019, 03:03:25 pm
content: I'm not sure if this will help but I'll throw it out there in hopes that it will.https://www.sevenwatt.com/main/wh1080-protocol-v2-fsk/
Understanding Delta Electronics DTC1000 parameter? - Page 1
Forum section: electronics / beginners

author: Vindhyachal.takniki
author_rank: Frequent Contributor
timestamp: May 14, 2020, 05:25:45 am
content: We ran Autotune test for temperature oven using Delta Electronics DTC1000, it retunns below paramters. Any one know what :1. Pb: THis is proportional band, How it is calculated ?2.What is I offset?

author: MosherIV
author_rank: Super Contributor
timestamp: May 14, 2020, 07:17:03 am
content: PID - proporitional, integral, differential.So Pb is likely to be the proporitional part of the PID.Not sure what the I offset does.
Understanding fuse size in a power supply - Page 1
Forum section: electronics / beginners

author: ghulands
author_rank: Contributor
timestamp: April 29, 2016, 07:12:22 pm
content: Hi,I'm still going through chapter 1 of learning the art of electronics, so I'm still a green newbie so bear with me.I was watching some youtube videos on power transformers and then feeding it into a bridge rectifier and then into a switching and/or linear power supply. So I thought i would try to design one based on my knowledge from the videos and then gleaning info from the data sheets of some of the selected components. I was also watching some of Dave's tear downs on multimeters and how important input protection is. So I started googling MOV's for power supplies and came acrossthis threadwith a design for one.Now what I couldn't quite figure out was the sizing of a 600mA fuse on the AC side. Now that I've started on the basics in the textbook I think I understand now and just wanted to get some validation of it.DC Side:Vmax = 25VImax = 4AP = IVPmax = 100WSo taking the max power consumption on the DC side and working back on the AC side100W = (115 / 0.707) x II = (100 x 0.707) / 115I = 614mASo the 600mA fuse would be about right. Is this correct? Any help is greatly appreciated.

author: Karlo_Moharic
author_rank: Regular Contributor
timestamp: April 29, 2016, 07:22:43 pm
content: Short answer: yes

author: ghulands
author_rank: Contributor
timestamp: April 29, 2016, 08:07:58 pm
content: Thanks! I know I still need to know more about power factor and correcting for it.

author: Fank1
author_rank: Regular Contributor
timestamp: April 29, 2016, 09:34:56 pm
content: The 115 V is already RMS not peak.

author: Ian.M
author_rank: Super Contributor
timestamp: April 29, 2016, 10:24:31 pm
content: You are already over the fuse's nominal rating without any allowance for regulator and transformer losses, or inrush current. This is likely to result in premature fuse failure as it will be overloaded at every switch on. An 800mA or even 1A anti-surge fuse would be a far better choice, and if its a toroidal transformer you may also need a series NTC thermistor for inrush current limiting.

author: TimFox
author_rank: Super Contributor
timestamp: April 29, 2016, 10:39:06 pm
content: Also note that there are two different authorities for rating fuses, and their specification differs.3AG (1.25 inch) fuses are rated by Underwriters' Laboratory (US), the US organization.5 x 20 mm fuses can be rated by UL rating, or IEC (European) rating, and UL can verify both ratings.Please check out the fuse manufacturers' data depending on which style of fuse you intend to use.Also, both styles are available in different speeds of blowing (including the normal fast blow and slow blow).

author: jitter
author_rank: Frequent Contributor
timestamp: April 30, 2016, 10:44:11 am
content: Guide.
Understanding impedance - Page 1
Forum section: electronics / beginners

author: Richardcavell
author_rank: Contributor
timestamp: December 09, 2024, 11:37:15 am
content: I once saw a stereo that had attachments for the speakers. Each attachment was marked “Impedance 50 Ohms“.1. Does that “50 Ohms” refer to the vectoral sum of resistance and reactance?2. Is it referring to the impedance inside the stereo, or the impedance of the speaker that should be attached?3. May I assume that the reactance in question is inductive reactance?

author: Zero999
author_rank: Super Contributor
timestamp: December 09, 2024, 12:32:24 pm
content: 50 Ohms is a common impedance for RF. It probably refers to the aerial/antenna input, for the RF, normally FM/digtial audio broadcast, input. It has nothing to do with the speakers.

author: CaptDon
author_rank: Super Contributor
timestamp: December 09, 2024, 01:28:56 pm
content: Impedance is the algebraic sum of resistance and reactance. The reactance can have a leading or lagging phase angle. Impedance is sometimes calculated as the square root of (R squared) + (X squared). In a right triangle illustration resistance is the bottom, Xc or Xl is the adjacent side and of course the hypt or slanted part is 'impedance'. As for 50 ohms, that is oddball for speakers but does agree with some headphone specs. What do you mean by the words "attachments for speakers"? I would understand that with component stereo the speakers are the attachments?

author: Analog Kid
author_rank: Super Contributor
timestamp: December 09, 2024, 09:59:20 pm
content: Quote from: Richardcavell on December 09, 2024, 11:37:15 amThose two things (output impedance of the stereo amp and impedance of load--speaker) are supposed to match.Usually 8 ohms output, 8 ohms speaker impedance.50 ohms? Unless it's for headphones, never heard of that as an audio-amp impedance. More like for RF as someone else mentioned.

author: Zero999
author_rank: Super Contributor
timestamp: December 09, 2024, 10:56:29 pm
content: Quote from: Analog Kid on December 09, 2024, 09:59:20 pmIncorrect. The output impedance of a an audio amplifier should be as low as possible, in order to damp the resonance of the speaker. Look updamping factor.

author: Analog Kid
author_rank: Super Contributor
timestamp: December 09, 2024, 11:01:50 pm
content: True dat: the actual impedance of the amp might even be in sub-ohm territory.I guess I was referring to the nominal (rated) impedance.In any case, the impedance of theloadshould never be less than that of thesource.
Understanding induced voltage - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: January 27, 2019, 07:22:18 am
content: Case 1I have a 100 turn coil wound on a rod of soft iron and I feed it some current such that one pole of the rod is now N pole and the another S pole. And I have another coil wound on this rod to read the induced voltage.So firsta)I move a soft iron rod towards the N pole and away from it in a reciprocating manner.and then for another testb)I move a soft iron rod in a side by side like a clock arm moving past a hour digit. So the soft iron rod will first be aligned with the N pole then swing to the left then back to the center and then to the right and again back to the center.Should I expect AC to be generated in this way or only pulsed DC? In both the sub cases (a) and (b)Is there any way, any program that I can visualize this?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 27, 2019, 07:37:02 am
content: You are unlikely to see any voltage generated.For generating voltages you need a moving coil in relation to an externally generated magnetic field.Moving iron, (or any conductor) in a magnetic field will generate eddy currents, but that is another phenomena.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: January 27, 2019, 07:46:06 am
content: Quote from: Doctorandus_P on January 27, 2019, 07:37:02 amWouldn't the moving rod change the flux linkage in the fixed rod there by change the flux in the fixed rod as it is approaching it and so there is an induced voltage?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 27, 2019, 08:55:34 am
content: I'm having some doubts here, you might see some small effectHowever, when you're looking at electric motors they always have 2 separate coils, or a coil and a permanent magnet. (The cage in a squirrel cage induction motor is also a coil).10 minutes later...So I grabbed an electromagnet with a 120Ohm coil and a 120Ohm resistor and put 20V over it, which resulted in a current of 80mA. I used the resistor to have a higher (and symmetrical) impedance from the 20V power supply.Adjusted my rigol scope to AC coupling to see differences on the 10V over the coil and then I started pulling the iron core out of the magnet like a madman and pushing it back.Whenever I pulled the iron core out of the magnet the voltage over the electromagnet dipped by 4V, and when the iron core was pushed back into the magnet I had a positive spike on the elctromagnet of also around 4V. Such an even has a duration of around 30ms to 40ms.So you are right. A change in the amount of iron within the inductor changes the flux and with that you see voltage & current spikes.
Understanding Isolation transformers  - Page 1
Forum section: electronics / beginners

author: WiredSmart
author_rank: Contributor
timestamp: December 04, 2022, 07:20:09 pm
content: HelloI found a old looking power supply in a workshop.I would like to know if a variac can ever be a isolated transformer.Thanks for any help in advance

author: TimFox
author_rank: Super Contributor
timestamp: December 04, 2022, 07:34:03 pm
content: There are "isolated" variacs with two windings on the same toroidal core, but they are quite rare.From the photo, it looks like the box contains a variac driving a step-down transformer to get a lower output AC voltage range.

author: David Hess
author_rank: Super Contributor
timestamp: December 04, 2022, 07:55:53 pm
content: Quote from: TimFox on December 04, 2022, 07:34:03 pmThe old Lionel train transformers are an example of a variac with two windings. I have never seen a line voltage output variac with two windings, but I assume they exist.
Understanding laser diode power consumption.  - Page 1
Forum section: electronics / beginners

author: thescientist1064
author_rank: Newbie
timestamp: September 07, 2024, 10:39:37 pm
content: Hello everyone!I have a laser diode which generates a 65 W peak power when driven with a 20 A (at 9.5 V) pulse (100 ns, 0.01% duty cycle). Given that the diode is rated for a total electrical power of 20 A x 9.5 V = 190 W, will it draw this much power when operated at a pulsed mode?FYI, chat GPT claimed that the current needed would only be 2 mA (= 20 A x duty cycle). So I am confused.Any help would be greatly appreciated. Thank you!

author: Xena E
author_rank: Frequent Contributor
timestamp: September 08, 2024, 05:05:11 am
content: Quote from: thescientist1064 on September 07, 2024, 10:39:37 pm2mA is the average current over time. It's irrelevant to the operation of the laser.The total power consumed over time will be ≈ 19mW.The PS will still need to be able to pulse the laser at 20A/9.5V. For the rated maximum output power.Regards,X

author: thescientist1064
author_rank: Newbie
timestamp: September 09, 2024, 03:00:55 am
content: Thank you Xena, for the comment!Yes, my driver (LDP-V 240-100, manual attached here) is capable of providing up to 240 A pulses.Its super counterintuitive that the driver only takes in a low electrical power (19 mW) to deliver such a high pulse power (65 W). So if I am understanding it right, I still need apeak currentof 20 Amps (at 9.5 V) to drive my diode but theaverage currentdrawn from my power supply will be close to 2 mA (20 A x Duty cycle)?

author: Xena E
author_rank: Frequent Contributor
timestamp: September 09, 2024, 06:46:38 am
content: Yes! You got it.Making that large current available isn't a great problem of course.The secret sauce is in the delivery to the laser but that's a well trodden path, and all that's needed is to follow the prior art, and use the components as directed.A few years back, we used a similar component in a targeting system for a mining application. The emitter was driven almost fully, but the device was powered from a 1604/6F22 disposable battery.
Understanding LO signal in mixer circuit - Page 1
Forum section: electronics / beginners

author: anvoice
author_rank: Frequent Contributor
timestamp: May 13, 2020, 11:39:23 pm
content: Hello, I'm trying to build a quadrature amplitude modulator (also a demodulator which is roughly the same thing in reverse) using two mixer circuits and a combiner/splitter circuit. I am planning to use an AD9959 signal generator, and an AD831 mixer. The AD9959 can produce two In phase and two quadrature signals (also has a clock port), and from what I understand, for the transmitter, up-conversion is performed by connecting a single I and a single Q signal to the IF ports on each AD831, and a signal will be available on the RF port.What I have trouble understanding is the function of the LO (Local Oscillator) port. I understand it is typically driven by a sinusoid or a square wave, but where do I source this signal? Do I generate it via a microcontroller (in which case I'm guessing it has the form of a square wave) or does it come from elsewhere? Is it the same for both mixers in the modulator (and the two in the demodulator on the receiver side), or do I generate it individually for each?