10% current bleed resistor on voltage divider, how?

[FriedMule]

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?

[bdunham7]

Where did  you get that circuit diagram??   

[Terry Bites]

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).

[FriedMule]

The circuit comes from this 10-hour long video about basic electronic: https://youtu.be/nb4ovfwqup8?t=26078

Terry 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.

[fourfathom]

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.

And it will, if you disconnect the loads.  As said upstream, the loads become part of the divider so you need to allow for that.

[Ian.M]

The circuit comes from this 10-hour long video about basic electronic: https://youtu.be/nb4ovfwqup8?t=26078

That 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_Bill

Skip to: https://youtu.be/nb4ovfwqup8?t=26824 where 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

[FriedMule]

Thanks, this makes a lot of sense, does that mean I add like this (math not my strong side):

(1/R3 + 1/Rx)^-1 = T3
T3 + R2 = T2
(1/T2 + 1/Ry)^-1 = T1
T1 + R1

How do I then calculate the optimal bleed, why the same word for two different functions?

[BeBuLamar]

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.

[Ian.M]

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 radio
• 24V 200W, allegedly a radar

Let *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 exceed 15V at the lower tap and 30V at 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 of 10.5v and 21V respectively they will thus draw 6W/10.5V = 0.571A peak^*, and 200W/21V  = 9.52A peak.

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 = 10.1A in 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 ohms Thévenin 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!

[FriedMule]

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!?

[Ian.M]

|| 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 as 1Meg||20pF and I'm using it here as an operator invoking the resistors in parallel formula. i.e. if used in the expression Rpar = Rx||R you'd expand the RHS to either Rx*Ry/(Rx+Ry) or 1/(1/Rx+1/Ry)
 

[BeBuLamar]

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.

[fourfathom]

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.

[FriedMule]

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?

[FriedMule]

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.

[bdunham7]

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?

As 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.

[Ian.M]

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).

[polarKaung]

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. (Attachment Link)

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).

So, 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)?

[Ian.M]

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 R1

However 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.

[BeBuLamar]

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?

If you have a 48V power source like a 48V battery and want to power the 2 loads in example I would use 2 of these
https://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=1

Very little wasted power and the voltages are steady when load current draw changes.

[Ian.M]

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!

[fourfathom]

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.

[BeBuLamar]

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!

I 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.

[FriedMule]

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?

[Ian.M]

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.