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I am now learning about voltage dividers and bleeder current.I read article on this website http://www.tpub.com/neets/book1/chapter3/1-35.htm

there is example of voltage divider,it goes like this "Figure 3-64 is used to illustrate the development of a simple voltage divider. The requirement for this voltage divider is to provide a voltage of 25 volts and a current of 910 milliamps to the load from a source voltage of 100 volts"

Its simple two resistor voltage divider with single load,it says 910 miliAmps and 25 volts so I calculate it needs 82 ohm + 28 ohm resistor for total resistance of 110 ohms and voltage between resistors of 25 volt like the requirment needs.

Then I look at the curcuit on that website and there is 75 ohm and 25 ohm resistor,so that is 1 amp and not 910 miliAmp but then I remember it said something about bleeder current and that I should give the circuit more current than it needs,about 10% of the load rule of thumb.

Ok,that make sense,but then the website says that actualy since the resistance of load creates parallel resistance with that bottom 25 ohm resistor,we need to increase it to 275 ohm from 25.

Now the fact that load creates parallel circuit with the second resistor so the second resistor value should be such value that when connected to the load it will create total resistance of 25 ohm is simple,but here is something I dont understand,where did he come up with 275 ohms?!

Not a single time did he mention what resistance the load is! I read that load resistance is usualy in the mega-ohm range.When I calculated total resistance of 25ohm resistor in parallel with 10 mega ohm resistor I got 24.9999! That is so close that it makes the parallel effect almost non existant,doing what he says is correct way to do it would give the load 78 volt instead of 25,and the current to load will be 285 miliAmp instead of 910 required by the load!

Again I want to stress and remind you that at no point in the article does it mention resistance of load,it would only be correct if the load was 28 ohm,since 28 ohm in parallel with 275 ohm is 25 ohm.I ask,is the knowledge that load is 28 ohm somehow so common,that it isnt even worth mentioning when writing articles for absolute beginers? I dont know what to think about it,do I not understand something properly,or does the author of that article truly think that potential reader would automaticly know that the load is 28 ohm so he thought that it doesnt even need to be mentioned?

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The load is defined to be 27.5ohms. That's in figure 3-64 of your link. It doesn't say it anywhere, but it does give you the voltage and the current.

This is the problem with voltage dividers supplying any significant load, they have no voltage regulation. If the load changes, then the delivered voltage changes.

The extra 275ohm resistor in shunt with the load is one of those things I call 'unnecessary and insufficient'. It's not sufficient to regulate the voltage, and it's not necessary to get the right voltage given that you have to muck about calculating correct resistor values anyway. Your 82ohm would have been just fine.

What the article should really say is that 'you would never drive a load this low with a voltage divider, but if you did, and if you wanted to divert 10% of the load current round the load just for the heck of it, then these are the resistor values you'd use.'

Now if you diverted (say) >100x the load current round it, then you'd have reasonable regulation, but you'd need your load to be very low current to be even remotely efficient. You did the sums for a 10M load.

Not every tutorial on the net is actually noob friendly, not necessarily even if it claims to be.

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  • \$\begingroup\$ ahhhhhhhhhhhhhhhh ( facepalm ) Thank you Neil_UK... well,that could have hit me,but I still think its cruel trap for noob like me. \$\endgroup\$ – wav scientist Mar 7 '17 at 10:13

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