2 volts from 5

Discussion in 'General Electronics Chat' started by jgessling, Oct 25, 2009.

  1. jgessling

    Thread Starter Active Member

    Jul 31, 2009
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    I have a regulated 5v source that I need to get 2v from to control an oscillator. My first thought was to make a voltage divider like on the right side of my attached diagram (ltspice). I never was comfortable with this since about 8ma goes through R4 for no purpose. So I thought to use a voltage regulator, like an LM317. That's on the left side of the diagram. That gets the two volts and only 1ma goes wasted through R2. I have a couple questions.

    1) on the right, if there is some path with resistance from Vout to ground won't that mess up the voltage divider and change the 2v?
    2) on the left, there are lots of combinations of R1/R2 that will get 2v output. I chose one with a large R2 value to lessen wasted current, are there other reasons to choose these resistors?
    3) on the left is the 2v going to be more steady because there is no effect like I wondered about above?

    Thanks, Jim
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Do you know how to calculate the Thevenin Equivalent of a voltage divider circuit? After picking R1 and R2 in the divider, the Thevenin equivalent resistance is the parallel combination of R1//R2= R1R2/(R1+R2)=240*360/(240+360)=144Ω. Only you can tell if your 2V oscillator will do if it is fed from a 2V volt source which has such a high output impedance? I'm guessing that it will behave badly...

    By comparison, the LM317 will have an output impedance of a few thousandths of an Ω.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, it will.
    You would be putting the load resistance (Rload) in parallel with R4.

    Yes. In the case of the LM317T, R1 should be 120 Ohms, unless there will always be some load current. In the case of the LM317L, R1 should be 240 Ohms.
    An LM317T needs at least 10mA current flowing from the output to provide guaranteed regulation. Vref (measured from OUT to ADJ) is kept at a nominal 1.25v by sourcing current from the OUT terminal.

    Vref is typically 1.25v, but it may be anywhere from 1.2v to 1.3v, and still be within manufacturer's specifications. If R1=120, 1.2v/120 Ohms=10mA; so the minimum 10mA current flow is satisfied.

    The LM317L is a 100mA max current regulator available in a TO-92 package. It only requires 5mA current flow from the output to provide guaranteed regulation; this is why you can use a 240 Ohm resistor for R1.

    However if you know that your load will always draw a certain minimum current, you can subtract that from the R1 current requirement.

    Note that there is also a small amount of current that will flow out of the ADJ terminal; nominally 50uA, but may be anywhere from 20uA to 120uA. This usually has little effect when using low-value resistors, but becomes increasingly significant when higher value resistors are used for R1 and R2.
    The LM317 is an active voltage regulator. It will source current from the OUT terminal to keep Vref at a nominal 1.25v, up to about 1.5A current.

    You must have a 0.1uF capacitor on the output to reduce the possibility of oscillation. You should also have a capacitor on the IN terminal to ground if the wiring to the voltage supply is more than a couple of inches long.
     
  4. RimfireJim

    Member

    Apr 7, 2008
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    MikeML,
    I understand the importance of having a high input impedance for a load so that it doesn't unduly load down a supply. Can you help me understand the importance of the supply output impedance to the function of the oscillator? Is your point that the oscillator won't really see 2V because of the voltage drop across the 144Ω Thevenin resistance due to whatever current the oscillator is trying to draw?

    Not challenging your statements at all, just trying to get a better understanding as a mech engr who dabbles in electronics.
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    In addition to the information contained in the earlier post regarding the LM317 (and all 3-terminal linear regulator for that matter), it is important to consider the power dissipated by the LM317 in the design. You will notice that whatever voltage you set the output of the LM317 to provide, the difference between the input voltage to the regulator and the output voltage of the regulator is dropped across the regulator. That voltage difference mulitplied by the current drawn by the load yields the amount of power that is basically being wasted in the form of heat built up in the LM317. A quick calculation can permit you to determine if this heat is going to place the 3-terminal regulator in danger of overheating. Most 3-terminal regulators (including the LM317) have internal heat sensing that will shut the regulator down if the heat gets too great. If you ever have an LM317's output drop to zero unexpectedly under load, it is likely that its internal overheat protection circuitry has been triggered. The remedy is to provide an adequate heatsink.

    hgmjr
     
  6. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Yes, exactly that. For example, look at the following: I load the voltage divider with a current that varies from 0 to 1mA, much as the oscillator might do. Note that the voltage varies from 2V down to 1.855V, which is very poor regulation. The load current is ~1/8 of the current that is shunting through the voltage divider. To improve the regulation, you would have to increase the ratio of the current to ground through the divider to the load current to a hundred to one or so...
     
  7. jgessling

    Thread Starter Active Member

    Jul 31, 2009
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    The oscillator in question is a 74s124 VCO that I need to put 2v on the RNG and FC pins. With a 470uF external capacitor I get about 1hz, this is for breadboarding a VFD tube clock. For the real thing I will need a more accurate time source but this will do for now. The datasheet says that the 74s124 will take a max of about 20uA on the 2v. I'm thinking that would be the equivalent of a 100,000 ohm resistor at 2v. Putting that resistor in for a load does drop the divider voltage a bit, like to 1.997v.

    I wonder if that load will change as MikeM suggests, the datasheet is mum, maybe this will be motivation to fix my scope. It's bothered me that in the books they show a voltage divider as so simple, then when I went to really make one I started thinking how the load is going to change everything so maybe it's not so simple. Anyway, thanks all for helping me think about these circuits.
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You are correct. This is one of those threads where it would have been much simpler if you simply told us what your application is in your original post. Of course, you would have missed the introduction to Thevenin equivalents...:)
     
  9. RimfireJim

    Member

    Apr 7, 2008
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    Thanks, MikeML - excellent explanation.
     
  10. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    The LM317 has a minimum operating current between 5 and 10 mA so that won't help you.

    You can use a high impedance resistive divider (like resistors in the 100k range) and buffer that with an LM358 or other cheap op amp.
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    With a 20uA max load, the original voltage divider should work fine if the resulting wasted current is acceptable. Even scaling the divider up by a factor of 10 would only result in a 30mV max error, still less than that potentially caused by resistor tolerances.
     
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