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Topic: How potential/voltage dividers compare to voltage regulators? (Read 2 times) previous topic - next topic

Bobu

Hello Arduino community,

Me again with what I'm sure is either another basics question or a misunderstanding on my part.

As I understand it, linear voltage regulators will take a higher voltage input and then output a specific voltage.  For instance 9v input can be clipped to 5v with a 5v regulator.  The excess voltage is converted to heat to reduce the voltage which means you have to attach a heatsink depending on how many watts you're burning off.

From what I read, a potential divider set up does exactly the same, but with 2 resistors instead of a regulator.  It also doesn't need a heatsink, as it doesn't generate any heat to clip off the voltage.  Is that right?

If it is correct, this sounds like a cheaper way to take a higher input voltage and output a lower voltage (in the sense it costs 2 resistors which costs maybe 50p, while a regulator costs about £1.50 + some capacitors to smooth out the Vin Vout).  Is the voltage regulator just easier to setup for reducing voltage (as you don't have to work out the 2 resistor values) or is there something that the regulator does that a potential divider set up won't do (where the potential divider setup is 2 resistors in series)?

Does a voltage regulator allow you to draw a higher current for instance?

Thanks, hopefully this isn't a completely ridiculous post :P

Nick Gammon

The load on the divider will alter the amount dropped, according to Ohm's Law.
http://www.gammon.com.au/electronics

Nick Gammon

Read here, for example:

http://www.diystompboxes.com/smfforum/index.php?topic=69911.0

Try other searches for "voltage divider vs voltage regulator".

eg.

http://www.wd5gnr.com/basiccir.htm

Quote
Practical Voltage Division

Don't fall prey to the temptation to use a voltage divider instead of a voltage regulator. Voltage dividers are totally dependent on their supply voltage, so don't try to "regulate" 12V to 5 with a divider. However, there are cases where you might want to use a divider. For example, perhaps you have a 5V A/D converter, but you need to measure 0-10V with it. Of course, you will lose precision, but maybe that's OK.


http://www.gammon.com.au/electronics

KeithRB

Wrong.

A voltage divider would waste just as much heat as a regulator. The main difference is that the regulator has a low output impedance. That is, the output voltage does not change very much if the load changes. Your voltage divider has a relatively high output impedance so that if the load resistance changes, the voltage will change.

A voltage divider will only work for low-current, very constant loads (like the base resistance of a transistor).

Far-seeker

#4
Oct 24, 2012, 11:47 pm Last Edit: Oct 24, 2012, 11:53 pm by Far-seeker Reason: 1

From what I read, a potential divider set up does exactly the same, but with 2 resistors instead of a regulator.  It also doesn't need a heatsink, as it doesn't generate any heat to clip off the voltage.  Is that right?


As others have noted, current flowing through any non-superconductor will generate at least a little heat due to innate resistance.  In fact, an easy way to make a small  electric heater is to wire-up low value resistors in series and run a few amperes of current through them.

Also there's the fact that a voltage divider has a fixed proportion the input voltage is reduced by, not a fixed value it reduces the input voltage to.  So if for some reason the 9 VDC is increased significantly, e.g. to 12 VDC, the output voltage will be too high to power an Arduino.  By contrast a linear regulator will give you a known output for any voltage within the range of rated input voltages.

Edit: I forgot to mention that the resistors in the heater example would be 0.5 or 1 W resistors.  :smiley-red:

DVDdoug

#5
Oct 25, 2012, 12:12 am Last Edit: Oct 25, 2012, 12:15 am by DVDdoug Reason: 1
Quote
The excess voltage is converted to heat to reduce the voltage which means you have to attach a heatsink depending on how many watts you're burning off.
Actually power (Watts) is converted to heat.   Power is calculated as Voltage x Current.   If you are not drawing any current, there is only a tiny amount of current used to "power" the regulator, and the regulator will remain cool.

A voltage divider will typically draw more current and generate more heat, given the same amount of current through the load, because you have current through the "bottom" resistor in the voltage divider, as well as the current through the load.  If you don't want the output voltage to change very much when you add (or change) the load, you have to use lower-value resistors and draw more current through the voltage divider.   Typically, you want 10 times as much current through the voltage divider as through the load.  That's fine with "signals", but it's very-bad for power supplies!

If you "cheat" and use the load as the bottom resistor, the top resistor will generate the same amount of heat as a regulator.  This can be done in rare cases, such as with an incandescant lamp, but it usually wastes too much power and generates too much heat.

And if the load changes (such as an LED turning-on, etc.) the voltage will change.  (The voltage is unregulated.)

A voltage regulator has two kinds of regulation...   There is line regulation, which means that you'll get 5V out of your 5V regulator, even when your 9V battery gets weak and drops to around 6V.   And, there is load regulation, which means that the 5V doesn't change when the load current/resistance changes (as long as you stay within the specs).

A switching regulator is more complicated than a linear regulator, but they can be almost 100% efficient, and generate very little heat.  You generally get more current out than you feed-in.  So the power-out (voltage X current) is nearly to the power-in (voltage x current) and very little power is wasted as heat.

fungus


Thanks, hopefully this isn't a completely ridiculous post :P


You've missed something important...whatever device you connect to the center of the resistor divider will alter the resistances of the divider and therefore change the voltages. eg. If you connect something with a low resistance which goes to ground it will be like shorting out the lower resistor.

Resistor dividers are only really useful when the thing you connect to the middle has a very high resistance, ie. it can't draw any amps.

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

Bobu

Thanks guys, this post has been really educational and I'm sure you guys have saved me from burning my house down!  Looks like for my purpose in this particular circuit, I'll be using a regulator instead of a divider as the input may be 7.5v-9v and it will be reduced to 5v.  There will be 6 servos in the circuit so potentially the current they will try to draw could be a few amps depending on the load.

@DVDDoug: I've heard switching regulators are more efficient than linear regulators (most posts claim 84% efficiency whilst linears are only 50%), but had some trouble understanding how you would set one up in a circuit.  My understanding is you should add some capacitors to the Vin and Vout for a linear regulator to help smooth out the voltage, would you need to do the same for a switching regulator (I was even wondering if I could remove the Vin cap all together and just have one on the Vout because of the way it works)?

A quick search turned this up, so I was thinking of ordering this as it looks pretty affordable in some shops:
http://pdf1.alldatasheet.com/datasheet-pdf/view/8660/NSC/LM2576T-5.0.html

Of course, I have to figure out how to wire it all up before I order it :)

Bobu

#8
Oct 25, 2012, 01:25 am Last Edit: Oct 25, 2012, 01:26 am by Bobu Reason: 1
Just realised the datasheet shows a typical set up for setting up a switcher regulator :S

I did have some questions though about why you set it up this way (sorry for going off the original topic slightly).  Could someone confirm my understanding of why it's set up this way and answer some questions on it (again sorry for having to ask questions that I assume are basics, but my background is on the software side so I'm pretty much a beginner when it comes to hardware):-

  • Pin 1 is the input voltage.  The capcitor that's 100?F, is to help smooth the input voltage

  • Pin 2 is the output.  It's connected to an inductor that stores the output charge?  The capacitor that follows (1000?F) is to smooth out the output voltage?  I'm not sure what the last symbol represents (solid black triangle pointing upwards with a sine wave line over the top)?

  • Pin 3 is ground

  • Pin 4 monitors the current amount 'charged' so it knows whether is should activate or deactivate pin 5

  • Pin 5 controls whether it needs to charge or shut off the voltage depending on the state of whether it has charged 5v



Cheers

Bobu

Just wanted to follow my last post up with this link:-
http://www.radio-electronics.com/info/power-management/switching-mode-power-supply/regulator-fundamentals.php

Helped me understand why you set up a swithing regulator in this way.  Really helped me understand the setup, so hopefully it helps someone else.

Now all I need to do is buy the components to give all of this a whirl as a linear regulator in my circuit may generate a lot of heat and I don't particularly want to add a bulky heatsink to it :S

dhenry


Bobu

Sorry if that wasn't clear.  By "why you set it up this way" I meant why would you set up a circuit like it says in the switching regulator datasheet (the inductor, capacitor etc...).  I wanted clarification on what the additioanal components roles were. 

This isn;t really a valid question anymore as the link I posted afterwards cleared up my questions for me.

MarkT


Just realised the datasheet shows a typical set up for setting up a switcher regulator :S

I did have some questions though about why you set it up this way (sorry for going off the original topic slightly).  Could someone confirm my understanding of why it's set up this way and answer some questions on it (again sorry for having to ask questions that I assume are basics, but my background is on the software side so I'm pretty much a beginner when it comes to hardware):-

  • Pin 1 is the input voltage.  The capcitor that's 100?F, is to help smooth the input voltage

  • Pin 2 is the output.  It's connected to an inductor that stores the output charge?  The capacitor that follows (1000?F) is to smooth out the output voltage?  I'm not sure what the last symbol represents (solid black triangle pointing upwards with a sine wave line over the top)?

  • Pin 3 is ground

  • Pin 4 monitors the current amount 'charged' so it knows whether is should activate or deactivate pin 5

  • Pin 5 controls whether it needs to charge or shut off the voltage depending on the state of whether it has charged 5v



Cheers


The inductor stores energy as magnetic field (the energy is associated with current flowing).  Capacitors store energy associated with a voltage.  Pin 5 is the on/off control for whole regulator.  Pin 4 senses output voltage.  Pin 2 is the switched output (there's a MOSFET or transistor on the chip switching the current from the input at a high frquency).

When the switch is on current flows from input (thus from Cin) to the output (Cout), when the switch is off the current still flows (that's the inductor's job) but has to go through the schottky diode D1 instead.

When the switch is on the voltage across the inductor causes the current to rise, when the switch is off the reverse voltage across the inductor causes the current to fall (but not reverse).  If you switch fast enough the current through the inductor is mainly DC (with ripple) but sometimes the current comes "magically" from ground via the diode rather than from Cin.

This is because of the stored energy in the inductor being released as the current falls.
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