hybridfree/eevblog-posts huggingface dataset:
Vintage Pulse Generator - Page 1
Forum section: electronics / beginners

author: Selectric540
author_rank: Newbie
timestamp: October 05, 2025, 06:31:30 pm
content: I have an old vintage Pulse Generator, bought at an antique store. I need to know its voltage and amp output, as well as its impedance, but it came with no manual or any kind of documentation. The only info printed on it says, "Digital Electronics, Westbury L.I., NY. Model 521, ID: 1140, WRAIR Property 6P. A barely visible Serial Number says either 4H4 , or 4A4, or 4B4 or 4R4. Other numbers on labels pasted on it are: 5820-607-0014, also 03 6p 5820k070018 and 076047. AI says,"The unit was likely manufactured for the Walter Reed Army Institute of Research (WRAIR) as military surplus equipment, likely during the 1960s or 1970s, and is not associated with the modern company Micro Center." AI says I could use an oscilloscope to find its output, but presently I do not have one, only a handheld multimeter. The range of the pulse generator is 5 cycles - 500 Kilocycles. That's what the dials say. It does have amplitude, spacing, width and polarity control dials. That's it. Any info on its voltage and amps output and impedance would be greatly appreciated. Thanks, from a newbie !

author: ftg
author_rank: Frequent Contributor
timestamp: October 06, 2025, 12:14:44 pm
content: The 5820k070018 is likely 58206070018. Both 58206070018 and 5820-607-0014 sound like NSN's.And them being NSN's would track with the thing having been owned by Walter Reed Army Institute of Research (WRAIR).5820 is apparently the NSN numbering for "Radio and Television Communication Equipment, Except Airborne"https://www.nsndepot.com/FSC/5820I thought that 076047 could be a CAGE code, but it does not seem to be.Googling also revealed this cross-post on the Electronics Stackexchange:https://electronics.stackexchange.com/questions/655412/i-need-information-about-a-vintage-pulse-generator-model-521Where folks had found and advertisement for the model 522 from the same company.Which has these features:"Output pulse width may be varied from less than 0.5 µsec to greater than 0.1 sec.Rise time is less than 50 nsec for a negative pulse and 200nsec for a positive pulse.Fall times are the same.Amplitude of the output may be varied from 0 to 15 V and is 5 V into a load of 100 ohms."Maybe someone knows how the NSN and CAGE codes have changed between 1963 and now.Seeing a photo of the markings could also potentially help.

author: Selectric540
author_rank: Newbie
timestamp: October 10, 2025, 06:09:15 pm
content: ftg: I have sent an email to WRAIR and am awaiting for their reply. Thanks for you helpful reply!
Virtual ground and reference ground - what is the differenace? - Page 1
Forum section: electronics / beginners

author: Evy
author_rank: Contributor
timestamp: August 13, 2017, 09:38:13 pm
content: Hello!I have stumbled upon different "grounds" in my book and wounder if anyone could explain the difference to me?I have done some exercises with "signal grounds" where I could reset DC sources in the circuitry. I could replace DC voltage source with a short circuit and a DC current source with an open circuit. Is that a virtual ground? Could virtual grounds be in other places than reference grounds?Sorry if you find this question to be to broad.

author: Zero999
author_rank: Super Contributor
timestamp: August 13, 2017, 10:20:08 pm
content: Yes this question is a little vague.Can you please post some schematics showing examples of what you're referring to?Generally it's better to have all the power supply rails connected together, with a large copper plane on the PCB and keep digital and analogue circuitry, away from one another.

author: Evy
author_rank: Contributor
timestamp: August 13, 2017, 10:30:42 pm
content: I am working with amplifiers wight now so its all analog. The circuitry can be seen inthisthreadI realized that that type of signal ground I was talking about is when you move from bias circuit to a signal graph.

author: Zero999
author_rank: Super Contributor
timestamp: August 13, 2017, 10:41:44 pm
content: So you mean a virtual ground, in the context of a differential amplifier?It actually isn't 0V, just very close to 0V because the amplifier has a very high gain and uses negative feedback, to keep its inputs at the same voltage, which is normally 0V. In reality, the voltage at the virtual ground is simply the amplifier's output voltage divided by the gain.

author: Evy
author_rank: Contributor
timestamp: August 13, 2017, 11:19:48 pm
content: Oh, so you can call the input terminal of a differential amplifier a virtual ground then since we assume that its 0V?

author: julian1
author_rank: Frequent Contributor
timestamp: August 14, 2017, 01:03:15 am
content: Dave explains the concept of a virtual ground in this video,https://youtu.be/7FYHt5XviKc?t=17m15sIn the example, the non-inverting input (ie. +) is tied to reference ground. Then, given a negative feedback configuration, a virtual ground is created on the inverting input (ie. -) because the op-amp changes it's output in order to try and make the two input voltages equal. Note, that the op-amp has both +ve and -ve supplies, so ground (reference and virtual) is the common supply voltage - eg. 0V.

author: bson
author_rank: Unknown
timestamp: August 14, 2017, 02:51:57 am
content: A virtual ground is a point in a circuit that wants to be at the same potential as ground, without an actual connection. You often find them in feedback circuits and they add a known quantity that can help with circuit analysis.
Virtual ground and shielding. - Page 1
Forum section: electronics / beginners

author: rgawron
author_rank: Contributor
timestamp: January 02, 2017, 06:55:28 pm
content: I want to build a device (amplifier for a photodiode), that would need dual power supply +3V (for itself) and -20V (bias voltage to polarise the diode), current drawn would be very small, couple of mA for the amplifier, almost nothing for a bias voltage. The output will be connected by the BNC to the scope. There will be a grounded metal enclosure to reduce noise electromagnetic interference.I don't want to use dual power supply (because I don't have one), batteries, or DC/DC converter (I want to avoid potential sources of a noise, because the signal is small).I came with this:- I will use power supply with not grounded output- set its output to roughly 25V, my device will have one socket to plug power in, inside the device, I will connect in parallel two linear stabilisers: one set to output voltage 20V, one set to output voltage 23V.- I will use 20V rail as a mass, 0V as a -20V rail, and 23V rail as a 3V rail. Then according to my new mass I will have +3V and -20V.- the metal enclosure will be connected to the new mass as well as the mass of the BNC for the osciloscope.- differences between input and output voltages will be small so the stabiliseers won't heat upI know that if some day I will use grounded output from a power supply and connect a scope, it will be bad, but as long as I won't, should it be ok? In addition, the voltages are too low to be dangerous (metal enclosure that can be touched) I think so with being careful I think it is acceptable.I think this should work, or did I make a mistake somewhere?I won't use neither LM317 nor 80xx family here.Thanks for sharing your advices.

author: bson
author_rank: Unknown
timestamp: January 03, 2017, 10:13:48 am
content: Quote from: rgawron on January 02, 2017, 06:55:28 pmThe 20V rail won't be able to sink current if it's the output of some kind of regulator, so won't function as a ground. Ground needs to be able to both source and sink. In fact, hooking it up to a rail 3V higher than its output is likely to destroy a linear regulator.I'd just start with a transformer with suitable outputs, say 18VAC and 4.5VAC and rectify and regulate those into whatever I need.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 03, 2017, 02:23:07 pm
content: Using a 20 and 23 V regulator will not work well, as the 20 V regulator may not be able to sink much current. One could add a minimum load though - still a strange way. Better, and only slightly different would be using on regulator for -3 V and one for -23 V, as there is very little current at the 20 V. The 20 V between the -23 and -3 would need extra filtering anyway.
virtual ground in multiple output amplifiers - Page 1
Forum section: electronics / beginners

author: kokodin
author_rank: Regular Contributor
timestamp: April 22, 2018, 10:59:27 pm
content: I post it here because i am just embaresed that i don't know itsituation is simple:single power suply, ground at 0V and positive rail at 24Vi was thinking about making 4 tda2030 output stages for 4 separate 4 ohm speakers ~10W eachi made first power stage, tested it, and i begin to think. can i use virtual ground from this power stage to bias other 3 or should i make 4 separate resistor deviders to bias each tda chip biasing as its own thing?not that i want save 2 resistors per chip, but i was wondering how much would single common virtual ground drift and distort inputs offor instance silent at this moment channel?

author: paulca
author_rank: Super Contributor
timestamp: April 23, 2018, 07:01:09 am
content: Take a look at my recent thread "Circuit Review?" before committing to the resistor only based v-ground. It has serious draw backs. There is a link to literally dozens of designs for v-ground circuits for when you need much more power handling.If the v-ground is stable enough I would expect it would be fine for biasing multiple outputs. Mine does two.A few tests I would do.... connect your circuit to a noisy power source, or power something noisy and digital from the same power source as your amp. I found I got a lot of noise pick up with the resistor only divider. The other is to measure the v-ground with a scope in AC coupling mode to see how much drift you have at high power outputs. Personally I take that as a percentage of the full output and subjectively determine if it's good enough. So 10mVpp drift outputing 10Vpp is 0.1% and fine. 100mV is 1% and probably too much.
Virtual ground mixer  - Page 1
Forum section: electronics / beginners

author: funk1980
author_rank: Contributor
timestamp: January 19, 2013, 11:10:58 pm
content: I'm working on a parallel effects loop for a bass pre-amp. I know about tubes, but opamps are a bit new territory. The schematic below works perfect for my intends, except that the mixing of the two signals is not linear. I tested this by providing the same signal (1khz sine wave) to both inputs of the mixer. The two pots are actually a single stereo pot and is used to blend the output signal from wet to dry.The differende in output signal when the pot is fully turned one way or at 50% is 2.15dB (3.6V vs 2.8V).I'd expect the output signal to be constant, which is desired because the difference is audible.Can anyone provide me with some tips or insight why this effect occurs and perhaps how to resolve it?Thanks

author: David_AVD
author_rank: Super Contributor
timestamp: January 20, 2013, 12:00:21 am
content: Is the "stereo pot" (dual gang) linear or log?

author: funk1980
author_rank: Contributor
timestamp: January 20, 2013, 12:24:08 am
content: Quote from: David_AVD on January 20, 2013, 12:00:21 amGood question, I totally forgot to mention that. It's linear. The lowest output is achieved at a 50% turn.I did some simulations. As it turns out, replacing the 10k pot with a 1k fixes the problem for the most part. My guess is when the pot is at lets say 50%, the parts of the pots is series with the 10k input resistor become part of the feedback equation, thus lowering the amplification of the whole.Is my assumption correct?? If so, that means the pot needs to be a fair amount smaller than the input resistors. In that case, would it be better to decrease the pots value, or increase the feedback/input resistors.

author: David_AVD
author_rank: Super Contributor
timestamp: January 20, 2013, 03:32:20 am
content: Sounds like the source impedance of the return input is higher than the buffer. Ideally you'd buffer the return so it's a known impedance (same as other buffer) for the mixer section.
Virtual grounds on overdrive pedal - Page 1
Forum section: electronics / beginners

author: Fantasma25
author_rank: Regular Contributor
timestamp: February 11, 2014, 04:20:05 am
content: Hello. I'm trying to make an overdrive effect for a guitar, but I want it to operate out of a single supply input. This is a basic clipper (overdrive) circuit with bipolar supply using a 741.In the simulation it works perfectly. Then, looking a little bit at an article by TI about single supply op amp design, I came up with this single supply circuit using a 318 op amp.The thing about this is that I'm placing a virtual ground (Vref) on the input and I'm not really sure that is a good idea. The simulation works perfectly as well with this one. Can anybody tell me if I'm doing the correct thing or something really wrong? Thanks!Here is the simulation. Theredone is the original and thegreenone is the modified one with the virtual ground. Thanks!!

author: Paul Price
author_rank: Super Contributor
timestamp: February 11, 2014, 07:42:18 am
content: The proof of the pedal is in the pushing.The circuit works fine but the 7805 chip has a quiescent current of 6-8mA. A simple voltage divider can create the virtual ground with just two resistors and a large valued bypass electrolytic cap, end up using much less battery current.
Vishay TSOP38238 application circuit question - Page 1
Forum section: electronics / beginners

author: GreyArea
author_rank: Contributor
timestamp: January 16, 2018, 09:28:35 pm
content: In the attachment I'm a little confused about the wording underneath the circuit.It says "components should be in the range of 33R < R1, <1K, C1 > 0.1uF"Should the value of R1 be 33R or 1K or simply not important and just somewhere between the two? I'd imagine the value is not all that important but I just find it written a bit strangly.The 33R < R1 is the most perplexing, how can it be less than R1 if R1 is not yet specified, or is it 1K-33R=967R.Stared at it for more than 5 minutes so thought I'd ask

author: floobydust
author_rank: Super Contributor
timestamp: January 16, 2018, 10:27:08 pm
content: It is an RC filter for power for the IR receiver module. It must be quiet power or the range goes down.Usually 100R-1K and 0.1uF or larger used, it depends on how noisy your 5V rail is to begin with.

author: drtaylor
author_rank: Regular Contributor
timestamp: January 16, 2018, 10:30:23 pm
content: It just means that R1 should fall between 33R and 1kohm. You wrote a comma where there was just a subscripted 1. So, reading it out would be 33 ohms is less than R1 and R1 is less than 1k. This is just components for filtering. I usually use about 100 ohms for this type of circuit with a 22nF cap. If you make the resistor too large, you limit speed...too small you reduce protection.
Visiting Shenzhen - Page 1
Forum section: electronics / beginners

author: taitny
author_rank: Newbie
timestamp: October 22, 2024, 02:07:52 pm
content: I'll have a few days the first week of January 2025 in Shenzhen to visit HQB, SEG market, visit Seeed Studios. What else might be of interest? Anyone want to meet in Shenzhen that week, let me know.

author: BILLPOD
author_rank: Frequent Contributor
timestamp: October 22, 2024, 06:31:25 pm
content: Good Morning Taitny, Shenzhen is sometimes called China's Silicon Valley, so there should be lots of high tech everywhere. Also, Shenzhen has a thriving art district and an impressive range of museums, galleries, theaters, and attractions have earned Shenzhen its place on UNESCO's Creative Cities list. They must have a visitors center, so check there.HAVE FUN, and take LOTS of pictures and when you get back, come back to the forum and tell us all about it. ENJOY, and stay aware of your surroundings.

author: taitny
author_rank: Newbie
timestamp: December 10, 2024, 09:05:16 pm
content: Thanks for the suggestion! I do plan to take lots of pictures. I was there 12 years ago, and took lots of pictures, I will have to get them posted. I will take lots of pictures this time as well. Last time we visited the Minsk, a Russian aircraft carrier that had been converted to a museum. It's gone now. Should be a fun trip. Thanks!

author: artag
author_rank: Super Contributor
timestamp: December 10, 2024, 09:39:36 pm
content: You might find this usefulhttps://www.crowdsupply.com/machinery-enchantress/the-new-essential-guide-to-electronics-in-shenzhen
Visual inspection of tantalum caps - Page 1
Forum section: electronics / beginners

author: rdagger
author_rank: Contributor
timestamp: September 16, 2018, 01:09:06 am
content: I’m troubleshooting and old PCB from the 70’s and I’ve narrowed the problem down to several tantalum caps. I noticed 1 has a small crater. All the tantalums on the board have small circular depressions, but this one is jagged (see pic). Is this indicative of a failure or is it just a blemish in the casing from the manufacturing process?

author: GadgetBoy
author_rank: Regular Contributor
timestamp: September 16, 2018, 01:17:07 am
content: I can see it's insides. That's probably not a good thing.

author: DC1MC
author_rank: Super Contributor
timestamp: September 16, 2018, 05:18:06 am
content: This one is already dead AND rotten, the problem is that it's like in the zombie movies, you look on it, looks normal, you turn your back on it, it's a zombie now and it bites you with flames.That another way to say that the tantalums at the end of their working life show no external signs of damage, like the elcos, they are just look and behave normal and then they go full short-circuitKick them all out,DC1MC

author: SeanB
author_rank: Super Contributor
timestamp: September 16, 2018, 06:42:41 pm
content: It's 40 years old, it is a moulded case tantalum, it is suspect anyway. Holes that are like that means it popped a bit from pressure to let out the magic smoke, and replacement is in order. Any more of ther same batch on the board replace as well, they will be close to going bang as well. Looks like it has a 10n ceramic chip capacitor in series, so easy to replace with a regular low ESR electrolytic, probably 22uF 25V, as that will be as low ESR as the 10/15 uF 16v capacitor, and will work well on that 5V rail.Don't know how many i changed as they were EOL, though i was getting them in in packs of 100, and going through packs a week doing so. Must have thrown around 10kg of tantalum slug away at the end of it.
visual studio  - Page 1
Forum section: electronics / beginners

author: abdullahseba
author_rank: Frequent Contributor
timestamp: May 29, 2014, 09:31:23 am
content: in visual studio express 2012 there is a graphical programing where you can drag & drop buttons menus etc.is there a way to make buttons respond to clicks with out code. I want the open dialog box to open when you click a button

author: 8086
author_rank: Super Contributor
timestamp: May 29, 2014, 09:40:25 am
content: It's very simple code, if you use Google you should have the code working within 10 minutes. Which language are you using?

author: abdullahseba
author_rank: Frequent Contributor
timestamp: May 29, 2014, 10:27:00 am
content: Quote from: 8086 on May 29, 2014, 09:40:25 amc#/vb

author: 8086
author_rank: Super Contributor
timestamp: May 29, 2014, 11:31:39 am
content: http://www.techrepublic.com/blog/software-engineer/let-users-pick-a-file-to-open-via-openfiledialog-in-vbnet/This seems like what you're after. 3rd result on google searching for "visual basic open file dialog" (no quotes).

author: abdullahseba
author_rank: Frequent Contributor
timestamp: May 29, 2014, 11:50:06 am
content: Quote from: 8086 on May 29, 2014, 11:31:39 amthank you
Visualising USB 2 High Speed (480 Mbit/s) on an oscilloscope - Page 1
Forum section: electronics / beginners

author: JustAnotherGuy999
author_rank: Contributor
timestamp: February 25, 2016, 07:24:21 pm
content: What is the minimum specification (bandwidth, sample rate,memory depth, number of channels) of an oscilloscope to visualize USB 2 High Speed Signals.PS: i posted this question on many forums, and i either got blocked, or i don't get an answer.Is there an international law preventing people from visualizing USB 2 High Speed signals on oscilloscopes?its just usb 2 not usb 3, come on

author: marshallh
author_rank: Unknown
timestamp: February 25, 2016, 07:46:57 pm
content: at least 2-4x the bandwidth (0.48gbps) so 1ghz to even see the signaling, 2ghz or 4ghz preferred.

author: JustAnotherGuy999
author_rank: Contributor
timestamp: February 25, 2016, 07:55:11 pm
content: thanks

author: free_electron
author_rank: Super Contributor
timestamp: February 25, 2016, 08:25:26 pm
content: 1ghz bandwidth, 4gs/s and deep memory. and get a machine that can run protocol analysis.

author: Jeroen3
author_rank: Super Contributor
timestamp: February 25, 2016, 08:33:27 pm
content: An "if you have to ask the price, you cannot afford it"-grade oscilloscope.USB is differential, so either a diff probe, or a math channel. A protocol analyzer that is able to trigger would be helpful too.Maybe try renting the equipment.

author: John_ITIC
author_rank: Frequent Contributor
timestamp: February 25, 2016, 10:47:17 pm
content: Do you want to visualize the analog voltage or the packets sent on the link? For the latter, you'll need a USB protocol analyzer. I design and sell such units.

author: JustAnotherGuy999
author_rank: Contributor
timestamp: February 25, 2016, 11:01:10 pm
content: Yeah i want to visualize the analog voltage

author: gslick
author_rank: Frequent Contributor
timestamp: February 27, 2016, 01:33:04 am
content: USB electrical specification compliance testing is probably much more involved than what you are looking to do, but you still might find some useful information in some of the compliance test procedure documents available here for some of the major test equipment manufacturers:http://www.usb.org/developers/compliance/electrical_tests/
Visualizing Pythagoras' Theorem, and an Electronics Related Fact - Page 1
Forum section: electronics / beginners

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 26, 2018, 07:31:02 am
content: Highly recommended channel for math facts, by the way.An interesting geometric fact appears at 15:58. Here it is:Did you catch the electronics related fact?If the lengths of A and B represent the resistances of two parallel branches in an electric circuit, then D is their parallel equivalent resistance!Tim

author: LeoTech
author_rank: Regular Contributor
timestamp: February 26, 2018, 07:50:56 am
content: Wow, I never thought of it that way, but it actually makes perfectly sense!Resistance is directly proportional to distance, hence the resistance of something could also be expressed in meters - or any other distance unit. (Think of series resistance)Wonder how many other electronic 'laws' can be expressed with other units/systems?LeoSent from my E5603 using Tapatalk
Vivado can't start - Page 1
Forum section: electronics / beginners

author: blueskull
author_rank: Unknown
timestamp: June 10, 2017, 08:57:36 am
content: A real beginner's question here...I just got my CMOD-A7 and I'm trying to get some weekend fun project done on it.The first problem I encountered is Vivado. After downloading 3GB of archives, Vivado finally installed, with an error reporting VC redist is not properly configured.I ignored it and finished the installation. On first launch of Vivado, it reported the same bug and can't be fixed.I then uninstalled all same version of vcredist (2015) from my computer and let Vivado install its own, which succeeded.Upon next startup of Vivado, it still requested me to repair vcredist, which I did a few times, and still, no help.After following a post on Xilinx website, I renamed vcredist installer and this time there's no nagging anymore, but Vivado still can't start.A few tens of seconds later, a window popped up with "Error when launching 'C:\Tools\Vivado\2017.1\bin\vivado,bat': Launcher tine out".Now, what to do? Any help is greatly appreciated.
Voice Dictation IC Suggestion - Page 1
Forum section: electronics / beginners

author: Cydget
author_rank: Newbie
timestamp: August 23, 2020, 07:19:44 pm
content: Hello, I just got my first oscilloscope yesterday, and want to try a slightly ambitious project to have real time voice dictation appear in a led matrix. I wasn't sure if there are any ICs that already handle dictation, or if its only possible to do it through a micro controller with some libraries. I'm mainly asking to see if such an IC exists and if so where to order them.Thanks for any help/links. I'm looking forward to starting this project.

author: MosherIV
author_rank: Super Contributor
timestamp: August 24, 2020, 07:40:32 am
content: HiWelcome to the forum.QuoteIf you mean convert speach to text, then no single ICs exist. This is generally done by software in general purpose microprocessors.No, those Amazon speaker (and other similar) devices not not decode your speach in the device, they digitize it and send the audio clip to a data centre where software will convert the sound to words, then more advanced software will work out what you said/asked and work out the response.There are a few open source projects for speach recognition.

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2020, 12:41:56 pm
content: If found the following two by Googling for speech to text IC. It's not an easy thing to do. I suspect a lot of beginners don't appreciate the complexity of such a project.http://en.waytronic1999.com/ProductsStd_195.htmlhttps://www.rhydolabz.com/components-other-ics-c-172_180/hm2007-speech-recognition-ic-plcc-p-1922.html

author: radiolistener
author_rank: Super Contributor
timestamp: August 24, 2020, 08:27:58 pm
content: Since voice recognition is a complicated problem, I would suggest to build "keyboard recognition" project as a beginner first project
Voice pitch changing while tunning an AM receiver - Page 1
Forum section: electronics / beginners

author: fubar.gr
author_rank: Unknown
timestamp: January 13, 2015, 01:31:45 pm
content: Why voice pitch is altered when the receiver is slightly off frequency relative to the transmitter? Does this happen in SSB too?

author: Howardlong
author_rank: Super Contributor
timestamp: January 13, 2015, 02:27:09 pm
content: SSB is just the baseband audio shifted in frequency and transmitted. So, if when you receive it, and it's shifted back to baseband and you're a bit off frequency, it sounds like the pitch is wrong.AM is different, because a standard AM demodulator simply looks at the power envelope around a given frequency. If you're a bit off frequency the filter skirts will make the tonal quality sound less than good, perhaps it will sound a bit more trebly for example, but it won't change in pitch.You can however demodulate AM perfectly well with an SSB receiver, in which case the first paragraph applies for reception.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 13, 2015, 05:52:49 pm
content: Surely the two modulations have been confused? SSB is not baseband shifted (as that would include negative frequencies -- only an analytical signal is non-hermitian, and that's not a real-valued signal), but it is also filtered to remove all extraneous information (namely, the sideband which used to be the negative frequencies -- or positive as the case may be). SSB detection requires a BFO to reconstruct the original waveform, which means the apparent pitch can be all out of whack.A tightly filtered AM signal can end up in similar circumstances, but is unlikely to appear pitch-shifted as long as the carrier is present. The signal may sound distorted or filtered if it is off-center, but shouldn't be pitch shifted.Tim
Voltage and current display - Page 1
Forum section: electronics / beginners

author: TorqueRanger
author_rank: Frequent Contributor
timestamp: October 26, 2014, 03:51:59 am
content: I am working on a constant current load like Dave's Jones design and I want to add a 2x16 led backlight display .. With the display I want to be able to see the voltage,current,and the total watts being generated .. I have never dealt with being a display like to known where to begin ?? I have read a couple display voltage and current articles but I need the watts too so I don't overload the heatsink with too much heat..

author: Richard Crowley
author_rank: Super Contributor
timestamp: October 26, 2014, 03:55:39 am
content: That would seem like an ideal case for a microcontroller solution (Arduino, et.al.). Pretty easy to measure voltage and current, calculate the power, temp, or whatever, and display the values on most any type of display(s).

author: sleemanj
author_rank: Super Contributor
timestamp: October 26, 2014, 04:30:08 am
content: Quote from: TorqueRanger on October 26, 2014, 03:51:59 amFirstly, Watts = volts multiplied by current, since you know both volts (use a voltage divider on the DUT input (DUT = Device Under Test) to get a value suitable to feed into an ADC on your micro controller) and current (measure the voltage at the top of your sense resistor(s) and calculate from it's known resistance using Ohm's law, then you can calculate the power the DUT is delivering.Secondly, it is probably better to stick a thermistor on your heatsink (or the tab of the FET) and use that to get an actual reading of the temperature, you can use this to turn on a fan when it's "warm" and shut it down when it's "too hot", the values of which are up to you, and your datasheets.If you have some spare op-amps (eg you've used a quad op-amp and have one or two left over) you can do the fan and shut down using those (use the op-amp as a comparator, with some hysteresis, calculate the voltages which will be present at your resistor+thermistor divider for the appropriate temperatures you want, and construct a set-point divider to match, feed the resistor+thermistor divider and the setpoint into your comparator, and use the output to drive a fet to turn on the fan, or to pull the gate of the main load FET low to shut it down).
Voltage and Frequency Transformer - Page 1
Forum section: electronics / beginners

author: NDV
author_rank: Newbie
timestamp: October 18, 2016, 03:51:34 am
content: I have kitchen equipment which runs on US standard of 110v at 60hz while I live in Australia with 220v at 50hz. Most of them have motors which are not rated for 50hz and some of the machines take around 700 watts.. I was wondering if there was any easy solution to convert both the Voltage and the Frequency preferably in a single and relativity cheap unit?Thanks

author: johansen
author_rank: Super Contributor
timestamp: October 18, 2016, 06:47:54 am
content: you can run regular 120v 60hz induction motors at 100 volts, 50hz.you can probably get away with running them at 120v, and certainly 110v is fine at, 50hz, but for largish motors 120v 50hz is too much.most refridgerators compressors should be fine at 120v 50hz because the motor is small enough, that the extra volts won't saturate it much more than it already is, but if the line amps are more than 20% more than they are in the states on 120v 60hz i would drop the voltage.that said most kitchen equipment is series wound motors.. mixers, electric knives, blenders. the frequency for these things does not matter. anything with a heating element won't matter either.
Voltage at the end of an open transmission line Stub - Page 1
Forum section: electronics / beginners

author: Gecko
author_rank: Contributor
timestamp: August 27, 2015, 10:57:51 pm
content: A question for the RF gurus:I have (simplified) a transmission line circuit like in the attached image: Just a simple open ended stub on a transmission line. In my case the stub has a length of lambda/8. But I think for my question that doesn't really matter.My question is: How can I calculate the Voltage (Amplitude) at the end of the stub (node B)?My first Idea was to use the equationV(z) = Vf* (1+r(z))where Vfis the amplitude of the forward Voltage wave, which should be 0.5 Volt in the example if I'm not totally wrong. V(z) is the voltage at a point "z" on the transmission line. r(z) the reflection coefficient at point "z".But with "z" chosen at the end of the stub, where r=1, I would get a result of 2V. But the simulation differs from that(0.633V with -26° phaseshift).What am I doing wrong. How can I get the correct result?

author: rfeecs
author_rank: Frequent Contributor
timestamp: August 28, 2015, 12:17:17 am
content: First, since you have an extra 50 ohm termination, you are driving the transmission line at point A with an equivalent 25 ohm source of amplitude .25V at point A. (This would be the voltage if you terminated point A with a 25 ohm load instead of the open circuit line.)So with the open circuit line, you are driving a 50 ohm line with a 25 ohm source, you have a standing wave along the line. The current at the open circuit end equals zero.There are lots of ways to solve it. You have boundry conditions defined on each end of the 50 ohm line. You can then take the transmission line equations and solve for the voltage and current at any point on the line.Another way is: calculate the impedance looking into the open stub: -jZ0cot(bl). With that you can calculate the voltage at point A. Now you have a 50 ohm open circuit line with a standing wave and you know the voltage at one end, so calculate the voltage at the open circuit end.Of course the easy way you did already. Throw it into a circuit simulator.

author: Gecko
author_rank: Contributor
timestamp: August 28, 2015, 08:38:54 pm
content: Thanks rfeecs, you pushed me into the right direction and I finally got it!One of my problems was, that I neglected the phase of the reflection coefficient in my calculations, but (1+|r|) not= (1+r).So with that error ruled out I could calculate the Voltage at Node A. Here is how I did it. Maybe it helps someone:VA=VP*(1+r) = 0.447 exp(-j26°) with VP=0.5V and r=0.447 exp(-j116°) (r can either be calculated from the values with ((R1||ZX) - Z0)/((R1||ZX) + Z0) with ZXbeing the impedance when looking into the stub. Or simply read it from a Smith Chart)With VAthe voltage at Node B can be calculated with VA=V_B * cosh(gamma *L) + IB*ZL*sinh(gamma L). Because the stub is open IB=0.So VB=VA/cosh(gamma L).With a lossless line gamma=j*beta and henceVB=VA/cos(beta L).In my example L=lambda/8 and so: VB=VA*sqrt(2) = 0.633V exp(-j26°)
voltage balancer - Page 1
Forum section: electronics / beginners

author: tubejim101
author_rank: Newbie
timestamp: October 08, 2011, 06:02:57 pm
content: I have two strings of LEDs in a single circuit.They have the same number of LEDs in each string.One string is brighter then the other.I am guessing this is do to the resistance being different for each string, making the voltages different.Is there a way I can automatically balance the voltages, so both strings of LEDs shine the same brightness?

author: armandas
author_rank: Frequent Contributor
timestamp: October 08, 2011, 06:52:49 pm
content: You should supply constant current to the LEDs. There are plenty solutions for that, just search for LED drivers.Check out thevideo here: exactly what you need.

author: IanB
author_rank: Super Contributor
timestamp: October 08, 2011, 09:03:40 pm
content: Quote from: tubejim101 on October 08, 2011, 06:02:57 pmDo you have any resistances in series with each string? Current regulation is always required for LEDs and resistors are the simplest way of providing a measure of current regulation. If you are connecting your LED strings directly to a regulated voltage supply you are doing it wrong. Remove some LEDs and put a suitable resistor in their place.If they are power LEDs running at 300 mA or more then resistors are not so suitable. In that case follow the advice of armandas and get a proper current regulated LED supply.

author: tubejim101
author_rank: Newbie
timestamp: October 08, 2011, 11:31:13 pm
content: Ok. Here is what I did.Put a POT in series with the bright string.Then dialed it till both LED strings looked the same.Then took an ohm reading on the POT. Came out to about 97 ohms.So I jumpered in a 100 ohm resistor. Looks great.Now I am just running the circuit with the resistor soldered in, to make sure it does not heat up.So far so good.Thanks for all the tips.

author: IanB
author_rank: Super Contributor
timestamp: October 08, 2011, 11:49:36 pm
content: You still may have a problem if you don't have a resistor on the other string. What is to stop them burning out with over current? (You have not said anything about your power supply, which could be one of the most important factors in the equation.)
Voltage between two points - Page 1
Forum section: electronics / beginners

author: sm263
author_rank: Newbie
timestamp: January 04, 2019, 01:10:46 am
content: Hello!I'm currently working through Practical Electronics for Inventors and still a bit stumped on an example I found early in the book. Attached is figure 2.13a from page 18 with the example stating:QuoteFor a. I understand that the voltage between points A and B would be 12v since it's across the voltage source. But for b, c, and d is where I get lost. What I don't understand is in figure b where Vac = 3V. I'm taking a wild guess here, the only thing I can think of is the 9v source is negative giving the 3v reading between the two points. Then for c., Vad = -21V. Looking at that I can only think of the -12V flowing to the -9V side and adding it together giving -21V. I feel like I'm approaching this incredibly wrong and missing something really important. How would you guys approach solving this?

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: January 04, 2019, 01:26:17 am
content: I guess this is trying to teach you that voltage is a potential difference. Without interconnection or a common reference like ground, there is no potential difference.Like if you have two batteries, and measure between the plus pole of one battery and the minus pole of another one (without any other connection): what would you expect?

author: ataradov
author_rank: Super Contributor
timestamp: January 04, 2019, 01:30:22 am
content: For (c), Vad = +21V, if you take (a) Vab = +12V. The situation is no different, just a higher voltage.Your thinking looks correct on all counts.

author: Brumby
author_rank: Unknown
timestamp: January 04, 2019, 04:56:43 am
content: Quote from: 0xdeadbeef on January 04, 2019, 01:26:17 amThat is part of it.I see this exercise as getting the student to understand the role of thereference pointand to understand therelativenature of voltages.Like this example I've cited before:The only difference is where you put the black lead of your meter.

author: Ian.M
author_rank: Super Contributor
timestamp: January 04, 2019, 06:59:45 am
content: In theory, using an infinite input impedance voltmeter, all the voltages asked for in O.P's Fig. 2.13a are undefined. Given a real near infinite impedance voltmeter, *if* the insulation of the batteries is good enough, you'll measure an essentially random voltage determined by the static charge on each battery (not charge *in* the battery), and the capacitance between them and to ground. However the capacitance is extremely small, and most real voltmeters have an input impedance of under 100 Meg, (typically 10 Meg for DMMs and 20K or 50K per V of FSD for analog ters) so any static charge present will equalise by flowing through the meter very quickly, resulting in a reading of zero volts from that time forward.As all the other figures include a ground connection to each battery, all voltages between any two points in each figure wont be influenced by static charge.
Voltage Booster - Page 1
Forum section: electronics / beginners

author: skillz21
author_rank: Frequent Contributor
timestamp: April 16, 2016, 11:22:02 am
content: i was trying to build a voltage booster. and i was failing pretty badly, check out my schematic for proof. the capacitor was 10uF and it is an arduino uno (Rev3). i am sorry for the messiness, i am in a bit of a hurry. so when i did this, the arduino started to make a weired noise. like it had a buzzer... What? it started making a faint beep. why is this happening? is this harming my arduino?And if you guys can, can you link in the simplest voltage booster you can find? i need to step around 4.7v to around 9v.Thanks

author: danadak
author_rank: Super Contributor
timestamp: April 16, 2016, 12:03:28 pm
content: Good chance you have fried the Arduino.Basics -https://www.maximintegrated.com/en/app-notes/index.mvp/id/2031Regards, Dana.

author: Seekonk
author_rank: Super Contributor
timestamp: April 16, 2016, 12:24:07 pm
content: I could tell you how to do it, but you are probably in too much of a rush to read it. You schematic likely has nothing to do with how you actually wired it.
Voltage booster: voltage drop and high pitch noise - Page 1
Forum section: electronics / beginners

author: Rapidice
author_rank: Newbie
timestamp: January 31, 2021, 12:45:40 pm
content: Hello,I am currently working on a project, which I want to power with 4 rechargeable AA batteries. The max voltage of those is about 1.4V. For my project, I would need a constant 6V at about 2A. On the internet, I have found a voltage booster circuit, for which I have ordered a PCB and all the parts.The first time I tried to use it, I have noticed a significant voltage drop and a quiet high-pitch noise. The circuit is based on an MT3608. I have read that it should be able to output 2 amps. Did I use the wrong inductor?https://lcsc.com/product-detail/Power-Inductors_SXN-Shun-Xiang-Nuo-Elec-SMMS0630-220M_C128694.html.Or is it something else?

author: iMo
author_rank: Super Contributor
timestamp: January 31, 2021, 01:04:13 pm
content: With larger output currents usually a smaller inductor value is used with switchers. There should be a formula in the datasheet for L vs. output current calculation. Also 4x 1.2V to 6V conversion may lead to a low efficiency (the voltage difference is rather small)..

author: iMo
author_rank: Super Contributor
timestamp: January 31, 2021, 01:26:45 pm
content: With 2A output you have to push a lot of current into the L. The frequency is fixed - 1.2MHz - and pwm is used with that chip. I would start with that 4.7uH they mention there, or even smaller (like 2.2uH, 3.3uH) with larger saturation current.

author: Zero999
author_rank: Super Contributor
timestamp: January 31, 2021, 01:36:47 pm
content: The inductor's current rating should be higher.Quote from: imo on January 31, 2021, 01:04:13 pmActually the reverse is true: the smaller the voltage difference between the input and output, the higher the efficiency. The problem with the small voltage difference could be instability, if the controller doesn't work well with low duty cycles, but I did skim the data sheet and no mention of that was made.https://www.olimex.com/Products/Breadboarding/BB-PWR-3608/resources/MT3608.pdf
Voltage clearance - Page 1
Forum section: electronics / beginners

author: FrankT
author_rank: Regular Contributor
timestamp: May 05, 2013, 10:46:59 pm
content: I muffed up the design of a board. At the last minute I changed the mcu from an atmega to an atxmega, and forgot the atxmega only works up to 3.6v while the board was designed to work at 5v.I'm replacing the rs485 driver (another forum question), and the other part in question is an MC33887 H-Bridge. It is powered from the motor power supply, but I'm concerned about the logic inputs. The datasheet says it has VIH=3.5v logic high threshold, and I'm driving it at 3.6v. Is this adequate clearance over the high threshold? One signal is the PWM running at 10kHz.I am running it at the moment, and it appears to work. What dangers am I facing running just 0.1v above logic high threshold?Thanks,Frank

author: zehome
author_rank: Newbie
timestamp: May 06, 2013, 11:04:42 am
content: Take a look at jeelabs blog:http://jeelabs.org/2013/04/24/what-if-i-mix-3-3v-and-5v/http://jeelabs.org/2013/04/25/what-if-mixing-3-35-0-part-2/http://jeelabs.org/2013/04/26/what-if-i-mix-3-35-0v-part-3/It's explained very well what you can expect in most cases.
Voltage comparator in the 555 timer, trying to understand the circuit - Page 1
Forum section: electronics / beginners

author: Teza
author_rank: Contributor
timestamp: March 13, 2017, 10:07:11 am
content: I am currently breaking up the 555 timer into the different components, and trying to learn how they in theory work and interact with each other.Project notes:http://www.terjeandersen.no/forum/evvblog/Project_AAB-555_timer_LED_dimmer.pdf(how can I hide the URL so only the "Project notes" shows as a link?)When it comes to the voltage comparator 1 (green) and 2 (blue), I have a problem indentifying what components in the schematic that belongs to what comparator.Don't understand why comparator 1 is connected to 1 transistor when the comparator 2 is connected to 4 transistors.

author: w2aew
author_rank: Super Contributor
timestamp: March 13, 2017, 03:47:20 pm
content: In short, output(s) of comp 2 (the threshold comparator) are connected to a pair of current mirrors. When the THRES voltage is above 2/3Vcc, the tail current of the circuit will be mirrored by the mirror on the upper left (2 PNPs on upper left) of the schematic. When the THRES is below 2/3Vcc, it provides the tail current for the TRIG comparator (comp 1). In other words, the output of the THRES comparator is a current that is fed to the subsequent circuits.

author: dnwheeler
author_rank: Regular Contributor
timestamp: March 13, 2017, 03:51:19 pm
content: If you want to see the signals inside the 555 timer, you should check out the discrete 555 kit from Evil Mad Scientist:http://www.evilmadscientist.com/2013/555-kit/

author: iainwhite
author_rank: Unknown
timestamp: March 13, 2017, 03:53:57 pm
content: Dave's video number 555 comes to mind:

author: Teza
author_rank: Contributor
timestamp: March 14, 2017, 09:55:41 am
content: Quote from: dnwheeler on March 13, 2017, 03:51:19 pmNice one!I have considered it, but now I think I will try the different parts on my breadboard.I guess I can use what ever type of PNP and NPN transistors for this, as long as the voltage specifications are higher than the voltage in the circut?

author: Teza
author_rank: Contributor
timestamp: March 14, 2017, 09:59:29 am
content: Quote from: iainwhite on March 13, 2017, 03:53:57 pmThis one is just perfect, could have saved me for hours on other sites.
Voltage control bidirectional constant current source - Page 1
Forum section: electronics / beginners

author: DanioIO
author_rank: Contributor
timestamp: July 03, 2014, 07:30:06 pm
content: Hi,I need to make a voltage control bidirectional constant current source for impedance measurement. Unidirectional circuit is simple one op-amp and a transistor, but i cannot pass direct current through my stuff (corrosion and other stuff) is there any circuit that can provide bidirectional current and still allow control using voltage?

author: Andy Watson
author_rank: Super Contributor
timestamp: July 03, 2014, 07:33:37 pm
content: Does your current source float with respect to the load? If so - place the load in the feed-back loop of the op-amp. If not, look-up Howland current source.

author: retrolefty
author_rank: Super Contributor
timestamp: July 03, 2014, 07:50:29 pm
content: So basically you are asking for an AC constant current source? Is their a specific frequency you require?

author: DanioIO
author_rank: Contributor
timestamp: July 03, 2014, 08:15:13 pm
content: I would need about 100kHz.

author: David Hess
author_rank: Super Contributor
timestamp: July 04, 2014, 05:28:33 am
content: You did not say what current levels are required.High frequency operation is going to make it more difficult to get high output impedance because of shunt capacitance.Basic ways include the Howland current pump which you can search for online however the Howland design requires matched resistors.Page 65 of Linear Technology application note 47 shows a simple and fast design. A slower but more precision version may be found in the LT1102 datasheet.For the highest performance, I would use a pair of singled ended current sources. If one fixed current source is placed in parallel with a variable current source of the opposite polarity that goes from zero to double the current, then the output will be a bidirectional current. Cascode output stages can then be used to raise the output impedance and operation into the 10s or 100s of MHz will be possible.