Voltage Dividers for voltage regulation

Discussion in 'General Electronics Chat' started by Transatlantic, Apr 13, 2016.

  1. Transatlantic

    Thread Starter Member

    Feb 6, 2014
    35
    0
    (I know there are better ways to create voltage regulators, but I'm going through the basics)

    How do you determine the resistors used for a voltage divider?

    Say I have a 6v supply and I want a 5v output, where my load draws 30ma. For this, the lower half of the divider would need to be 6 times the resistance of the upper half.

    I've read that you want to keep the divider as 'Stiff' as possible. One way to do this is to keep the values low, this way, changes in the load won't affect Vout much. But not too low as to have the supplies internal resistance come into the picture. So I might use for example, a 60ohm and a 10ohm.
    upload_2016-4-13_13-18-41.png

    The problem here is that it's not very efficient. I'm drawing 0.7w, where as my load is only using 0.15w. So I'm wondering how you calculate the optimal values for R1 and R2?

    Here are my thoughts.

    - The load has a resistance of 160ohms
    - The circuit needs to be drawing around 30ma
    - The lower half needs to be 6 times higher than the upper half

    To achieve this, I can calculate it as a single resistor.

    6v / 0.03ma = 200ohm
    As the load resistor is 160ohm, that leaves 40ohm for R1.
    To keep the lower parallel part at around 160ohms, I can set R2 to be something very high, like 100k

    upload_2016-4-13_13-28-33.png


    This is now more efficient. Instead of drawing 0.7w, it's now 0.18w

    Is my reasoning/logic correct? .. is there a more accurate way to do this?

    This raises another question for me though. Why bother with R2 in the first place? why not keep things simple and have R1 and RL in series?

    upload_2016-4-13_13-32-22.png

    I assume the answer is do with the first circuit being better at handling changes in the load resistance, at the cost of excess power draw? So the circuit above would only ever work if the load had a very stable resistance.
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
    3,242
    619
    If you want to use a resistor divider, current in the divider needs to be at least 10X the load current. A 30mA load would require 300mA in the divider.

    Total divider resistance R=V/I=6V/0.3A=20Ω
    Vout = Vin*R2/(R1+R2)

    Too much power wasted; use a more efficient method.
     
  3. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,797
    1,103
    You can use a divider with a buffer stage to reduce the power wastage, but there is a small drift in the buffer output voltage with temperature. The buffer allows load current to be changed over a reasonable range without changing the output voltage as much as would happen without the buffer.
    Divider+buffer.PNG
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,534
    1,251
    Nope. 5 times.
    Conceptually correct, except for the math error.

    And as you have determined, the result is not "regulated". That is, one output characteristic is not (relatively) independent of the other. A voltage regulator holds the output voltage constant over a range of output current values. What you have is a voltage divider.

    ak
     
    Last edited: Apr 13, 2016
  5. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,789
    You ask about how to find "optimal" values but for that question to have any meaning, you have to define what the metric is by which you compare one solution to another. As you've noted, you have a tradeoff between load regulation and power efficiency. But which of those need to be given more weight depends on what your needs are. The place to start is with your specifications: (1) how much can your load change by and how much are you willing to accept your load voltage changing by? and, (2) what is the minimum power efficiency that you can accept? Both of these place constraints on the resistor values in your divider. If you can meet both sets of constraints with a single set of resistors, then you have a viable circuit that meets your needs (may not be the best circuit possible, but if it's good enough, then it's good enough). You may well find, however, that you can't meet both sets of constraints at the same time. If that's the case, then it's time to look for a different circuit.
     
Loading...