hybridfree/eevblog-posts huggingface dataset:
Voltage Controll - Page 1
Forum section: electronics / beginners

author: bopstar
author_rank: Regular Contributor
timestamp: April 02, 2018, 06:38:01 am
content: Hi im in the process of making a modular synthesizer and would like some links to websites that have schematics.What i cant get my head around is making a Control Voltage Keyboard to control CV inputs on modules.If someone can lead me the correct direction i would be great full.

author: BrianHG
author_rank: Super Contributor
timestamp: April 02, 2018, 09:54:19 am
content: Over the year, I've seen much on this forum about your topic. Search VCO of VFC and synthesizer and a ton of topics will come up with example schematic and links to hobbyist websites with more schematics.

author: bopstar
author_rank: Regular Contributor
timestamp: April 02, 2018, 01:11:37 pm
content: ok thanks

author: BrianHG
author_rank: Super Contributor
timestamp: April 02, 2018, 01:50:46 pm
content: OOOOOOPPPPPPSsssss, make that search 'VCO Audio'....

author: Lee Leduc
author_rank: Regular Contributor
timestamp: April 02, 2018, 05:17:32 pm
content: MFOS (Music from outer space) DIY sectionhttp://musicfromouterspace.com/index.php?MAINTAB=SYNTHDIY
Voltage Controlled Oscillator Circuit Explanation - Page 1
Forum section: electronics / beginners

author: johnnyblaz
author_rank: Newbie
timestamp: May 28, 2016, 09:08:14 am
content: Hi there!Attached below is a VCO circuit. Can anyone explain how the components work (especially the resistors and capacitors) and why the resistor and capacitor values are selected so.Any kind of help would be greatly appreciated. Thanks!!P.S (This is not a homework question)

author: Ammar
author_rank: Regular Contributor
timestamp: May 28, 2016, 09:34:12 am
content: Big question. You could start by looking at relaxation oscillators:https://en.wikipedia.org/wiki/Relaxation_oscillator

author: danadak
author_rank: Super Contributor
timestamp: May 28, 2016, 09:54:27 am
content: IC2, R5, R7 act as a comparator with hysteresis set by R5, R7.Tool to determine valueshttp://sim.okawa-denshi.jp/en/compkeisan.htmIC1, R1, C1 act as an integrator, to integrate the output of the comparator, tochange the comparator pulse into a ramp.Q1, R4 used to reset the integrator, ramp it down.Regards, Dana.

author: Marco
author_rank: Super Contributor
timestamp: May 28, 2016, 09:57:16 am
content: It doesn't really work like a relaxation oscillator.The left opamp is an integrator, the right opamp is a schmitt trigger. The integrator charges the capacitor with a current equal to (Vin-Vin/2)/R1 when the output is low and discharges the capacitor with a current equal to (Vin*R4/(R1+R4)-Vin/2)/(R1//R4) (// meaning the parallel resistance, also I'm ignoring Q1 resistance for a moment ... it's about 1 Ohm when on). The voltage at the output of the integrator is Vin/2+Vc1, as that voltage reaches either 1/3*Vref or 2/3*Vref (the Schmitt trigger threshold voltages) the output changes from 0 to 1 and vice versa.Why they say f is independent of R4 I don't know.

author: ebclr
author_rank: Super Contributor
timestamp: May 28, 2016, 10:55:54 am
content: The 1st stage is on integrator and the second stage is a comparator who triggers the FET for inverse integration.The key componets is C1 and R1 ( they define the frequency

author: JPortici
author_rank: Super Contributor
timestamp: May 28, 2016, 06:56:25 pm
content: but adjusting R1 means you have to adjust also R4 to mantain the simmetry of the waveform
Voltage Controlled Oscillators - Page 1
Forum section: electronics / beginners

author: Unknown
author_rank: Unknown
timestamp: March 29, 2016, 11:47:20 pm
content: Hello!I am endeavoring to build a synthesizer and related bits.I have done some looking into VCOs but I find many differing ways of doing it, so I am just gonna ask.What is your favorite or preferred way to constructing a VCO and why?What is the benefits and drawbacks of this design?

author: tautech
author_rank: Super Contributor
timestamp: March 30, 2016, 12:20:20 am
content: 4000 series IC like 4046, why? cause it's in the chip and easy to use

author: retrolefty
author_rank: Super Contributor
timestamp: March 30, 2016, 12:24:51 am
content: First one needs to know the frequency range you are interested in covering. Lots of different choices for audio ranges Vs RF LF/HF/VHL/UHF

author: dentaku
author_rank: Frequent Contributor
timestamp: March 30, 2016, 01:32:51 am
content: If your VCO is for musical purposes you probably want something that will stay in tune and work using the standard V/Octave that most analog synths use (if you care about that kind of thing).To start, it's always fun just messing with the 40106 Schmitt Trigger Inverter or even an LM13700 OTA. The complicated part of making basic noisemakers more musical comes later because it involves building exponential converters, which I haven't gotten into yet.These sites have lots of information about voltage controlled oscillators and filters and envelope generators.http://www.electro-music.comandhttps://muffwiggler.com/forum
Voltage converter for 110V USA model Metcal soldering iron - Page 1
Forum section: electronics / beginners

author: musicloverdenon
author_rank: Contributor
timestamp: August 02, 2014, 09:03:35 pm
content: Sorry if the topic is discussed before. I couldn't find through search.I am looking at buying a Metcal MX500 or PS2E model from eBay US which are rated at 100V-110V and 60Htz. My country supports 220-240V at 50Htz. I can buy a voltage converter. But it won't convert the 50Htz to 60Htz.Has anybody experienced any problem running USA model of these Metcals to European countries using a voltage converter due to difference in Htz frequency?

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: August 02, 2014, 10:03:04 pm
content: Frequency shouldn't be an issue.Some European Metcals have internal links to select voltage, but I've seen a US ne that doesn't

author: musicloverdenon
author_rank: Contributor
timestamp: August 03, 2014, 03:30:14 pm
content: Thanks for the quick revert Mike.I got worried as it seems some 110V musical instruments do have problem operating in a 50Htz frequency countries.Just out of curiosity, is it possible to convert a 110V/60Htz Metcal into 220V/50Htz machine by changing transformer or other parts. That would be a more sane solution than using voltage converter if achievable within reasonable budget.I am sure many others in 220V counties must have faced this issue as procuring a used Metcal is easier from USA than other countries.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: August 03, 2014, 03:56:29 pm
content: Quote from: musicloverdenon on August 03, 2014, 03:30:14 pmYes, if you can find an equivalent transformer the right shape/size
Voltage detector circuit - Page 1
Forum section: electronics / beginners

author: Mugiwara303
author_rank: Contributor
timestamp: June 19, 2021, 07:33:57 pm
content: I have a question about the role of the diode between the base and emitter this is a voltage detector circuit, schematic in attachement

author: Manul
author_rank: Super Contributor
timestamp: June 19, 2021, 09:40:18 pm
content: To protect the base-emitter (BE) junction from significant reverse bias. For NPN transistor, positive base potential relative to emitter cause the junction (which is like a diode) to be forward biased and start conducting. With forward bias, transistor works as a current controlled device. More current between BE means more current between CE. But BE junction may be damaged if reverse bias (base negative relative to emitter) is applied. BE junction of a typical small transistor starts to breakdown at around 5-7V of reverse bias voltage. This breakdown may damage the BE junction, reduce the hfe (basically gain) or cause total failure. Because in this circuit input signal from antenna is unpredictable, the diode is used to clamp any excess negative voltage up to safe limits, so base does not go too much negative in respect to emitter.

author: ledtester
author_rank: Super Contributor
timestamp: June 19, 2021, 10:46:23 pm
content: For fun I made this schematic...I used Multisim Online -- multisim.com. The "100A/A" is simply the beta of the transistors.

author: bob91343
author_rank: Super Contributor
timestamp: June 19, 2021, 11:10:17 pm
content: Not just for protection. Without the diode, the antenna will develop a charge which prevents the first junction from conducting.

author: Manul
author_rank: Super Contributor
timestamp: June 20, 2021, 12:33:07 am
content: Quote from: bob91343 on June 19, 2021, 11:10:17 pmFair point, however I don't think that it would cause a problem in this voltage detector circuit, because base has a tendency to float slightly positive. That is because of CB and BE leakage being present. Because the input capacitance is presumably very small I would not expect base DC point to float negative for extended time to disrupt the work of circuit.

author: Mugiwara303
author_rank: Contributor
timestamp: June 20, 2021, 10:05:13 am
content: Thank you, one more thing to add is when I put a shotky diode instead of the 4007 the circuit doesn't work at all and without any diode the range of detection decreases

author: Gyro
author_rank: Super Contributor
timestamp: June 20, 2021, 10:13:38 am
content: A Schottky diode has a higher reverse leakage current than a normal silicon diode, this will most likely be biasing the transistor hard off.You may want to try substituting a normal signal diode for the 4007, something like a 1N4148 which will have lower junction capacitance. The 4007(1kV) might have been chosen for very low leakage at very low voltage though.
Voltage divider biasing NPN transistor solution. - Page 1
Forum section: electronics / beginners

author: leftek
author_rank: Regular Contributor
timestamp: October 15, 2018, 03:41:46 pm
content: Hello,I would like to solve one exercise on the schematic in attached file.The request is what is the value of Vcollector if the resistor R2 is OPEN.I have make this in simulator and the values are as showing. But i can't find HOW i can calculate these if the R2 resistor in open.Can anybody help me?

author: Zero999
author_rank: Super Contributor
timestamp: October 15, 2018, 04:35:38 pm
content: The impedance seen at the base is a diode in series with R3*Hfe, which will be very high, compared to R1, so can be ignored.The transistor will just turn on as much as it can, giving a saturation voltage of a couple of hundred mV. The voltages across R3 and R4 can be calculated using Ohm's law.I = (V1-VSAT)/R3+R4)V = I*RWhere VSATis the saturation voltage of the transistor, which will be around 200mV. If you're unsure, presume it to be zero, calculate I, then look up the value on the data sheet and recalculate. This will give as gooder result as you're going to get, as the saturation voltage isn't 100% predictable and can vary from device to device.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: October 15, 2018, 06:16:29 pm
content: Quote from: Hero999 on October 15, 2018, 04:35:38 pmI think he wanted to take into account the extra base current . Thus when operating in the saturation region there is insufficient Ic to cutoff the base emitter diode then the b-e diode now just looks like a low impedance diode and more current will flow into the base until the b-e diode reaches cutoff (or the base drive can supply no more current).In saturation that circuit can be roughly approximated (ignoring Vce(sat) and re') by this, then just apply superposition or your favorite method to find the currents and voltages.

author: Zero999
author_rank: Super Contributor
timestamp: October 15, 2018, 07:28:26 pm
content: Quote from: Kevin.D on October 15, 2018, 06:16:29 pmYes you're right. I missed that. Not only do you have the potential divider consisting of R3 and R4, but there's also R3 in series with a diode, in parallel with R3.

author: leftek
author_rank: Regular Contributor
timestamp: October 16, 2018, 04:46:25 am
content: Thank you very much. You helped me to become understood.
Voltage Divider for RF Switch control - Page 1
Forum section: electronics / beginners

author: Nikos A.
author_rank: Regular Contributor
timestamp: May 14, 2019, 02:28:58 pm
content: Hi everyone,I am going to use a RF SP4T switch for a project and the control input voltage for the high state is 2.5 - 3.3V.I will send the signals through an Arduino which will be fed with 5V and thus the digital pin output will be 5V.Can I use a voltage divider to step down the signal to 3V for this purpose?Thanks in advance

author: pwlps
author_rank: Frequent Contributor
timestamp: May 15, 2019, 04:30:00 pm
content: Yes but you might need to take into account the current drawn by the switch to dimension your voltage divider. Check the high and low-state currents of the control inputs of your SP4T. If you can afford having a divider drawing much more current (say a factor of 5-10), then loading the divider by the control input will have negligible effect, otherwise recalculate the divider. For example if the current is 1mA at 3V it is equivalent to have a 3.3k resistance in parallel with the low-side resistance of the divider.Edit: you will find the formula for a loaded divider here:http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

author: Nikos A.
author_rank: Regular Contributor
timestamp: May 16, 2019, 08:07:30 am
content: The datasheet does not provide such an information for control current.. Maybe this is negligible?

author: pwlps
author_rank: Frequent Contributor
timestamp: May 16, 2019, 09:10:44 am
content: Quote from: Nikos A. on May 16, 2019, 08:07:30 amThey sayQuoteThis is of course misleading because the max input voltage is 3.3V which is LVTTL and not TTL, at any rate if these are CMOS inputs then the current is negligible indeed.NB. some useful refs. on CMOS/TTL specs:https://forum.allaboutcircuits.com/threads/cmos-and-lvttl-voltage-levels.62014/https://www.classe.cornell.edu/~ib38/teaching/p360/lectures/wk09/l26/EE2301Exp3F10.pdf

author: Nikos A.
author_rank: Regular Contributor
timestamp: May 17, 2019, 04:39:34 am
content: Thank you very much for your help!!
Voltage divider or level shifting? (Analog sensor signal optimization) - Page 1
Forum section: electronics / beginners

author: ghuebo
author_rank: Newbie
timestamp: March 12, 2021, 11:12:09 am
content: Hi there! totally newbie here with his first post.I'm currently working on the design of a portable device that acts as an interface for a precise distance measuring sensor. I've made some tests with ESP+ADS1115 ADC on a breadboard and got some good results but I'm having some issues handling the output voltage from the sensor (from 0 to 10V) so that the system working at 3.3V can read it accurately.As the datasheet states, the sensor outputs a signal when sensing mild steel in a 0 to 3mm range, going from 0 to 10v. I've managed the non-linearity of the output via software, but having trouble with what I call "over range". The sensor, also pointed in the datasheet, actually outputs more than 10v (something around 10.4) and thus, measuring more than 3mm. So my approach has been limiting the max output with a zener diode and dividing the resultant voltage via resistors as follows:But this solution is limiting the operational range of the sensor, as the zeners tolerance cuts the output voltage at around 9.7V, loosing about .5mm of the maximun working range , which being only 3mm, is quite a lot.Is there any way to get the full range voltage (even over 10v) shifted down to 3.3v and then just ignoring via software the values over 3mm? I know that I could just add a voltage divider from whatever the output voltage is and reduce it to 3v3 using just a resistor and a trimmer, but as I am working with several sensors and each outputs different Vmax (allways more than 10.3) , I was looking for a more plug and play solution that does not need adjusting the trimmer every time I connect other sensor.The sensor data, if it helps, can be found here:https://www.baumer.com/cn/en/product-overview/distance-measurement/inductive-distance-sensors/miniature/ir06-d03s-q22-ua1z-7scv/p/24527Thanks in advace!!

author: mvs
author_rank: Frequent Contributor
timestamp: March 12, 2021, 02:57:11 pm
content: ADS1115 has built in PGA and voltage reference, so full scale range is not 0-3.3V as you have assumed. Make a divider 10V->2V and just use 2.048 full scale range setting. If your sensor outputs more then 10V, you will just get ADC max value.

author: Terry Bites
author_rank: Super Contributor
timestamp: March 13, 2021, 06:06:15 pm
content: 62k+16k10V>>2.05VRload =78K
voltage divider problem - Page 1
Forum section: electronics / beginners

author: khatus
author_rank: Regular Contributor
timestamp: May 05, 2020, 06:52:36 pm
content: Hello guys I have some bulbs like this.Which takes about 3.8v,0.3A current. Now i wanted to drive 21 bulbs in parallel.But i have only 12v volatge source.So i decided to use a voltage divider . when i calculate the resistor value required for voltage divider i get the following result.Since each bulb takes 0.3A current and voltage across each bulb is 3.8V.But when i simulate it in proteus i have seen that increasing the load causes decreasing the output voltage across the resistor R2.i.e, the output voltage drops from 3.8V to 3.76V.But my requirement is 3.8V. My question is there anything wrong in my equation or solving method??

author: bob91343
author_rank: Super Contributor
timestamp: May 05, 2020, 07:08:37 pm
content: I will not critique your simulation. But your circuit is flawed in that R2 is superfluous. Probably the best way to drive so many lamps would be with individual dropping resistors. Or put the lamps in a few series strings.Should one lamp open, due to filament failure or poor connection, the voltage on all the other lamps will increase, shortening their lives. Soon there will be another failure, exacerbating the situation, until all have failed.Making each lamp independent of the others is a better solution, as mentioned in my first paragraph. Otherwise use a voltage regulator.

author: magic
author_rank: Super Contributor
timestamp: May 05, 2020, 07:11:47 pm
content: Why burn so much power? Have 7 strings of 3 bulbs and some small resistor in each string so that the bulbs see 3.8V rather than full 4V each. The resistance needed would be 0.6V/0.3A, whatever that is.

author: Manul
author_rank: Super Contributor
timestamp: May 05, 2020, 07:20:25 pm
content: Quote from: khatus on May 05, 2020, 06:52:36 pmProbably not, but the idea itself is very wrong. My suggestion will be 3 lamps in series with a series 2 Ohm resistor. And 7 of these strings in parallel. Actually you may use 3 ohms resistors to lower voltage a bit. Lamps will work a lot longer if you decrease voltage just a little.

author: james_s
author_rank: Super Contributor
timestamp: May 05, 2020, 09:31:57 pm
content: Voltage dividers are only useful for developing a potential, not for supplying power to a load.You could use a resistor but as others have said, any bulbs failing or loose connections will cause the voltage to rise. Why not use a voltage regulator?

author: StillTrying
author_rank: Super Contributor
timestamp: May 05, 2020, 10:30:42 pm
content: Did you not think it was becoming a bit impracticable when it draws over 400 Amps from the 12V battery, and needs 0.02 ohm and 0.009 ohm 5kWatt resistors to light a few 0.3A bulbs.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 06, 2020, 10:57:02 pm
content: Quote from: StillTrying on May 05, 2020, 10:30:42 pmThat is a usual newbie mistake - TS have calculated I and R, but forget to calculate P (dissipated power).
Voltage Divider vs. Op-Amp - Page 1
Forum section: electronics / beginners

author: luissizzle
author_rank: Newbie
timestamp: January 05, 2022, 07:23:01 pm
content: Hey all,I needed to read a 5V signal with a 3.3V MCU. I used an inverting op-amp with a VCC bias to map 0 to 5V to 3.3V to 0, respectively. The circuit works great (picture attached). Someone asked me why not just use a voltage divider, and I didn't have a good answer to that. A voltage divider with a zener voltage clamp would protect the input and accomplish the same thing. In this application, are there advantages to using the op-amp circuit that i'm not seeing?

author: Benta
author_rank: Super Contributor
timestamp: January 05, 2022, 08:23:10 pm
content: Well, you're inverting the input signal, so there must be a reason for that.

author: luissizzle
author_rank: Newbie
timestamp: January 05, 2022, 09:55:55 pm
content: Quote from: Benta on January 05, 2022, 08:23:10 pmthat's a consequence of the op-amp circuit. The inversion is handled with the MCU

author: GerryR
author_rank: Frequent Contributor
timestamp: January 05, 2022, 10:06:00 pm
content: A simple divider would have worked, and you wouldn't have had to handle the inversion in the processor, but your solution is fine, just more work.

author: Kim Christensen
author_rank: Super Contributor
timestamp: January 05, 2022, 10:28:48 pm
content: QuoteNone that I can see. A voltage divider would do just as well.The only advantage I can see is future flexibility. ie: In the future, if you wanted to change a 0 - 1.5V signal to 3.3V - 0V all you'd need do is swap out some resistors on your PCB and thus avoid a PCB redo/mod. (Or some other ratio where you wanted to increase the voltage swing)

author: rstofer
author_rank: Super Contributor
timestamp: January 06, 2022, 02:25:04 am
content: When you use an op amp, you are likely using high value divider resistors, Perhaps on the order of 100k or so. But the output impedance of the op amp will be quite low and this is sometimes required to drive the ADC.The alternatives: Low value resistors consuming excessive power sized to meet the ADC's input requirements or modest value resistors with a capacitor creating a low pass filter on the ADC input that limits high frequency response but provides a relatively low impedance to drive the ADC.It's a tradeoff. I would probably use the op amp if I had room and power was a consideration.

author: ledtester
author_rank: Super Contributor
timestamp: January 06, 2022, 02:37:28 am
content: Quote from: luissizzle on January 05, 2022, 07:23:01 pmYou haven't said what other rails are available or what guarantees there are about the input signal, but it seems that if you need to use a zener voltage clamp with the voltage divider you would also need some sort of input protection with the op-amp.
Voltage Divider - Page 1
Forum section: electronics / beginners

author: hotpointer
author_rank: Newbie
timestamp: February 11, 2015, 06:37:48 pm
content: Hi All.. and thanks in advanceI have written a programme for the Arduino thats uses a 10k resistor and a 10k ntc voltage divider.on the breadboard using the arduino, a lcd screen shows the temp . this works fineon a purpose made Pcb using a atemega 328 etc the temp on the lcdscreen fluctuates and is unusablei dont really know what i am doing but i have put a 10Uf capacitor across the 10k resistor and this stops the Flucation.my questions ouris this the correct thing to do ?is there a better way if not?why did it do this ?and why didn't it do it when using the arduino and breadboardthanks againrgds John

author: IanB
author_rank: Super Contributor
timestamp: February 11, 2015, 07:04:25 pm
content: I think I would have put the capacitor across the NTC, but no matter.I guess your fluctuation is from external noise that is getting into the MCU through the A/D converter. Adding the capacitor bypasses or filters the noise out of the way. Since temperature changes slowly a low pass filter is needed on the temperature sensor to block high frequency noise. The capacitor is doing this.There are two ways to reduce unwanted noise in a system. One is to stop the noise getting in to begin with. The second is to filter out the noise that makes it through.To stop the noise getting in you need to know and identify what the possible sources are and use good design to eliminate them.To filter the noise you can use various filter arrangements, but an RC filter is the simplest.

author: hotpointer
author_rank: Newbie
timestamp: February 11, 2015, 07:09:12 pm
content: thank you

author: Vgkid
author_rank: Super Contributor
timestamp: February 12, 2015, 12:59:04 am
content: Why an ntc used in a voltage divider?

author: Nerull
author_rank: Frequent Contributor
timestamp: February 12, 2015, 02:58:49 am
content: A voltage divider is the standard way to use a thermistor. It outputs a temperature correlated voltage which can easily be read with an ADC.

author: Vgkid
author_rank: Super Contributor
timestamp: February 12, 2015, 03:26:23 am
content: That is what i was thinking you were hoing to use it for. A rather innaccurate rtd, unless calibrated.

author: Nerull
author_rank: Frequent Contributor
timestamp: February 12, 2015, 03:39:22 am
content: Thermistors have a much smaller temperature range but can be more accurate than class B RTDs when used properly. Also a hell of a lot cheaper and simpler to use.
Voltage Doubler Issue - Page 1
Forum section: electronics / beginners

author: lesabb
author_rank: Contributor
timestamp: February 20, 2015, 01:46:16 am
content: I've drawn 2 schemo in LTSpice and performed simulations. Can anyone explain to me why one of them gives me the correct 34V and 68V but the other gives me the double voltage twice?ThanksL

author: Zero999
author_rank: Super Contributor
timestamp: February 20, 2015, 01:20:18 pm
content: That's what you should expect.Trace the circuit out, using conventional current flow charging the capacitors and you'll come to the same conclusion.Don't forget the 0V nodes are joined together.

author: Jolle
author_rank: Contributor
timestamp: February 20, 2015, 01:59:38 pm
content: In the second schematic you have connected the "negative" terminal of the voltage source to the node between C1 and C2 which will be at approx 34 V.When the voltage source goes positive the final voltage at the "positive" terminal of V1 will therefore be the 34V plus the actual voltage of the source - giving you the 68V you simulate in LTspice...Does that make sense?

author: lesabb
author_rank: Contributor
timestamp: February 21, 2015, 01:45:45 am
content: I understand the doubling part. What I don't get is why the bottom portion also gives 68v when it's not in a "doubler" configuration.

author: SeanB
author_rank: Super Contributor
timestamp: February 21, 2015, 08:14:10 am
content: It is being fed voltage on the common ground connection by the upper negative doubler, so floats down ( the bridge rectifier diodes are reverse biased fully there so do nothing to prevent this) to be the -34V from the doubler. As your measurement is referenced to this it looks like it has doubled to 68V.Basically the top section is a half wave rectifier with positive and negative outputs, and you are then trying to connect the negative rail to the bridge rectifier.

author: lesabb
author_rank: Contributor
timestamp: February 22, 2015, 12:43:20 am
content: I get it! Thanks!
Voltage drop - Page 1
Forum section: electronics / beginners

author: paulca
author_rank: Super Contributor
timestamp: May 18, 2020, 08:07:19 am
content: I intend to run 5 meters of 12V LED Strip (WS2811) totallying 150 LEDs.Theoretical max amps will be (150 * 3 * 20mA) / 3 (series leds) = 3 AmpsI'm going to guess that the 12V line could fall as low as 11Volts and as high as 14.40V. I'm going to also guess the LEDs will be fine down to 10 Volt though this might be extreme.From the power source to the LED start is going to be around 4 meters and the length of the LED strip is 5 meters (unless I trim it).So my finger in the air was I need cable running 10m with less than 1 Volt drop. A quick online calculator told me this would require 14AWG or 2mm2cable. This seems to be available in flat twin core, thin wall for automotive cable at reasonable price.Note due to the low current of the strip - and the unlikely hood of it actually pulling 3Amps - I am only powering it at one end. I include the length of the LED strip PCB in the total length to try and account for the drop along the PCB, although it will not be 14AWG obviously.I could power it at both ends, it would be about twice the price in cabling though and one more hold drilled in the garage wall.Do my finger in the air sums sound about right?

author: paulca
author_rank: Super Contributor
timestamp: May 18, 2020, 08:11:35 am
content: Of course I think 2.5mm twin and earth mains installation cable could turnout cheaper than possibly plated aluminium automotive cable.
Voltage drop across fuse - what to expect? - Page 1
Forum section: electronics / beginners

author: FlashEF
author_rank: Contributor
timestamp: August 03, 2012, 03:37:28 pm
content: I'd like to use a fuse in an circuit powering a mobile robot (MicroMouse) from a LiPol battery. The fuse is going to be a protection device in case of a hard short of power lines, in order not to blow up the battery (which has no protection what so ever - its an RC hobby battery).Fuse name:Littelfuse 0154005.DRDatasheet:http://www.littelfuse.com/data/en/Data_Sheets/Littelfuse_fuseblocks_154_155.pdfThere is nothing about resistance/voltage drop of this fuse. Is is supposed to be so insignificant? My battery nominally has 7,4V so say half a volt of drop on the fuse would be really much.

author: Short Circuit
author_rank: Frequent Contributor
timestamp: August 03, 2012, 03:48:56 pm
content: Look for the datasheet of the furnised fuse. That contains additional data, including "nominal resistance, cold ohms".

author: FlashEF
author_rank: Contributor
timestamp: August 03, 2012, 04:40:57 pm
content: Thaks - knowing what to search it was easy to find. It has cold ohms specified in the fuse family data:http://www.littelfuse.com/products/fuse-blocks-fuseholders-and-fuse-accessories/fuse-blocks/154/0154005_drt.aspxFor this particular one cold ohms are: 0.0136R. So the voltage drop should be really in the millivolt range.
Voltage drop across resistor - Page 1
Forum section: electronics / beginners

author: medical-nerd
author_rank: Regular Contributor
timestamp: June 21, 2018, 12:50:10 pm
content: HiyaAaaarghhh....I'm probably having a face palm moment but am having a blank at the moment.I know how to calculate basic voltage drops using ohm's law by calculating current flow first. But what do I do if current is a known fixed value?If I have a module that uses e.g 275mA at a supply of 5V, if there is a resistance of 10ohm after the voltage regulator, what is the voltage drop across the resistor?Cheers

author: capt bullshot
author_rank: Super Contributor
timestamp: June 21, 2018, 12:51:25 pm
content: U = I * Rmultiply the current by the resistors value

author: mikerj
author_rank: Super Contributor
timestamp: June 21, 2018, 12:59:43 pm
content: Quote from: medical-nerd on June 21, 2018, 12:50:10 pmAt 275mA a 10 ohm resistor will drop 0.275*10=2.75volts. The problem here is that with the resistor in circuit your load is now only seeing 5-2.75=2.25v, so is unlikely to be taking 275mA any more (unless it's a constant current sink). Current sensing resistors usually have low values to minimise voltage drop, which may then require amplification to get a useful signal.

author: Audioguru
author_rank: Super Contributor
timestamp: June 21, 2018, 01:17:07 pm
content: Quote from: medical-nerd on June 21, 2018, 12:50:10 pmIs the 10 ohms resistor in series with the 275mA module? Then the module is not getting 5V and probably is not drawing 275mA anymore.
Voltage Drop for an op-amp circuit - Page 1
Forum section: electronics / beginners

author: Dellyjoe
author_rank: Regular Contributor
timestamp: November 29, 2020, 04:49:16 pm
content: Hello everyone,I'm working on a home project during these COVID times.The project is a Distortion effects peddle for a bass guitar.Here is the output circuit, when I have an 8ohm speaker connected to it.Does anyone know why the voltage drops? Does it have to do with the output impedance drops to 8ohms when I connect the speaker?

author: Dellyjoe
author_rank: Regular Contributor
timestamp: November 29, 2020, 04:49:49 pm
content: This shows the circuit without an load on it

author: Dellyjoe
author_rank: Regular Contributor
timestamp: November 29, 2020, 04:51:43 pm
content: Here is the circuit in full.Sorry for all the posts wasn't sure how to attach more then one attachment at a time.Thank you for reading,Joe

author: schmitt trigger
author_rank: Super Contributor
timestamp: November 29, 2020, 05:00:15 pm
content: The amount of bias current in the output buffer is at least 4 orders of magnitude too small to drive an 8 ohm load.That, and the 2N3904 cannot handle the load even with proper bias.

author: Paul Rose
author_rank: Regular Contributor
timestamp: November 29, 2020, 05:04:34 pm
content: That final emitter follower isn't enough to drive an 8 ohm load accurately.(plus it will put DC across your speaker without either a blocking cap or transformer).You really want a full power amp stage if you want to drive a speaker.You could just use amplified computer speakers, or use something like an LM386 or TDA7267 for a low power small speaker amp.

author: Dellyjoe
author_rank: Regular Contributor
timestamp: November 29, 2020, 05:15:48 pm
content: Quote from: Paul Rose on November 29, 2020, 05:04:34 pmHmm ok that makes sense,Lucky I have a LM386 that I randomly bought two weeks ago.I will look into how to run this as a buffer and hope it will give me the power I need to drive the speaker.Now if I can do that I'm going to plug it into a head my dad has down stairs and see if that works b/c atm it doesn't and I think it has to do with the output buffer not being able to handle the load.Thanks for the help both of you.Joe
Voltage Drop on Battery Side during the servo operation - Page 1
Forum section: electronics / beginners

author: elefurtronik
author_rank: Regular Contributor
timestamp: October 08, 2023, 12:14:42 pm
content: Hello,I have two servos in my hand. One is brand1, and the other is brand2. When I apply the same signal to both of them, I see the following results on the power supply. My question is, what is the difference between these two servos, causing the brand1 servo not to achieve the smooth operation like the brand2 servo and causing a drop in the power supply as seen in the picture. I would be very happy if someone could answer this question.Summary:Everything is the same, except the servos are different.

author: BillyO
author_rank: Super Contributor
timestamp: October 08, 2023, 02:29:47 pm
content: My guess is they are just different servos. Once you get beyond interpreting the control signal (should be pretty universal) all bets are off on how the actual motion is controlled. Basically it sounds like Brad2 is doing that better than Brand1. Stick with Brand2.It might be fun to tear them and look at the circuitry both visually and with a scope.

author: RoGeorge
author_rank: Super Contributor
timestamp: October 08, 2023, 03:17:13 pm
content: Can be many causes, most probably one doesn't have enough filtering capacitors, such that it produces big current spikes, visible as voltage drops in the power supply.

author: elefurtronik
author_rank: Regular Contributor
timestamp: October 09, 2023, 06:36:27 am
content: I am just trying to understand the difference and how I can fix brand 1. Because it is way cost competitor than brand 2.Filtering cap is interesting point for me that I would like to know how they are used in servo circuits.
Voltage Drop on MOSFET of A Buck-Boost Converter - Page 1
Forum section: electronics / beginners

author: Frodo Muggins
author_rank: Newbie
timestamp: May 21, 2023, 02:24:11 pm
content: Hello,I am trying to design a buck-boost converter on simulation programs Proteus and LtSpice. In both of the programs, there is a voltage drop on Mosfet equals to input voltage and I see 0V on the output of converter. What should i do to fix this issue?Thank you in advance.

author: jwet
author_rank: Frequent Contributor
timestamp: May 22, 2023, 01:34:14 pm
content: Provide a schematic- buck boost isn't a universal description, there are many variants.A hint in the meantime, If the voltage across the FET is V+, its off. The voltage should very low when on- Iout times Ron and Ron is usually fractions of an ohm. You also have to wait a while in simulator to have a converter make an output- often setting initial conditions at vout can help speed simulations.
Voltage Drop Over Length of Wire - 24VDC/12VDC - Page 1
Forum section: electronics / beginners

author: Yamaha32088
author_rank: Newbie
timestamp: April 13, 2015, 06:10:36 pm
content: I am looking at cleaning up the CCTV system at work when we move to the new office because in the current state it is a complete rats nest. I found a CCTV Hub that allows you to power and extract video from CAT5 cables cleanly. Our current setup uses several Baluns on each end so this would help eliminate it on one end and keep it looking nice in a rack-mount format.My question is they are stating in the documentation that in Power-Thru Mode using 3 pairs of UTP at 24VAC, 5VA the power signal is able to be carried a maximum of 518 FT. Our cameras are going to be a little more than half of that distance away which would be fine and dandy except the cameras we are using have a 12VDC power requirement. So I am wondering if I will still get that kind of range with our current power supplies. The power supply we are using outputs 12VDC with a max of 1.1A per Channel.I have linked the spec sheet and the webpage of our current power supply and the CCTV Hub I am thinking about purchasing.https://www.mrsupply.com/winic-power-supply-12vdc-9p-5a.htmlhttps://www.elvessupply.com/assets/pdfs/VENC/61/Datasheet/VENC-500131.pdf

author: cosmicray
author_rank: Frequent Contributor
timestamp: April 13, 2015, 06:30:12 pm
content: What you are describing is a standard PoE injector. In fact, I use an 8 port PoE injector to power some remote UBNT radios (but only use 2 ports). The unit was originally sold for CCTV, but works well for my needs. The one I use has no internal power source, but lets you connect your own power.

author: Christopher
author_rank: Frequent Contributor
timestamp: April 13, 2015, 07:07:40 pm
content: Cat5e has a resistance of <0.188Ohms per meter.518 ft = 160m.Let's say you are going for 100m:total resistance per wire = 0.188Ohms * 100meters.You have 3 wires in parallel = (0.188Ohms * 100meters) / 3Multiply by two for the return: Total resistance = ((0.188Ohms * 100meters) / 3) * 2 = 12.53 OhmsVoltage Drop at 1.5A:1.5A * 12.53Ohms = 18.8V drop across the cables!Try a PoE boost at 48Vdc then regulate at the end of the cable to 12Vdc.
Voltage drop over USB Ground to Power Supply Ground (Green connector) - Page 1
Forum section: electronics / beginners

author: golivera
author_rank: Newbie
timestamp: March 10, 2016, 09:23:11 am
content: Helllo,I wanted to learn how some op amps works, and I found a small problem with my understanding of "ground" but must be pretty simple.Between my usb powered board earth ground and my power supply's earth connection there is a voltage drop, I measured 0.1mA flowing in between these two with my multimeter.So my op amp design uses my power supply to power the op amp and my input signal (DC square wave) comes from the USB powered board. So if I earth ground, my power supply (connect the green and black connector) had the "same" reference as the input source (from USB), but because of the voltage drop I measured something tells me this cant be right. Should I just connect the negative source of my power supply to the earth ground of the USB board and don't give much more thought about this voltage drop across different earth connections?Sorry in beforehand if the question isn't well described. Thanks!

author: ElektronikLabor
author_rank: Regular Contributor
timestamp: March 10, 2016, 09:33:05 am
content: Every time when current flows thru "ground" it causes a voltage drop.For analog stuff, especially when op amps are used for measurement, it is better to use a star grounding point as a reference.By connectung the PSU and USB ground you could cause an ground loop (in your case an earth loop) which could cause noise.

author: tszaboo
author_rank: Super Contributor
timestamp: March 10, 2016, 09:56:50 am
content: When you connected the two grounds together, you made a ground loop. And the USB cable has some resistance, so some of the current will flow other ways, so through your power supply-> mains cabling-> computer ground-> computer USB power supply. Say the resistance of the USB cable is 100 times less than the other way, but you will still have some current going that way.Disconnect power supply earth,!!!! After verifying that what you are about to do is safe!!!!And it should fix the issue.

author: Simon
author_rank: Unknown
timestamp: March 10, 2016, 10:15:20 am
content: I once made an amplifier when I was a kid and was very prowd of it an my nice earthing with a nnice thick wire from one point to another on my veroboard mess. Well it's turned out i didn't make an amplifier I made an oscilator. when i followed my dad's advice and did a star ground the amplifiers magically worked again.

author: golivera
author_rank: Newbie
timestamp: March 10, 2016, 10:22:23 am
content: Thanks for your answers, I hadn't heard of the term star ground, thanks for that!So if I understand correctly I shouldn't test my op-amp like this, but could a dirty trick be in this case to ground my power supply to the ground of the USB board directly? Any reasons not to do this? Sorry I cant test stuff until tonight!And yeah Simon, I was getting very weird behaviour when I got lost in ground
Voltage drop power supply LED circuit - Page 1
Forum section: electronics / beginners

author: Jr.Maxwell
author_rank: Contributor
timestamp: April 23, 2020, 06:54:35 pm
content: Hello,I'm setting up my first simple led circuit, yet something strage is happening and I just can't explain the reason. I came here in hope to find answersI've set up this simple circuit: See attachment1attachment1.PNGAs seen in the figure I've set the bench power supply to 3.3v/ 20mA. connected it to a Led spec'd at 2V/20mA with a simple current limmiting resistor of 47ohm.The moment I connect the power supply to the led and resistor, the power supply voltage drops to 2.84. I don't get it. why does this happen? Shouldn't the power supply stay steady at 3v?Edit: I think I found the reason: The power supply was set to constant current. that's why the power supply voltage drops.

author: ArthurDent
author_rank: Super Contributor
timestamp: April 23, 2020, 07:09:10 pm
content: The supply probably goes into current limiting. Having the supply limit set the same as the load current causes them to 'compete' for which has control.
Voltage Drop to a Readable signal  - Page 1
Forum section: electronics / beginners

author: Instro_Spark
author_rank: Contributor
timestamp: March 20, 2014, 09:01:21 am
content: Hey just wondering if someone would be able to help me measure voltdrop across a resister. At the moment im trying to measure the voltage drop with a voltage divider setup. Im actually using a trim pot so ive connected the two end pins across the resister and then measured the voltage of the wiper pin.What i cant understand is that when i measure with my multimeter across the resistor i get a range from 1-5v but when the trim pot is connected it gives me a reading from 23v to 19vSo how would i go with getting this high reading to a much smaller signalIve attached a schematic below (sense resister is R10)Circuit details+5v+24v0-10v input signal converts to a 4-20ma output which allows me inject a current into a loop powered instrumentation deviceHandful of opamps

author: KJDS
author_rank: Super Contributor
timestamp: March 20, 2014, 12:52:53 pm
content: The good news is that the circuit is doing what I would expect it to do.To get it to do what you want it to do, then connect one end of the 250R resistor to one op-amp input and the other end to the other input. Make these connections via a pair of 100k resistors to half the voltage, so that +24V becomes +12V at the op-amp, then set it up for a gain of x2. That way you avoid being outside the input common mode voltage range.
Voltage dropping for DC motors - Page 1
Forum section: electronics / beginners

author: bcross
author_rank: Newbie
timestamp: January 31, 2015, 07:23:58 pm
content: Hi guys,I am a complete newbie to electronics having brought an Arduino a few months ago and what can I say I'm hooked! Anyway forgive me for my low level of knowledge.I want to power two DC motors with a 5V source (powering a dual H-bridge IC controller connected to arduino) but according to the datasheet the DC motor is only rated 1.5-3.0V. The H-bridge must be supplied 4.5-36V. I was thinking I could drop the voltage across each motor by placing resistors in series. However (I can't get a decent resistance reading and) I am concerned that the resistors will overheat (burn out). Is it really a bad idea to supply 5v to a 3.0V max rated motor? If so what is the best (also cheap!) way of dropping the output voltage from the H-bridge?BenMotor Datasheet:https://www.pololu.com/file/download/fa_130ra.pdf?file_id=0J11H-bridge Datasheet:http://www.ti.com/lit/ds/symlink/l293d.pdf

author: Zero999
author_rank: Super Contributor
timestamp: January 31, 2015, 07:36:01 pm
content: Don't worry about it. The datasheet says the h-bridge drops a couple of volts so with 5V in, the voltage across the motor will be not much more than 3V.

author: langwadt
author_rank: Super Contributor
timestamp: January 31, 2015, 07:36:27 pm
content: Quote from: bcross on January 31, 2015, 07:23:58 pmAn L293 powered by 5V will give you less than 3V on the motorlook at Voh and Vol on page 5 of data sheet

author: bcross
author_rank: Newbie
timestamp: January 31, 2015, 07:39:07 pm
content: Brilliant! Thanks guys.
Voltage fault detection - Page 1
Forum section: electronics / beginners

author: Manx
author_rank: Contributor
timestamp: December 07, 2019, 12:36:24 am
content: 5V tolerant inputs of STM32 have an additional requirement thatVin < (VCC + 3.6 V), so I have to protect the MCU against a situation when 3V3 is lost when 5V is still supplied. I want to make a circuit that would assert INHibit pin of the 5V SMPS (L5973D) when 3V3 fault occurs. My idea is the following:R13 is for the situation when 3V3 fails as open circuit, R11 is for the situation when 3V3 gets shorted to ground. In any case, there will be more than 3.3V across the Zener diode, so the transistor will start conducting, thus making INH (volProbe1) HIGH.Does this make sense? What I don't like (based on EasyEda simulations) is that it can be finicky concerning component values. R14 is what I guess the internal INH pullup could be. R12 has to be low enough that it overpowers the stupid internal pullup, and at the same time high enough that it can be overpowered by the transistor. I guess a Darlington would be better here. But anyways, does my idea make sense? Or am I trying to reinvent the wheel the hard way?