What does "8 fV" mean? - Page 1
Forum section: electronics / beginners

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 07:21:22 am
content: While playing wth a super simple simulation of a equally simple power supply, did I add a extra circuit that should suppress ripples a bit extra but now do my simulation talk about "f Volts"

author: atmfjstc
author_rank: Regular Contributor
timestamp: April 07, 2020, 07:26:34 am
content: It might have computed a ripple in the femtoVolts, i.e. a quadrillionth of a volt. (seehttps://en.wikipedia.org/wiki/Femto-)I suspect you'll only ever obtain such a value in a simulation....

author: Ice-Tea
author_rank: Super Contributor
timestamp: April 07, 2020, 07:26:47 am
content: millimicronanopicofemto

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 07, 2020, 07:31:13 am
content: Your diodes D1-D4 are all pointing in the same direction, just guessing here but is that what you want?

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 10:47:40 am
content: Thanks for your answers.No the diodes were pointed in the wrong direction. :-)LOL yes if it means femto, then can it only be in a simulator, that's also why I asked. I'll try to see if it still show ridicules numbers afterward:-)

author: Zero999
author_rank: Super Contributor
timestamp: April 07, 2020, 10:55:09 am
content: Yes it means 8 femtovolts, which is 8✕10-15V. Please don't remove the attachment to your original post. I appreciate you might be embarrassed by your error, but the thread makes less sense now and is bad forum etiquette.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 11:07:42 am
content: Sorry, it's reuploaded.But no I was not embarrassed, I have made fare worse mistakes then that:-)I removed it because on an told me that schematics and so on that contained errors, only cluttering the thread and were a big reason for a lot of confusion.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 11:57:21 am
content: I am wondering if my solution at reply #4, do work as a sort of "ripple blocker" or is it only the simulator who think it's great?If so is there a way to combine the two rails again to positive-zero-negative?

author: max_torque
author_rank: Super Contributor
timestamp: April 07, 2020, 12:13:51 pm
content: Simulation is great! The other day, in a complex 3d thermal sim i "measured" 27,532 degrees centigrade on a component! Turns out i'd loaded it with 5 MILLION watts, and not 500 watts....... In the real world, that could have been nasty.......

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 12:26:26 pm
content: LOL:-)Do you tell me that you work with anything below a couple of million Volts?!I do normally never go below 3! :-)

author: Zero999
author_rank: Super Contributor
timestamp: April 07, 2020, 12:31:53 pm
content: Quote from: FriedMule on April 07, 2020, 11:57:21 amIs that what it's supposed to do? Presumably R5 and R10 are meant to be the loads? They appear to be open circuit, as there are no blue dots where their wires go to U1 and J1. I don't see how it will work, any better than an ordinary RC filter, because that's what it is, except it's got two constant current loads on the output.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 02:33:31 pm
content: Thanks a lot for spotting that, it was connected but I have still moved the connection so there can be no doubt. :-)After edeting do the ripple look exactly the same.Yes R5 and R10 is just a load, do not know how I can use the Load component, so did use resistors instead.I am glad that it's a filter, that was my idea, were not sure since there are no inductor or capacitors.The question is if it do work great, if the fet's get heated up and if it can handle i.e. 40V-50V 6A-10AThe to Fet's is just two random fet's where I hope to find the right ones.

author: exe
author_rank: Unknown
timestamp: April 07, 2020, 02:35:46 pm
content: I wish LTSpice had at least some basic validation

author: Zero999
author_rank: Super Contributor
timestamp: April 07, 2020, 02:40:33 pm
content: Quote from: FriedMule on April 07, 2020, 02:33:31 pmWhy would the FETs get hot They're hardly dissipating any power. You've got 2k2 between the source and ground, with the gate also connected to ground. Not much current will flow through them. Please post the .asc file.

author: Kirill V.
author_rank: Regular Contributor
timestamp: April 07, 2020, 03:08:22 pm
content: Quote from: exe on April 07, 2020, 02:35:46 pmStill the basic validation in any simulator is your head with fundamental knowledgePeople have used the equivalent circuits of components and logarithmic rule. Then they assigned this math to a computer. But the result is still checked by you, as before

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 04:28:27 pm
content: Thanks a lot!I would also hope that LTSpice had some sort of validation, it's a great help for mee as a total noob. :-)No I do not know why it should get hot but they do get 3V ripple down to about zero, so thought that it had to do something.Here are the file:-)

author: Zero999
author_rank: Super Contributor
timestamp: April 07, 2020, 04:40:25 pm
content: Quote from: FriedMule on April 07, 2020, 04:28:27 pmI couldn't get that file to open properly because it uses non-standard models. You can import them into the .asc file by opening up the appropriate model file using a text editor, such as Notepad or the one built-into LTSpice, copying it and pasting it into a SPICE directive on the schematic.I substituted the J-FETs and as expected, hardly any current flows through the load resistors, which is why the ripply goes down to zero. LTSpice has invalidated that circuit pretty well. You just need to plot the current through the load resistors to see it doesn't work.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 07, 2020, 08:33:21 pm
content: I do only use the standard component in LTSpice.As a noob and almost first time user of LTSpice it's hard to locate such errors, you'll expect the simulator to "just not work" then "pretend to work" :-)I am not sure but often is the "face palm" sign, used when one think that the other one is an idiot. I was a bit taken back by that emoticon.I have just tried to make a simulation of the FET part of the circuit in falstad simulator, I chose to replace R5 and R10 with two LED's for visual "proof":http://tinyurl.com/sbrpknothere does it work for some strange reason!?

author: Zero999
author_rank: Super Contributor
timestamp: April 07, 2020, 09:19:03 pm
content: Quote from: FriedMule on April 07, 2020, 08:33:21 pmSorry, I forgot this is the beginners section. No offence was intended.All LTSpice does is calculate the voltages and currents in a circuit. It doesn't know what you intend the circuit to do. Even given today's level of AI it's not possible to just ask a computer "Does this circuit work?" because the computer doesn't know what you mean by work. Indeed, if it's an unusual circuit, a human won't know, unless you tell them what you expect it to do. If you had just posted that schematic, without any explanation, I would have guessed it's constant current bleeder circuit to discharge the capacitors, when the power is removed and would have assumed the circuit worked, because that's what it does: draw a roughly constant, around 567µA current from the power supply.In order to use SPICE to check whether a circuit is doing what you expect it to do, you need to already have a rough understanding as to how the circuit works. In this case, I knew it wouldn't do, as you described, without doing any in-depth analysis or simulation. Forget the transistors for now, just use Ohm's law. Focus on the output stage, just the positive for now. L1 is an inductor, with a low DC resistance, so can be ignored and R7 is a couple of orders of magnitude greater than R5 so can also be ignored. We have R8 = 2k2 forming a potential divider with R5 = 20. Without doing any proper calculations we can see that R8 will drop most of the voltage, leaving a small volt drop across R5 and the maximum current that part of the circuit can ever draw is roughly the supply voltage, divided by R8. We can now see that all it's doing is severely restricting the current to the load, which is why you noticed the ripple fall.In reality, it's a little more complex than that. What happens is U1 starts to turn on, as its gate is at 0V, a current flows through R8, lifting the source voltage up, thus biasing U1 more into its pinch-off region (J-FETs are on, when the gate-source voltage is zero and start to turn off, as the gate voltage falls below the source). The circuit will soon reach an equilibrium with a constant voltage across R8, thus a constant current flowing through the J-FET. With the model I used in the simulation, the current was 567µA, so the voltage across R8 would have been V = IR = 567*10-6*2200 = 1.25V. Different J-FETs, will give different currents.

author: Zero999
author_rank: Super Contributor
timestamp: April 07, 2020, 09:27:56 pm
content: Quote from: FriedMule on April 07, 2020, 08:33:21 pmYou've made a very nice constant current supply for powering LEDs. The 1k5 resistors can be removed, to give better current regulation through the LEDs. The downside is the current is dependant on the characteristics of the J-FETs, which isn't well-controlled. If you built a batch of these circuits, the current through each of them would be different. If you entered the exact same circuit into LTSpice, you'll get similar results.

author: Nerull
author_rank: Frequent Contributor
timestamp: April 07, 2020, 10:04:11 pm
content: SPICE simulated the circuit you told it to simulate and told you how it would behave. How is it supposed to know if the circuit does you expected it to do? That's up to you to verify.That's the point of doing simulations, so you can see how the circuit behaves. It isn't "I clicked run and it didn't crash so the circuit must work".

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 08, 2020, 12:30:01 am
content: Zero999, no problem at all, I do understand your reaction if the question had been on a thread under experienced people, that would clearly have been facepalm!! :-)I do really like your explanation, it teached me in fact a great deal! As I do understand it, would that small circuit be fantastic to control the "ON-LED" but be terrible to do anything where e.i. high and fast changing current is important.I know that I now am asking for a compleat meal, served on a silver platter but what do you think I have to search for and understand to build a simple-ish 45V p/p 10A out power supply?Here are my goals, reduce the ripples, even under suddenly changing power delivery. I would like to reduce the amount and sizes of capacitors. I would also really like if I can reduce the voltage drop and make the power supply fast!!Until now do I know that I shall protect against inrush current, that before the trafo do I have to make some sort of lightning protection and I know that you can reduce ripples by adding lots of capacitors to a certain degree:-)EDIT: Oh and forgot to ask, why do most use 4 diods, even in places where voltage drop is a problem? Why not just two diods, like in this circuit?

author: Zero999
author_rank: Super Contributor
timestamp: April 08, 2020, 10:34:51 am
content: The only sane way to build a 45V 10A power supply these days is to use a switched mode power supply. Most engineers simply wouldn't bother designing one for their product, they'd buy an off the shelf solution, unless they had very niche requirements which couldn't be met by a read made unit. I know that's probably not the answer you were looking for, because it's not fun.There is no passive way to get the ripple down without using massive capacitors, inductors or a three phase supply. In a large switched mode power supply, with a power factor correction stage, a boost converter is used to increase the voltage, as the input voltage drops, so smaller capacitors can be used. The input current and voltage are monitored to ensure the load is Ohmic, i.e. follows Ohm's law.Four diodes are used because it's cheaper and takes less space than using a transformer with a centre tapped secondary winding. A tapped secondary winding with two diodes was more common, back when diodes were expensive. You're right, the two diode, centre tapped transformer does have a lower voltage drop, so makes sense at very low voltages and Schottky diodes can also be used to minimise voltage drop.

author: FriedMule
author_rank: Frequent Contributor
timestamp: April 09, 2020, 02:34:49 pm
content: Thank you a lot for your reply, yes "buy myself" is far from as fun as "build myself" :-)I have tried to look what types there are and there are a lot, PWM, Flyback, Buck and a billion combination more of all sorts.What are the best to look fore when the application is 45V power amplifier?

author: Zero999
author_rank: Super Contributor
timestamp: April 15, 2020, 07:51:29 am
content: Quote from: FriedMule on April 09, 2020, 02:34:49 pmIt depends on what the 45V power supply is being used for? You said 10A, but what is it going to be powering? Is it just a large motor or an audio amplifier?If it doesn't need to be regulated and size/weight/space aren't constraints, then there's nothing wrong with using a transformer and big smoothing capacitor. Many audio amplifiers still use this configuration. The mains frequency ripple and poor voltage regulation don't matter because it's easy to design an audio amplifier which has good power supply rejection.A mains powered switch mode power supply is not a suitable project for a beginners because it involves capacitors charged to hazardous voltages and stringent requirements for separation/insulation between the mains and DC side and layout is critical, both for safety (adequate creepage and clearances) and functionality.