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

author: rodrigopires
author_rank: Regular Contributor
timestamp: June 04, 2015, 07:31:37 pm
content: Hi, i have two cores that i took out of some motors and i'm wondering if i can use them to make some transformers, or other things. What can i do with them? Maybe cutting the edges in the center just to have a regular surface? Any ideas?Thanks!

author: Cliff Matthews
author_rank: Unknown
timestamp: June 04, 2015, 10:02:00 pm
content: I think these formats will end with meager results and waist some good magnet wire.http://en.wikipedia.org/wiki/Magnetic_coreHack apart an old microwave oven and borrow a die grinder to break the welds - now that's a core.
Transformers in LTSpice: coupling coefficient - Page 1
Forum section: electronics / beginners

author: fmzambon
author_rank: Contributor
timestamp: May 31, 2024, 06:18:45 pm
content: Hello,as the title says I'd like to simulate a power transformer in LTSpice. I know I can do that by using two (or more) inductors and then by linking them with a K statement. However I'm a bit hesitant about the coupling coefficient value that I have to specify in the K statement.I know that many people suggest using 1, indicating perfect coupling, but this seems lazy to me. Unless someone can point me to a source where I can buy ideal transformers, that isIf I had a transformer in my hands (and an LCR meter) I could measure it and come up with a value, but I'm just looking for ballpark figures here.However my attempts at coming up with such typical values for the couplig coefficient for real transformers have yielded very little results. Inthis articlethe author lists some measured leakage inductance values for a few transformers. Higher up in the page are a few measured inductance values (though it appears that the measurements are not from the same transformers). For example a 300VA toroidal is listed as having 63H primary inductance and 1.63mH primary leakage inductance.The formula to calculate the coupling coefficient k should be:k = sqrt((Lpri - Lleak)/Lpri)where Lpri is the primary inductance and Lleak is the primary leakage inductance. With the above values this gives a value of k = 0.999987Is this value reasonable for a 300VA toroidal transformer? Does anyone know what typical values for various types and sizes of transformers are?I know the coupling coefficient alone cannot accurately simulate a transformer in LTSpice, far from it. But if inputting 0.999987 (or some other value) instead of 1 increases the accuracy of the simulation, why not?Best regards,Andrea

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: June 01, 2024, 01:03:45 pm
content: ltspice can give you convergence issues when using k<1. i always use k=1 and add discrete elements as desired. makes life much easier when taking measurements from a real specimen into a simulation.on a tangent subject of your question re. a 300 VA transformer: for a power transformer, you will often find the "short-circuit voltage" in the specification, which is the required primary voltage to achieve the secondary nominal current when shorting that secondary winding. that voltage should be in the 1% ballpark of the nominal Vprim for a 300 VA transformer.

author: aliarifat794
author_rank: Regular Contributor
timestamp: June 02, 2024, 12:33:36 pm
content: You may find this article helpful:https://www.analog.com/en/resources/technical-articles/ltspice-basic-steps-for-simulating-transformers.html
Transformers - specs - Page 1
Forum section: electronics / beginners

author: alin_im
author_rank: Regular Contributor
timestamp: January 08, 2015, 06:53:28 pm
content: Hi guys,I am new here and also a newbie to electronics. How can i determine the characteristics of a transformers if i can not find any data sheet or circuit diagram of the old pc power supply which had the transformers?? The big one has WIN-39-600W YC and SES-TR written on it and the other WIN-19-01 YC

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 08, 2015, 11:37:29 pm
content: You can estimate based on wire size, core size, operating frequency, and you'll determine more from the surrounding circuit, since you still have it.Tim

author: Evil Lurker
author_rank: Regular Contributor
timestamp: January 09, 2015, 04:50:21 am
content: Those are ferrite switch mode transformers the electromagnetic characteristics of which are totally different in operation than the typical clunky iron plate linear transformers. The big one is the primary transformer and the smaller one is for the +5V standby circuit. Also you will not find any datasheets on completed switch mode transformers as they are custom wound by/for manufacturers and will more or less only work in the PSU that they were designed for. Reason being the ferrite material from which the core is made out of has been specially selected for things such as switching frequency and saturation along with the pinouts for the windings have to fit the traces on the PSU PCB.That is not to say you can't yank them off the board and disassemble them to see how they are made and or tinker with them. The trick is the cores come in two halves, and they are epoxied together. Only practical way to get them apart intact is to put them in the oven at around 400-500 degrees and then using some heavy rags (socks work good too) to gently pry the cores apart as quickly as you can while scorching hot. And then you have to be VERY careful not to break the plastic bobbin, not just during the core removal but while unwinding them as well. Oh and the tape is some nasty sticky stuff kapton or some weird high temperature darn near flameproof polymer. You just about have to unwind a little bit, stick it back in the oven to soften the tape up, unwind some more and rinse and repeat until you get it all off.
Transient time of opamp integrator to reach stability - Page 1
Forum section: electronics / beginners

author: zzlong94
author_rank: Contributor
timestamp: November 23, 2017, 07:32:31 pm
content: The higher the value of the Rf (feedback resistor), the longer the time it take to reach its stability for it output voltage.whats the concept of it?it is because the larger the resistance, it need much time n energy to charge up?

author: danadak
author_rank: Super Contributor
timestamp: November 24, 2017, 01:22:43 am
content: Schematic ? Including driving source....Regards, Dana.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: November 24, 2017, 02:16:43 am
content: Hi,I true integrator will never settle.If you are talking about this circuit:(From your other thread)The answer output is an exponential describedVout = -vin x R2/R1 (1 - e-t /(R2 x C1))It is normally accepted that this type of circuit settles in 5 time constants or 5 x R2 x C.But, it will continue to settle for ever.5 time constants gets you (1- e-5) equals 0.993 x final valueRegards,Jay_Diddy_B

author: rstofer
author_rank: Super Contributor
timestamp: November 24, 2017, 04:54:39 am
content: Realistically, you don't have much of an integrator left once you add Rf. That is, the result of operating on the signal for some period of time is not the integral of the signal over time because the capacitor has been constantly discharged by Rf. There are simply too many terms in the transfer function.At best, the circuit is a low pass filter.http://www.analog.com/en/analog-dialogue/articles/phase-response-in-active-filters-2.html
Transient Voltage & Current Suppression Options - Page 1
Forum section: electronics / beginners

author: Stupid Bear
author_rank: Contributor
timestamp: January 10, 2023, 05:58:32 pm
content: Recently I've been doing a lot of learning into transient voltage/current suppression and there are many options. MOV TVS polyswitch zener etc. While I understand the principles of it I'm having difficulty in figuring out when to use which (pros, cons, applications, etc). Can anyone recommend a good resource that gives a thorough rundown of them in relation to one another. Perhaps even real world example.Thank you.

author: jonpaul
author_rank: Super Contributor
timestamp: January 11, 2023, 04:27:01 am
content: see recent thread on same topic, re protection of 12V bus in vehicles.Jon

author: Stupid Bear
author_rank: Contributor
timestamp: January 12, 2023, 02:48:26 pm
content: Ive come across a great many articles. Do you have a link?

author: jonpaul
author_rank: Super Contributor
timestamp: January 13, 2023, 11:21:24 am
content: https://www.eevblog.com/forum/projects/surge-protection-design-and-calculations/MOVs are not used in our work, very high C and fire hazard. Limited surge life.Polyswithc is a fuse sub not really for transients.Zeners are not rated for the surge energy, TAZ= Transient Absorbing Zeners are.Check the IEEE papers and conf, and old General Semiconductor, Claire and other TAZ mfg app notes.Kind Regards,Jon

author: Terry Bites
author_rank: Super Contributor
timestamp: January 19, 2023, 04:38:32 pm
content: I'd start with app notes and guides from the major manfacturers. eg /www.littelfuse.com/products.aspxA lot depends on the market your end product is aimed at. The level of protection in a who cares LED panel will be a lot less demanding than a sued into infinity patient monitor.

author: Stupid Bear
author_rank: Contributor
timestamp: January 23, 2023, 03:04:48 am
content: I've actually been reading through the littlefuse literature. I was hoping to understand the thought behind the design but more figuratively like a flow chart of how one goes about deciding what kind of protection it needs at different stages and what design solutions work best and for what reasons. The actual calculation and component selection I can follow along. I'm just not sure what design topology works best for what.

author: David Hess
author_rank: Super Contributor
timestamp: January 23, 2023, 02:50:34 pm
content: Motorola application note AN-0843 - A REVIEW OF TRANSIENTS AND THEIRMEANS OF SUPPRESSION
Transimpedance amplifier: Simulating an unbiased photodiode - Page 1
Forum section: electronics / beginners

author: Bigman
author_rank: Regular Contributor
timestamp: October 18, 2015, 04:50:30 pm
content: Hello,The picture "normal_operation.png" shows the schematic of my transimpedance amplifier. At the moment, I cannot modulate the laser-light, but would like to test the circuit's amplitude response for different frequiencies. So, instead of a "light-modulation", I would like to use my waveform generator.In my first try, I connected the waveform generator directly with the photodiode cathode/inverting input. But then, I was advised to but a resistor Rsigin between, which as the same value as Rf(see picture testing.png). Reasoning: because the photodiode would be unbiased.Unfortunately, I don't understand, why Rsigshould help to "simulate" the unbiased condition of the photodiode. I would appreciate if somebody could help me to understand this.

author: awallin
author_rank: Frequent Contributor
timestamp: October 18, 2015, 05:58:01 pm
content: The transimpedance amplifier (TIA) is supposed to turn an input current (from the photodiode) into an output voltage.R_sig is there to convert the voltage output from your signal-generator into a sinusoidal input current for your TIA.R_sig doesn't have to be the same size as R_f, just choose something that gives the photocurrent you are expecting in your application.I would disconnect the photodiode while doing testing with R_sig and the signal generator.If your op-amp is single-supply you might want to add an offset to your sine-wave test signal, to have the output at some DC-level and not hit either rail.HTH

author: Bigman
author_rank: Regular Contributor
timestamp: October 18, 2015, 06:23:45 pm
content: Thanks for your kind reply. There is now one more question, arising from your answer. You wrote:Quote from: awallin on October 18, 2015, 05:58:01 pmindeed, I would like to see how the output-amplitude changes with different frequencies from the waveform generator. I assume, that the photodiode's capacity is influceing this amplitude response ... so I would like to keep the photodiode in the circuit, but I am also wondering, why did you advise to take the photodiode out.Thank again!

author: TimFox
author_rank: Super Contributor
timestamp: October 18, 2015, 09:13:09 pm
content: At zero bias voltage and no illumination (zero photocurrent), the first-order (linearized) impedance of the photodiode is a resistor in parallel with a capacitor.The resistor represents the slope of the current vs. voltage curve at the origin (0 V, 0 A), and the capacitance at zero bias can be relatively large: it falls off with applied reverse bias.Applying the external voltage through the resistor as in your diagram is equivalent to applying a current in parallel with the resistor (and therefore the capacitor) which is correct. The capacitance will also effect the noise of the transimpedance amplifier circuit.
Transimpedence amplifier photodiode - Page 1
Forum section: electronics / beginners

author: Vindhyachal.takniki
author_rank: Frequent Contributor
timestamp: April 23, 2015, 08:57:46 am
content: 1. I have to design transimpedence amplifier(TIA) for monitoring photodiode in Fiber Coupled Diode Lasers.2. I have checked for internet on how to design a TIA.3. I have attached a circuit diagram with formula mentioned.4. I have two different design approaches for selcting feedback capacitor(as in circuit attached) , one at 65 & other at 45 phase margin. Which one is most suited?5. Then I fed the signal to 24 bit differential adc.6. Any other comment or suggestions on it

author: TimFox
author_rank: Super Contributor
timestamp: April 23, 2015, 04:51:05 pm
content: 65 degrees is the conservative choice, but 45 degrees is probably adequate. If the waveforms look bad after building with 45 degrees, you can always change the capacitor for 65 degrees.
transistor abuse (?) in audio VCO based on w2aew circuit - Page 1
Forum section: electronics / beginners

author: Sasja
author_rank: Contributor
timestamp: December 23, 2013, 11:59:04 am
content: Hi everyone,I'm designing an analog audio synth, mainly for learning purposes.So inspired by w2aew's awesome video ()i designed, built and tested a voltage controlled sawtooth generator with a exponential voltage/frequency dependency.And it works pretty well over the necessary frequency range, only i did something dodgy(?) to make it work.Sorry for the messy schematics. The first attachment shows the relaxation oscillator (not showing the current source in top left corner). The second attachment is the whole schematic. Now it's the bottom transistor that is the problem, I realised i needed a low beta in order to make it work well but having no such transistors i found when using a regular BC547B backwards (swapping emitter/collector) i would get a fairly low beta = +-10. And it works great now, but will it still work great later or will it slowly degenerate somehow? I'm pretty much dumping 200nC from base to collector at a 1000Hz-ish rate. I could find some better components for it and change the circuit somehow, but you know...Please share your thoughts!

author: SeanB
author_rank: Super Contributor
timestamp: December 23, 2013, 06:51:29 pm
content: I would assume you do not zener the reverse biased junction, so it probably will have a pretty good life, though at high temps it will probably leak a lot.

author: c4757p
author_rank: Super Contributor
timestamp: December 23, 2013, 07:00:07 pm
content: What about adding a resistor base-emitter to reduce the effective beta?

author: mrkev
author_rank: Regular Contributor
timestamp: December 23, 2013, 09:25:11 pm
content: Quote from: c4757p on December 23, 2013, 07:00:07 pmYup I would use that one...Quote from: Sasja on December 23, 2013, 11:59:04 amThis was actually quite common, long time ago, as the number of transistors you could buy was much smaller... I don't think it will hurt that transistor, but I've never used it this way myself...

author: w2aew
author_rank: Super Contributor
timestamp: December 23, 2013, 09:59:42 pm
content: Check out Alan Yates' treatment on this design, and how to make it less abusive on the transistors (and to improve the performance in general):http://www.vk2zay.net/article/196

author: Sasja
author_rank: Contributor
timestamp: December 23, 2013, 10:24:37 pm
content: thanks for the suggestions everyone, that link looks promising w2aew!i'll study it tomorrow, update the circuit, and try to be less abusive on the transistors in the future
Transistor activation / deactivation voltage in - Page 1
Forum section: electronics / beginners

author: renzoms
author_rank: Regular Contributor
timestamp: October 16, 2019, 11:23:19 pm
content: The capacitor was allowed to discharge completely when the transistor was activated with the 3.4 V.The capacitor is allowed to charge again when the conduction from the 5 V line through the resistor to the ground of the IC is open circuited because the transistor deactivated.The transistor must deactivate at 0V to have the 5 second cycle. In the case that the transistor deactivated at 1.4 V across the capacitor, the 5 s cycle would be reduced to the time it takes to charge the capacitor from 1.4 V to 3.4 V.So my question!:If the transistor was activated at 3.4 Vin (3.4 V across the capacitor) and allowed the capacitor to discharge through it, then the voltage at the capacitor would be decreasing. V in of the transistor would not be 3.4 V anymore..There is the mystery.If The transistor activates at 3.4 V, then when does it close as it has current running through it from the capacitor to ground. The voltage across the capacitor would be 1 V at some time 't' , therefore the V in of the transistor would be 1V, which is less than 3.4 V, which is when it activates.Does it deactivate at 0 V in - ~0 V in AFTER it has activated (opened/activated), and only activate at 3.4 V in?A second question: If I replaced the cap with a resistor the IC would switch rapidly correct?The schematic on the top left is the one I redrew on the wide ruled paper.Super thanks.

author: atmfjstc
author_rank: Regular Contributor
timestamp: October 17, 2019, 12:17:24 am
content: The problem description doesn't show the circuit between Vin and the transistor's base, so you can't make any assumptions as to what the logic for turning it OFF is.The only thing that is clearly mentioned is that the transistor will be turned ON when Vin reaches 3.4V. We can safely assume that the circuit will not turn the transistor back off as soon as the voltage goes a bit under 3.4V, since then the circuit would just oscillate around that point and never discharge the capacitor. We can also conclude, given the wording used, that the transistor will be turned on fully (i.e. into saturation).There are many possible circuits that would achieve such an effect. Depending on which is used, you can make the circuit turn off the transistor at any point you wish... 1V, 2V, etc., or never turn it off at all.If you replace the cap with a resistor.. initially Vin will be the result of the voltage divider. If it's <<3.4V obviously nothing will happen and the circuit will stay that way forever. If it's more than that, the transistor will be turned ON and basically pull Vin close to 0 almost instantly. What happens then is again unspecified. As before, depending on the circuit, it might be turned off and restart the cycle at high speed (thus the whole thing becomes an oscillator), or it might just stay open and keep drawing current through the top resistor.

author: renzoms
author_rank: Regular Contributor
timestamp: October 17, 2019, 12:37:12 am
content: Thanks I think that is the best reply to a post I have made.
Transistor amplifier parasitic oscilation - Page 1
Forum section: electronics / beginners

author: ovidiu231
author_rank: Newbie
timestamp: September 13, 2019, 10:36:45 pm
content: Hello,I'm new here. I got a big problem, I build a class A amp with 2n2222a, biass point consist of 2x 10k resistors, on colector I have 4.7k and on emitor 2.2k. When I inject the signal from my function generator i get this...Anyone have an ideea what is going on?

author: bob91343
author_rank: Super Contributor
timestamp: September 13, 2019, 11:12:16 pm
content: I can't read the voltage scale but I suspect you are overdriving it.

author: Circlotron
author_rank: Super Contributor
timestamp: September 13, 2019, 11:45:05 pm
content: Is there a coupling capacitor between your signal generator and the base of the transistor? A schematic diagram would help.

author: ovidiu231
author_rank: Newbie
timestamp: September 13, 2019, 11:47:23 pm
content: It's 1V/div

author: ovidiu231
author_rank: Newbie
timestamp: September 13, 2019, 11:52:52 pm
content: I tried with all kinds of caps. When I increase the amplitude, the bump increases too.

author: kg4arn
author_rank: Unknown
timestamp: September 14, 2019, 12:00:35 am
content: Schematic please. Show the test point also.Sorry I missed it

author: ledtester
author_rank: Super Contributor
timestamp: September 14, 2019, 01:36:18 am
content: Check the DC biasing of the collector. With 10K / 16K biasing of the base LTspice says it is around 3.3V which is much too low.Change the 16K resistor to 2K and see what happens. The collector should be around 7.4V and you should see an amplification of about 1.75.Update: Attached the ltspice files.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 14, 2019, 03:34:25 am
content: That's what happens when the transistor saturates and the function generator is driving the emitter resistor. Draw out the voltages, you will see why this is so.Tim

author: bob91343
author_rank: Super Contributor
timestamp: September 14, 2019, 04:38:54 am
content: As I said, you are overdriving it. Back off on the input signal and it should clean up.

author: ovidiu231
author_rank: Newbie
timestamp: September 14, 2019, 10:27:17 am
content: It worked, thank you so much❤
Transistor amplifier voltage gain - Page 1
Forum section: electronics / beginners

author: king.oslo
author_rank: Frequent Contributor
timestamp: February 03, 2013, 02:32:03 pm
content: Hello there,I need to make an ultrasonic power amplifier. I have looked at class A and B and AB.I have 150V 7A adjustable PSU single supply, I need to amplify a sine from function generator to 70Vrms 5A or higher amplitude out. That's easy.My problem is that I want V_OUT to swing really close to V+ and GND, so that V+ can be as low as possible. That way i can dissipate as little heat as possible/high efficiency. How do I design the gain stage to achieve a voltage swing close within the limits of the supply?Thanks.M

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 03, 2013, 02:41:18 pm
content: Quotecan you share the schematics to us?

author: king.oslo
author_rank: Frequent Contributor
timestamp: February 03, 2013, 02:43:47 pm
content: So far I don't have any schematics. I will make the schematics when I know how I am going to make the gain stage.Thanks.M

author: SeanB
author_rank: Super Contributor
timestamp: February 03, 2013, 03:03:35 pm
content: You will need an auxiliary rail then to power the output stage, generally only about 10V or so. Look at bootstrapping audio amplifiers, as they make the auxiliary supply by using the output stage to derive this extra supply using a capacitor and a diode.

author: king.oslo
author_rank: Frequent Contributor
timestamp: February 03, 2013, 04:35:38 pm
content: Quote from: SeanB on February 03, 2013, 03:03:35 pmDear SeanB,I am certain that you are correct.My understanding of bootstrapping is taking the output of an amplifier and feeding it into the input to get more gain. Is that correct?I drew up an amplifier, where exactly would you wire in the capacitor to get the extra gain?Thank you for your time.Kind regards,Marius

author: SeanB
author_rank: Super Contributor
timestamp: February 03, 2013, 04:56:49 pm
content: http://en.wikipedia.org/wiki/Bootstrapping_%28electronics%29http://www.ti.com/lit/an/slyt077/slyt077.pdfWill help

author: olsenn
author_rank: Frequent Contributor
timestamp: February 03, 2013, 05:15:47 pm
content: QuoteYou're thinking of a darlington pair

author: king.oslo
author_rank: Frequent Contributor
timestamp: February 03, 2013, 05:19:46 pm
content: No, I was thinking of taking the output of an amp, and feeding it back into its own input. Positive feedback.M

author: SeanB
author_rank: Super Contributor
timestamp: February 03, 2013, 05:40:54 pm
content: In this application it is charging a capacitor to provide a rail higher than the supply so that a power device can be driven closer to the power supply, with the actual drive being able to briefly be higher than the supply rail so that it is still saturated.
Transistor analyser imformation required. - Page 1
Forum section: electronics / beginners

author: tech255
author_rank: Contributor
timestamp: December 29, 2017, 01:50:25 pm
content: I have purchased a full transistor analyser manufacturered by M.L Avation in the uk, but there was no manual with it and its not easy to figure out the sequence required for testing each stage.Just wondered if anyone has seen or used one or if anyone has a manual.I will try to upload some photos.

author: barry14
author_rank: Regular Contributor
timestamp: December 29, 2017, 10:16:39 pm
content: It would be better if your photos were not upside down.
Transistor as a swtich practical doubts - Page 1
Forum section: electronics / beginners

author: anishkgt
author_rank: Frequent Contributor
timestamp: March 06, 2018, 07:55:03 pm
content: Hi All,I am trying to power an ATmega328P via a USB when it is not powered by the 5v rail on the PCB. The Atmel will only be used to upload sketch when connected via the USB and nothing else. Would any one please advise if the schematic attached (VoltageSwitch)would work ?Another one that i had in mind based on the circuit of the Arduino R3 Schematic is also attached (ComparatorDesign)Thanks in advance.

author: fourtytwo42
author_rank: Super Contributor
timestamp: March 06, 2018, 08:59:32 pm
content: The first schematic does nothing as the transistor is always switched off.The second is some complicated affair doing something that can probably be acheived by a diode, from your description of powering the Atmega from USB when some other supply is not working a simple diode,anode to usb+ & cathode to Atmega 5V should work, of course I may have misunderstood your explanation ?

author: anishkgt
author_rank: Frequent Contributor
timestamp: March 06, 2018, 09:05:54 pm
content: To keep it simple, my idea is to power the avr via the 5v rail when the usbVcc is not available and power it with usbVcc when 5v rail is not available and also cut-off usbVcc when both usb is connected and 5v rail is available.Www.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D, Brymen BM869s

author: rhb
author_rank: Super Contributor
timestamp: March 07, 2018, 01:44:02 am
content: Use a SPDT 5 V relay with the coil connected to the 5 V rail.Relay down, power from railRelay up, power from USBThe way to solve such problems is to build a truth table.

author: anishkgt
author_rank: Frequent Contributor
timestamp: March 07, 2018, 04:44:42 am
content: The truth table is like this,V+ > V- = VccV+ < V- = veeI've been looking that the schematic of the Arduino R3 and modified it but am a little lost as how to supply the initial voltage to the divider and the opamp. The opamp wouldn't be any of any use when powering from the pcb +5v rail as there is no USB+ so the opamp should be powered by the USB+.

author: anishkgt
author_rank: Frequent Contributor
timestamp: March 07, 2018, 01:54:40 pm
content: I guess initially i wasn't thinking. The resistor wouldn't be necessary or would it ?

author: rhb
author_rank: Super Contributor
timestamp: March 07, 2018, 02:11:14 pm
content: All you need is the relay connected to the 5 V supply at one side, the USB 5 V at the other and your device in the center. The relay is energized if you have 5 V and the circuit uses that, otherwise it's not energized and the project uses the USB 5 V. One part for a buck or two.

author: anishkgt
author_rank: Frequent Contributor
timestamp: March 07, 2018, 02:33:31 pm
content: I had a couple of these opamp laying around and moreover I thought this would be good time to learn to use a comparator.The first design I thought was a really but I’d appreciate if somebody could guide me on the comparator design.Www.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D, Brymen BM869s
Transistor as buffer ( this time with pnp :D ) - Page 1
Forum section: electronics / beginners

author: fady232
author_rank: Contributor
timestamp: April 16, 2025, 08:23:00 am
content: Hello all, yesterday i posted the same buffer with npn transistor, and i guess im a good point now. i fully understand it.Today, im trying to simulate same circuit but with pnp transistor, but i had some questions to ask, and i guess i got some misunderstanding.Basically im giving a 9V AC sine wave into input ( to the base of the transistor ), the emitter is connected to vcc, via a resistor of 1k ( so around 9V-the drop on resistor).In this condition,the voltage on emitter is lower than base( so it shouldnt conduct, remember the fact that a pnp conduct when Base < Emitter by 0.7V )so my question is why do i see the AC sine wave still in output? the diode like that shouldnt be forward biased, in fact it shouldnt conduct no? cus if i understood well, what happens is that we use the D.C voltage for polarization ( and D.C wins against A.C), at this point the diode is forward biased and the signal can be seen in output+0.7V. But as i said in our case the emitter is lower than base , soit shouldnt conduct. Can someone explain me please? sorry for all of that.

author: PGPG
author_rank: Frequent Contributor
timestamp: April 16, 2025, 09:13:20 am
content: Have you tried to view picture you have attached?It has 120x53 pixels.

author: fady232
author_rank: Contributor
timestamp: April 16, 2025, 09:43:59 am
content: try this

author: PGPG
author_rank: Frequent Contributor
timestamp: April 16, 2025, 11:27:01 am
content: Quote from: fady232 on April 16, 2025, 08:23:00 amWhy you assume voltage on emitter being lower than base?This isn't so in any moment in your circuit.You see output being cut at tops - it is the moment when your base is not lower from emitter by 0.7V but still lower.Your transistor saturates (V emitter is close to V collecter) when you have at base about 4V.For all voltages below 4V (so from -9V to +4V) transistor is not working - you practically use only its B-E diode.

author: Ian.M
author_rank: Super Contributor
timestamp: April 16, 2025, 12:29:00 pm
content: Maybe plotting the transfer characteristic will help you figure out what's happening. You can do that with:Code:.dc V1 0 9Also, look at the base current, which will be much higher when the transistor is acting as a diode and isn't amplifying!Incidentally, you can attach LTspice sims here, and for simple circuits, using the built-in models, that's usually the easiest way of sharing them.

author: PGPG
author_rank: Frequent Contributor
timestamp: April 16, 2025, 01:24:20 pm
content: See what is going on at collector in your circuit.You will notice that sometimes signal from base is inverted at collector and sometimes not.I have made such mistake when in 80s was constructing 100MHz input amplifier for frequency counter with the result of doubling the frequency by input amplifier.
Transistor as diode - Page 1
Forum section: electronics / beginners

author: Simon123
author_rank: Regular Contributor
timestamp: April 08, 2018, 03:17:45 pm
content: Hy!A lot of times BJTs are used as diodes because of their small leakage current.But sometimes i see the BE junction being used and sometimes the BC junction is used.What are advantages/disadvantages of one over the other besides Reverse breakdown voltage of BE is smaller compared to BC.I have also read that BE has smaller recovery time, but i dont undestand why some people say BC junction is better?

author: KrudyZ
author_rank: Frequent Contributor
timestamp: April 08, 2018, 03:47:46 pm
content: The BE junction makes for a pretty good zener diode (~6-7V) in reverse.In forward direction with the collector tied to the base, you get the full advantage of the current gain of the transistor to get a much sharper knee and high current capabilities.The BC junction has lower leakage since that junction in a transistor is almost always in reverse bias, so that's what it is optimized for.

author: Zero999
author_rank: Super Contributor
timestamp: April 08, 2018, 08:40:07 pm
content: Another application is in a current mirror, where the same transistor type is used as a diode, purely in the hope its base-emitter voltage will be matched to the other transistor.https://en.wikipedia.org/wiki/Current_mirror

author: Cerebus
author_rank: Super Contributor
timestamp: April 08, 2018, 09:03:28 pm
content: Typical figures for a small signal NPN bipolar transistor:JunctionRecoveryTimeCapacitanceBreakdownVoltageReverse leakagecurrentCollector BaseEmitter Open200 ns3 pF30-60V4 pAEmiiter BaseCB Shorted1 ns5 pF6-8V10 pA

author: mrkev
author_rank: Regular Contributor
timestamp: April 09, 2018, 09:43:53 pm
content: Quote from: Hero999 on April 08, 2018, 08:40:07 pmNot purelyOnce oupon a time, there used to be a device with two pre-matched transitors in one package (those are pretty rare nowadays). But you can simply add emiter resistors.On the other hand, current mirrors like this are very often used in on-chip circuits, where it's much easier to match two transistors than to make one with precise parameters.And to the original post, exactly as KrudyZ wrote, the BE junction in reverse is often used as a low cost, very-low capacitance (and leakage current) zener for hi-frequency amateur circuits (usually as protection from overvoltage).

author: David Hess
author_rank: Super Contributor
timestamp: April 10, 2018, 10:06:35 pm
content: Most available small signal diodes are either gold doped switching diodes or schottky diodes. Low leakage and high conductance diodes exist but cost much more and have poor availability compared to a small signal bipolar transistor and fast low leakage high conductance diodes like a base-emitter junction have not been available in decades; if you want low leakage and fast recovery, there is nothing better.Of course, why would you need them? High impedance shunt protection circuits, analog waveform generation, analog processing, sampling, and RF applications can make use of them.
Transistor Base - Double Check My Maths? - Page 1
Forum section: electronics / beginners

author: w.ashcroft
author_rank: Newbie
timestamp: January 03, 2022, 03:07:06 pm
content: Hello,I am currently building a circuit that will utilise a PNP transistor to drive a high current LED from a current limited sink. When calculating the base resistor, I am getting a value of around4.7k Ωusing the hFE method or1.2k Ωusing the 1/10 method, assuming I have my maths right!Would appreciate somebody double checking this for me, and which method would be best to go with (transistor acting as a switch).Vcc = 13vVi = 12.5vIc = 100mALED If = 100mALED Vf = 9.2v3.8v Drop Resistor = 38.3ΩPNP SS8550 (datasheet)Vbe(sat) = 0.7vhFE @100ma = 120-400hFE(min) Method (/3 to guarantee saturation):Ib = Ic / Hfe = 100 / 120 = 0.83mAVb = Vi - Vbe(sat) = 12.5 - 0.7 = 11.8vRb = Vb / Ib(A) / 3 = 11.8 / 0.00083 / 3 = 4,739 ΩOr ignore hFE and use the 1/10 method:Ib = Ic / 10 = 100 / 10 = 10mAVb = Vi - Vbe(sat) = 12.5 - 0.7 = 11.8vRb = Vb / Ib(A) = 11.8 / 0.01 = 1,180 ΩMany thanks,Warren

author: rstofer
author_rank: Super Contributor
timestamp: January 03, 2022, 03:58:33 pm
content: Either method works and your numbers look fine but what you are really looking for is the VceSat to be around 0.2VSo, breadboard the circuit and measure Vce. Is the transistor saturated? Remember, HFEcan vary all over the map.I usually try for Ib= 0.1 * Icas in your 2d example.

author: Capernicus
author_rank: Frequent Contributor
timestamp: January 03, 2022, 04:23:11 pm
content: I just want to say, this is not even important if you were just making logic, if its just low and high theres no value even important imbetween anyway.And if your making a motor servo, its just full blast or nothing, so its just a louder bit, and theres no value importance to it either.So sometimes being an egg head with mathematics just slows you down.

author: TimFox
author_rank: Super Contributor
timestamp: January 03, 2022, 04:30:06 pm
content: Could you re-phrase that answer to make it understandable?
Transistor basics - falling curve on emitter side? - Page 1
Forum section: electronics / beginners

author: bitman
author_rank: Unknown
timestamp: January 28, 2018, 12:07:35 am
content: Hello, I have another newbie question. This time about transistors (a basic 2n2222). In the image below you can see the result of a simple curcuit using a toggling jk flipflop where the two output lines are each driving a 2n2222 transistor. I'm experimenting a bit to get better fanout as these clocks is driving 8-10 different ICs (7400 series).What I don't understand is the 4ms curve when the transistor turns off. Note - I'm using the NPNs where the load is on the emitter side - not the collector side. And that's my question. When used on the collector side I get fairly flat pulses - but this looks like a capacitor is in play and there's noe attached to the flip-flop. The input signal (the yellow one) is flat, and that's where the only capacitor is found.5v|2n2222 -- in signal from flip-flop|(load - in this case just an LED with a 330ohm resistor)|gndSorry, crude drawing. I wanted to measure 0->5v so I thought putting the load on the emitter side would allow me to plot that out, but this is what I get. I understand that there is some resistance in the transistor - but obviously it looks like there's some kind of inductance too? This also seems to be within tollerances on the switching characteristics on the basic 2n2222 (25ns raise time) - I'm using a whopping 2.0 ms way above that. And as I stated if it is the transistor I would expect to see the same curve on the collector side? Since I don't I doubt I need to keep looking at the transistor, but at something else.Basic question: What's causing this? Outside of putting the load "on the right side" is there anything that can be done to change this?

author: Audioguru
author_rank: Super Contributor
timestamp: January 28, 2018, 12:34:02 am
content: You said the emitter load is an LED. But it is a load only while it is conducting:

author: bitman
author_rank: Unknown
timestamp: January 28, 2018, 12:38:37 am
content: DUH - missing the obviously it looks like. That must be it - as I have to eat my words that this circuit on the collector side behaves differently. This is a single transistor, where I measure the base signal and the collector signal - and note the curve on the collector which again for test purposes is driving a LED. A cheap LED - so this really explains a lot. Thank you @Audioguru!

author: bitman
author_rank: Unknown
timestamp: January 28, 2018, 01:10:52 am
content: @AudioguruI'll end this very short thread by confirming that it was indeed the LED I was measuring - not the transistor. I replaced the load with a simple resistor and presto - it's perfectly square. Lesson learned. Thanks again.

author: David Hess
author_rank: Super Contributor
timestamp: January 28, 2018, 07:08:36 am
content: Base-emitter junctions are really good diodes with low reverse leakage so what you saw is actually the 1 or 10 megohm input resistance of the oscilloscope probe pulling the output down.
Transistor blows up - Page 1
Forum section: electronics / beginners

author: ALLelectroNIX
author_rank: Newbie
timestamp: May 01, 2020, 05:10:50 pm
content: Hello everyone!I have a little problem with my transistor..I'm trying to create a handsfree dispenser by using an arduino. To control my relais i use a transistor wich is controlled by my arduino.My problem is: when i plug in my transformer to the 240V and i turn on my transistor, everytime it blows up, i have no idea why! The current though my transistor is low enough when i calculate it.My first tought is: should i put a diode parallel to my relais? But i have no idea if that's the solution.Thanks!

author: Manul
author_rank: Super Contributor
timestamp: May 01, 2020, 05:19:29 pm
content: Well, sorry, but it is bad on many levels. Are you powering Arduino from resistor voltage divider? And why you have relay on emitter and not collector? I would not be surprised if you also damaged your Arduino with such a circuit...

author: WattsThat
author_rank: Frequent Contributor
timestamp: May 01, 2020, 05:24:37 pm
content: Several things wrong. Label the pins of the Arduino. Yes, perhaps you are powering it incorrectly, don’t know without pin labels.The emitter/collector connections are incorrect for a low side switch and yes, you absolutely need a diode across the coil. The base resistor is far too high at 15k.

author: mikerj
author_rank: Super Contributor
timestamp: May 01, 2020, 08:22:54 pm
content: What is part "A"? It looks like a switch, in which case why are your relay contacts wired in series with a switch across the 240V supply?

author: John B
author_rank: Frequent Contributor
timestamp: May 01, 2020, 11:42:21 pm
content: Are you sure the transistor even failed? The relay wouldn't even work in that configuration. You'll only apply about ~4.3V to the 12V relay. Powering the arduino with a resistor divider isn't the worst thing in the world, it probably draws about 30mA. At least the circuit could still conceivably function.

author: AVGresponding
author_rank: Super Contributor
timestamp: May 02, 2020, 07:57:56 pm
content: I'm sure it's an accident on the drawing, but that shows the mains 240V L and N being shorted together when the relay activates...

author: themadhippy
author_rank: Super Contributor
timestamp: May 02, 2020, 08:37:23 pm
content: QuoteOr its an arduino controlled fuse tester

author: thinkfat
author_rank: Unknown
timestamp: May 02, 2020, 09:19:09 pm
content: In that configuration the transistor has to drop about 8.4V. That's considerable. Even with little current, it can get pretty warm.
Transistor calculations (Art of Electronics exercise) - Page 1
Forum section: electronics / beginners

author: Aeternam
author_rank: Unknown
timestamp: January 13, 2016, 10:26:23 am
content: Guys,I'm working through The Art of Electronics and I'd like to double-check if my calculations are correct for a pretty basic transistor setup exercise. (Exercise 2.1 in the book, on page 77.)See attached circuit. The question is (a) what is the approx. LED current and (b) what minimum beta is required for Q1.So here goes:(a)Assuming a voltage drop of 1.8V for the LEDVoltage drop across R1 (collector resistor) = 3.3V - 1.8V = 1.5VThis gives Icollector = 1.5 / 330 = 4.5 mAHence LED current is approx 4.5 mA(b)Voltage drop on the base is 0.6 V (base-emitter diode of Q1 is in forward conduction)Voltage drop across R2 (base resistor) = 3.3V - 0.6V = 2.7VThis gives Ibase = 2.7 / 10k = 0.27 mAIf Icollector = beta * Ibase, then beta = Icollector / IbaseTo drive (terminology?!) 4.5 mA collector current with a base current of 0.27 mA we need a beta of 4.5 / 0.27 = 16.6 at least.After checking the datasheet of the 2N3904 the minimum gain of this transistor is 40, so we should be ok here.Does all this sound about right?

author: KM4FER
author_rank: Regular Contributor
timestamp: January 13, 2016, 03:34:25 pm
content: Aeternam, you have assumed that the voltage across the collector-emitter is 0V. Real transistors can't do this. When fully turned on there will be a small Vce present, maybe 0.2V, called Vsat in the datasheet.Taking this into account the KVL would be: V2 - Vd2 - Vr1 - Vsat = 0.See what effect this has on your calculations.Earl

author: Audioguru
author_rank: Super Contributor
timestamp: January 13, 2016, 05:01:31 pm
content: hFE is not used when a transistor is saturated. The datasheet for the 2N3904 (and all other transistors) shows that the hFE is used when the transistor is linear, not saturated and has a collector to emitter voltage (Vce) of at least 1V (more Vce voltage with other transistors). Its maximum saturation voltage loss is 0.2V or 0.3V when the base current is 1/10th the collector current. Since some 2N3904 transistors have more hFE than others then you might be lucky if you use less base current with some of them.So with your calculation with not enough base current to be saturated, some 2N3904 transistors would have a Vce of 1V and the poor LED would be fairly dim with a current of only (3.3V - 1.8V - 1V)/330 ohms= 1.5mA.LEDs have a wide range of forward voltage. Maybe your LED has a typical forward voltage of 1.8V but its max voltage might be 2.2V. Then its current will be only 0.3mA with some 2N3904 transistors.The resistors have a tolerance. What if the 330 ohms is actually 5% high and is actually 346.5 ohms and the 10k resistor is actually 10.5k?The power supply voltage also has a tolerance and might be 5% low at 3.135V.All these things must be considered if you are planning to produce many items with this circuit. When they are all occurring then the 2.2V LED will never light.The opposite tolerances might cause the LED to burn out. Then some of your items burn out the LED, others never light and maybe a few work fine.
Transistor Checking - Page 1
Forum section: electronics / beginners

author: Electro Fan
author_rank: Super Contributor
timestamp: August 21, 2023, 04:21:14 am
content: I don't measure transistors often but when I do I'd like to get it rightActually, I'd be happy to just get some consistent results, or different results I could explain.Anyone have any ideas on why the results shown below might be different on the hFE?Same transistor used for all the tests, of course.I didn't have a fifth device to break the tieThanks

author: Electro Fan
author_rank: Super Contributor
timestamp: August 21, 2023, 05:15:06 am
content: After reading a Peak manual, maybe the Peak (and each tester device) uses a different collector test current and a different CE voltage and therefore the tests being run aren't really apples to apples for hFE? In fact, Peak might use a collector test current of 2.5 mA for the DCA55 and 5.0 mA for the DCA75? Kinda strange and potentially confusing.So what these devices are showing are really just data points based on a set of fixed inputs that can differ from device to device, so it's really a nearly random glimpse of a larger picture? I guess on a larger Tektronix transistor checker the user could set test parameters and then look at curves, and here we have no control of test parameters and we get a data point instead of curves?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 21, 2023, 12:07:44 pm
content: Hfe is neither a tightly controlled, nor a constant parameter for a transistor.Hfe changes with base current, collector voltage ("early effect") and probably also (a little bit) with temperature (touching it to insert it), moon phases and planet alignment.The Atlas shows a higher base-emitter voltage, which strongly suggests a higher base current (A diode drop is also not a constant 600mV, but varies depending on the current). And with a higher base current, Hfe usually gets less.Hfe also varies a lot during production. Variation now is probably less then it used to be, but some 40 years ago small signal transistors were sorted in the factory in three Hfe ranges: "A", "B", and "C". and then Hfe ranging from about 150 (Lowest for "A") to over 900 (Highest for "C"). If you want to learn to design reliable electronics, then you have to allow for such changes. "The Art of Electronics" has a few chapters about this topic.

author: EPAIII
author_rank: Super Contributor
timestamp: August 22, 2023, 08:21:30 am
content: In a well designed circuit it is the circuit that determines the gain, not the hfe of the transistor in it.If you really need specific number for the hfe for your circuit to function properly, I would suggest you may need to redesign the circuit. I would. Or design a test circuit that measures the hfe under the circumstances in which it will perform in the circuit.

author: liaifat85
author_rank: Regular Contributor
timestamp: August 22, 2023, 09:02:23 am
content: Here is a similar discussion I found:https://www.reddit.com/r/AskElectronics/comments/hz40hm/3_dmms_3_different_hfe_readings_same_transistor/