10% current bleed resistor on voltage divider, how? - Page 1
Forum section: electronics / beginners

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 27, 2021, 03:07:55 pm
content: I thought I knew that a bleed resistor was a resistor made to dissipate power to avoid electric chock to the user, but now do I learn that a bleed resister can be the first resistor in a voltage divider.If I understand a voltage divider correctly, should the schematic be correct when looking at the voltage, I do just not understand how it is possible to get a 10% current bleed on R1?

author: bdunham7
author_rank: Super Contributor
timestamp: November 27, 2021, 03:35:40 pm
content: Where did you get that circuit diagram??

author: Terry Bites
author_rank: Super Contributor
timestamp: November 27, 2021, 04:23:22 pm
content: R1, R2 and R3 need to be recalculated. None of the values shown will work.Note that the first step is to convert the loads to their actual resistances.Use superposition:www.allaboutcircuits.com/textbook/direct-current/chpt-10/superposition-theorem/Calculate the currents that will meet the criteria of Kirchhoff's law(s).

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 27, 2021, 06:09:53 pm
content: The circuit comes from this 10-hour long video about basic electronic:https://youtu.be/nb4ovfwqup8?t=26078Terry Bites, thanks, I am truing to learn about that law, y values do come from another video that only told about how a voltage divider works. I.e. two resistors with the same value, gives ½ the voltage between the resistors, three similar resistors gives 1/3 of the value. That's why I thought that 48V / 4 = 12, so between 200 ohm and 100 ohm + 100 Ohm would give 24V and after 200 Ohm + 100 Ohm would be 12V.

author: fourfathom
author_rank: Super Contributor
timestamp: November 27, 2021, 06:28:36 pm
content: Quote from: FriedMule on November 27, 2021, 06:09:53 pmAnd it will, if you disconnect the loads. As said upstream, the loads become part of the divider so you need to allow for that.

author: Ian.M
author_rank: Super Contributor
timestamp: November 27, 2021, 09:48:00 pm
content: Quote from: FriedMule on November 27, 2021, 06:09:53 pmThat video is an example of everything that's wrong with American education!The instructor either doesn't understand the practical application of potential dividers or is being forced to teach a syllabus that's fundamentally broken. Its nearly as bad as:https://en.wikipedia.org/wiki/Indiana_Pi_BillSkip to:https://youtu.be/nb4ovfwqup8?t=26824where the instructor starts introducing a bogus rule of thumb: "The bleeder current needs to be 10% of the total load current." and he then goes on to say that's the current through the bottom divider resistor.Anyone who's ever done anything practical with dividers knows that the rule of thumb is the other way around. i.e the standing divider current needs to be ten times the load current (or more refined: ten times the difference between the minimum and maximum load currents), and that assumes you can tolerate a 10% voltage drop at full load!If you are going to do the math properly, you need to calculate the Thévenin equivalent resistance of the divider at each tap you are going to use and calculate the voltage drop due to your load current through the Thévenin equivalent resistance of that tap to see if its acceptable. If its not, scale the divider resistors down in proportion till it is.See:https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 28, 2021, 12:31:24 pm
content: Thanks, this makes a lot of sense, does that mean I add like this (math not my strong side):(1/R3 + 1/Rx)^-1 = T3T3 + R2 = T2(1/T2 + 1/Ry)^-1 = T1T1 + R1How do I then calculate the optimal bleed, why the same word for two different functions?

author: BeBuLamar
author_rank: Super Contributor
timestamp: November 28, 2021, 02:05:49 pm
content: The guy in the video chose different values for the resistors which made the divider works but unrealistic because first the would throw away more energy in the divider resistors than in the load. Besides the resistors will be very hot.With your resistor values the divider would give you 12, 24 V if you don't have the load connected. Once you have the load connected the voltage will go down a lot.

author: Ian.M
author_rank: Super Contributor
timestamp: November 28, 2021, 02:18:47 pm
content: That does look like how to calculate the equivalent resistance of the divider chain as seen by the source powering it. However that's of little use for calculating a divider that will 'work' for the instructor's problem.He's got two loads:12V 6W, allegedly a CB radio24V 200W, allegedly a radarLet *US* make a further assumption - that they can operate over a +25%, -12.5% voltage range from nominal 12V or 24V typical for automotive or marine equipment fed from a Lead Acid battery.Its obviously undesirable to overvoltage the loads and their current draw in the off or standby state may be negligible, so the unloaded divider mustn't exceed15Vat the lower tap and30Vat the upper tap.Lets further assume that the loads are modern ones with switching supplies so will draw more current at lower voltages. At their lower operating limit of10.5vand21Vrespectively they will thus draw 6W/10.5V =0.571Apeak*, and 200W/21V =9.52Apeak.Also, the fully loaded current 'from' the upper tap must include the lower tap load current, even though its drawn through R2 not directly from the tap, so we must use Iload1+Iload2 =10.1Ain further calculations involving the upper tap, not the Iload2 9.52A peak current.Lets get the upper tap done first:30V no load and 21V fully loaded is a 9V drop,9V/10.1A =0.891 ohmsThévenin resistance.That's R1||(R2+(R3||Rload1)).We've also got (R2+R3)/(R1+R2+R3)=30V/48V, and Rload1=10.5V2/6W.I'll let you do the math . . .Todo:solve the lower tap, for max. 15V when the upper and lower taps are unloaded and to supply 0.571A without dropping below 10.5V when the upper tap is fully loaded. Hint: calculate the voltage at the upper tap due to ONLY the radar full load current (no radio current) and use that when calculating the required lower tap Thévenin resistance.Conclusion:Its already obvious that the divider is going to be dissipating over one KW with both loads off, which is going to make a very effective wheelhouse heater, so the boat had better be fishing for Alaska king crab or similar where you *NEED* the heat!I've got another rule of thumb for you: If you are attempting to draw over 1mA from a potential divider tap your design probably *SUCKS*!* extremely unrealistic for a 12V CB radio unless its receive only, but we already suspected the instructor is a pig-ignorant ass-hat!

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 28, 2021, 04:25:22 pm
content: Thank you so much! I do not understand the "R1||(R2+(R3||Rload1))" part, what does the "||" mean?I was concerned that I had understood nothing of what I have tried to learn because what he is saying is somewhat conflicting with what I think I know.I thought how could you get 10% loss there, would that not be insane to get 20W trough that resistor!?

author: Ian.M
author_rank: Super Contributor
timestamp: November 28, 2021, 04:36:53 pm
content: ||is (or maybe was, as I think it was a USENET sci.electronics 'thing) 'in parallel with' used due to the limitations of describing circuits in a text-only medium. You might describe a typical scope input as1Meg||20pFand I'm using it here as an operator invoking the resistors in parallel formula. i.e. if used in the expressionRpar = Rx||Ryou'd expand the RHS to eitherRx*Ry/(Rx+Ry)or1/(1/Rx+1/Ry)

author: BeBuLamar
author_rank: Super Contributor
timestamp: November 28, 2021, 04:52:25 pm
content: Voltage dividers are generally used for very light load like transistor bias or speed reference for a motor drive. It's not good to supply power to any significant load. There are many other way like switching or even linear voltage regulation. But if you have to do it with only resistors then I would skip the resistor parallel with the load. It's useless unless it actually draws a lot more current than the load in which case way too much energy is wasted but does make the voltage stable when the load is varying.

author: fourfathom
author_rank: Super Contributor
timestamp: November 28, 2021, 05:00:18 pm
content: To emphasize what has already been said, that voltage divider is a terrible way to provide power to the loads in that example. You can calculate the mathematically correct resistor values, but nobody who know what they are doing would use such a circuit as the power loss and dissipation would be ludicrously large.An appropriate use of the voltage divider is to provide specific voltages to low-power loads. And that 10% business is just a rule of thumb, not particularly useful in most situations and can be ignored in most real-world cases where the loads are stable.

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 28, 2021, 05:02:58 pm
content: Thanks for the great explanation.I do just wonder if it is "smart" to run both 12V and 24V in the same circuit, if you have a 200W 24V load and an 8.33W 12V load, is it not better to run it separately with transformer and components build for that purpose, instead of budging it all on a resister?

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 28, 2021, 05:06:47 pm
content: I have only once used a voltage divider to a high load, but it was to build a sort of own for some components, did later change it to a transistor circuit.

author: bdunham7
author_rank: Super Contributor
timestamp: November 28, 2021, 05:15:31 pm
content: Quote from: FriedMule on November 28, 2021, 05:02:58 pmAs said previously, this is a terribly stupid example of an impractical circuit. I was astonished to realize, as I watched a bit of the video that you linked, that this seems to come from a textbook of some sort. I have stated before and will again in the future that there is no reason that you cannot teach the principles of electronics with actual, practical circuits. Teaching them with awkward, idiotic examples is just the result of laziness and lack of imagination and it confuses students.

author: Ian.M
author_rank: Super Contributor
timestamp: November 28, 2021, 05:21:54 pm
content: Its not smart at all, hence my suggested rule of thumb:If you are attempting to draw over 1mA from a potential divider tap your design probably *SUCKS*!The further idiocy was the instructor said the 48V was from a telecom power supply. -48V (with respect to Earth ground) is a common supply voltage in the telecoms industry inherited from the days of copper POTS circuits and Strowger electromechanical exchanges. However you'd never find it in anything much smaller than a cruise liner, and then only for an internal phone system.An electric powered boat might well have a 48V battery system as its a nice voltage for propulsion motors - keeps the current down while not having the electrical safety issues of exceeding SELV voltages. However you'd then have a SMPSU, probably float charging a 12V or 24V battery to provide backup power for emergencies, supplying your navionics, comms and 'house' circuits. You might also have one or more 20V-30V in 13.8V out buck converters to supply '12V' equipment if your main instrument &house supply was nom.24V. You certainly wouldn't have a power resistor divider - even in the days before power transistors it would have been a rotary converter (motor-generator set).

author: polarKaung
author_rank: Contributor
timestamp: November 28, 2021, 05:26:55 pm
content: Quote from: Terry Bites on November 27, 2021, 04:23:22 pmSo, to my basic understandings, Rx needs to be at least ten times that of R3 while Ry needs to be ten times that of combination of R2, R3 and Rx(so 10R3)?

author: Ian.M
author_rank: Super Contributor
timestamp: November 28, 2021, 05:50:59 pm
content: That's backward! Rx (Rload1) is what it is, and so is Ry (Rload2). You need to calculate R1, R2, and R3.*IF* you were going to use the divider current > 10x load current rule of thumb, and also design for nominal voltages at no load:Ry = (24V2/200W) = 2.88 ohms.=> R1=0.288 ohms, and so is R2+R3 as Vload2 is half Vsupply.As Vload1 is half Vload2, R2=0.144 ohms, and so is R1However that gives 83.3A through the divider no load and 4KW quiescent dissipation! That's not a viable power supply, its a large electric heater.Note that the 10x 'rule of thumb' gives a stiffer, higher standing current divider than my assumptions in reply#8.

author: BeBuLamar
author_rank: Super Contributor
timestamp: November 28, 2021, 06:06:41 pm
content: Quote from: FriedMule on November 28, 2021, 05:02:58 pmIf you have a 48V power source like a 48V battery and want to power the 2 loads in example I would use 2 of thesehttps://www.amazon.com/Adjustable-Programmable-Step-Down-Regulated-Converter/dp/B01N0VKTC2/ref=asc_df_B01N0VKTC2/?tag=hyprod-20&linkCode=df0&hvadid=241973970700&hvpos=&hvnetw=g&hvrand=3287969246773027355&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9026816&hvtargid=pla-399690329417&psc=1Very little wasted power and the voltages are steady when load current draw changes.

author: Ian.M
author_rank: Super Contributor
timestamp: November 28, 2021, 06:25:04 pm
content: That could be problematic as IIRC those have low side current shunts, and as both the radar and radio are likely to have other ground connections, the buck modules current shunts could get shorted out, resulting in no over-current protection and possible damage to the buck modules. Also good luck running a 200W 24V radar (8.3A) from a 5A buck module!

author: fourfathom
author_rank: Super Contributor
timestamp: November 28, 2021, 06:42:19 pm
content: The problem here isn't how to power those 24V and 12V loads from a 48V supply. If it were, the answer would likely be to use appropriate switching regulators.No, the problem is how to design a voltage divider. Unfortunately, the example is an impractical one, and this leads to unnecessary confusion. Let's help FriedMule (the OP) with voltage dividers. Actually, I think there is enough here in the replies that they can figure it out. If additional examples are needed, we can provide them.

author: BeBuLamar
author_rank: Super Contributor
timestamp: November 28, 2021, 08:09:44 pm
content: Quote from: Ian.M on November 28, 2021, 06:25:04 pmI don't think there would be a problem with grounding but yeah I would need a bigger unit for the radar. But that I would do to power 24 and 12V devices from a 48V power source.

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 29, 2021, 09:32:39 am
content: I know this is a large question when we are talking about the length of the video, but this voltage divider is not the only strange thing I have stumbled upon, i.e. that current moves through the wire with light speed or as he says "a matter of fact, instantaneously" I took this as a simplification that was to not complicate the basic course. He is also talking about "hole" movement contra electron movement. So in short, is he useful as a beginner course or does he "harm" more than help?

author: Ian.M
author_rank: Super Contributor
timestamp: November 29, 2021, 10:46:35 am
content: Good question. However expecting anyone competent to understand this s--t to wade through a 10H video looking for pig-ignorance is a bit of a stretch, so why don't you provide us with half a dozen timestamped links to bits you feel are dodgy, with a brief summery of what you felt was wrong with each and how long a waffle we have to suffer (i.e. time from start of clip to incidence of foot in mouth). Then those of us who are feeling masochistic enough can tackle each in a coffee break rather than giving up days to it.

author: BeBuLamar
author_rank: Super Contributor
timestamp: November 29, 2021, 12:14:52 pm
content: Why use voltage divider instead of a simple single resistor? If the load draws a constant current or the resistance of the load is constant like in the case of a heater then a single resistor in series would waste less power. But a load may draw varying current and thus the voltage divider would provide a more stable voltage but only if the voltage divider itself drawing significantly more current than the load. So voltage divider can really only be used in case the load draw negligible current.

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 29, 2021, 02:21:15 pm
content: Great idea, but I have almost seen it all, I do just think I will not recommend the video to anybody else and take a look for something better for the next part of my journey. :-)

author: gf
author_rank: Super Contributor
timestamp: November 29, 2021, 03:52:55 pm
content: There exist an infinite number of R1,R2,R3 combinations which result in Load1=12V/6W, and Load2=24V/200W.So an additional border condition needs to be specified in order to get a unique solution.This could be a condition like "among all possible sulotions, choose the solution which has minimum power dissipation", which leads to R3= infinite, R2= RLoad1, and R1= RLoad2|| 2 * RLoad1then.Of couse various other border conditions are imaginable, too, which lead to different solutions.

author: rstofer
author_rank: Super Contributor
timestamp: November 29, 2021, 09:23:47 pm
content: Quote from: FriedMule on November 29, 2021, 09:32:39 amThere's nothing wrong with hole theory. It is a common understanding in semiconductor devices where an electron moves from - to + into a hole leaving behind a hole for another electron to fill. As the electrons move from - to +, holes are apparently moving from + to -.The theory is useful in understanding flow in all electronic devices. It's just accepted by everybody except the US Navy that current flow is + to -. The US Navy wants to deal with electron flow from - to +. That's one of the reasons I don't mention the NEETS program. It just adds confusion to newcomers.https://energyeducation.ca/encyclopedia/Electron_holeHere's more:https://eng.libretexts.org/Bookshelves/Materials_Science/Supplemental_Modules_(Materials_Science)/Solar_Basics/D._P-N_Junction_Diodes/I._P-Type%2C_N-Type_SemiconductorsQuote

author: FriedMule
author_rank: Frequent Contributor
timestamp: November 30, 2021, 05:12:23 pm
content: Thanks, but he does not explain that at all, or at best does he hint some of it. I do not understand why he is not telling about how the normal convention is in reverse direction of reality.