What Am I Doing Wrong With This Voltage Regulator?

Discussion in 'General Electronics Chat' started by K-Young, Nov 7, 2009.

  1. K-Young

    Thread Starter Member

    Aug 22, 2009
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    I'm using the following voltage regulator from National, the LM338:

    http://www.national.com/ds/LM/LM138.pdf

    Now I'm trying to get get an output of about 11V-12V from this regulator. My input voltage coming from bridge rectifier is 11V. Now following the formula listed in the datasheet, I get completely wierd numbers for my resistance values. I haven't included any caps in my circuit yet, and I've been playing with potentiomers to adjust the voltage and still no luck :(

    However, I've been thinking if I even need the voltage regulator. I'm running a small DC motor and I intend to control it using PWM. If I'm planning to run it from anywhere to 6V to 11V, is there even a need for a voltage regulator(Since I'm getting 11V out of my bridge rectifier)? Can I just throw in a couple of caps after my bridge rectifier circuit to smooth out the voltage and run it straight to the motor, or is a regulator required no matter what?
     
    Last edited: Nov 7, 2009
  2. SgtWookie

    Expert

    Jul 17, 2007
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    OK, first thing to do is to look at Vref.

    Vref is the voltage at the OUT terminal in respect to the ADJ terminal. Nominally, it's 1.24v; but it could be as low as 1.19v or as high as 1.27v and still be within manufacturer's specifications (it's listed in the datasheet).

    Next, look at the line/load specifications; guaranteed regulation occurs when Iout >= 10mA; minimum Iout is 3.5mA.

    So, in order to be reasonably assured of guaranteed regulation specifications under most circumstances, use a 120 Ohm resistor as R1 (from OUT to ADJ terminals). Unless Vref is < 1.2v, the output will be within the specifications for line and load regulation.

    Note that Vref is the minimum voltage that the regulator can output (ADJ terminal connected to ground) unless you reference the ADJ terminal to a negative voltage.

    Next, go to the bottom of page 3 of the datasheet, and look at the graph labeled "Dropout Voltage". "Dropout voltage" is the difference between the IN and OUT terminals. Note that with a load of 1A at 25°C, the typical dropout is about 1.7v.
    So, if you had an input of 11v, and a load of 1A with the regulator at 25°C, the highest voltage output you could get would be 11v - 1.7v = 9.3v. If your output load requires 5A, your dropout voltage may go as high as 3v, depending on the temperature of the regulator.

    Let's go back to R1 for a minute. If Vref is 1.24v and you've used 120 Ohms for R1, then there will be 10.333...mA flowing through R1. So, for every 100 Ohms of resistance you place between the ADJ terminal and GND, you will get an approximate 1.0333...V increase out of the regulator over Vref - within the limits of the input voltage and keeping the voltage dropout in mind.

    Let's say you used a 500 Ohm resistor. 500 x 10.333...mA = 5.16v, +1.24v = 6.40666...V out.

    Note that this is an approximate number. One item that I have not accounted for yet is "Iadj"; or the current that flows out of the ADJ terminal to ground. For the LM138/LM338, this is nominally 45uA, but may be as high as 100uA. If you're using a 120 Ohm resistor for R1, Iadj has only a small effect on the output voltage. 45uA x 500 Ohms = 22.5mV; so instead of 5.16v, you'd have 5.1825v + Vref for the output voltage.
     
    Last edited: Nov 7, 2009
  3. beenthere

    Retired Moderator

    Apr 20, 2004
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    Page 2 of the data sheet shows that Vref is valid when the difference between the input voltage and the output is at least 3 volts. If you want to regulate 12 volts out, you will need at least 15 volts in.
     
  4. K-Young

    Thread Starter Member

    Aug 22, 2009
    27
    1
    Thanks guys, that answers a lot of my questions, but what about the second part? Can I just skip the regulator and run the the 11V straight from the bridge to the DC motor?
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
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    You should be able to. If it's a 12 volt motor, it may run a bit slow.
     
  6. studiot

    AAC Fanatic!

    Nov 9, 2007
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    A 338 may not be fast enough to supply PWM, depending upon pulse repetition rate (PRR).

    In any event you need to either cut back on your expectations or change your input to the bridge to get at least 14 volts basic supply. Whatever control circuitry you employ will drop at least a couple of volts.

    As a thought have you included a reservoir capacitor between this bridge and the regulator?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    If you do not have a filter capacitor on the output of your bridge yet, you will get a surprise when you install filter capacitor(s); instead of an average 11v output, you will suddenly see around 17v output with no load.

    Right now, you are just seeing rippled DC output. When you insert capacitance, the capacitor will tend to stay charged at the peak of the ripple voltage.

    What current will your load draw, and how much ripple voltage do you want to see?
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Have a look at the attached simulation of a voltage supply with a rectifier bridge, filter capacitor and LM317 voltage regulator wired for 11v output. Click on the attachment, and then click on the image that pops up so you see it full size.

    The LM317 is similar to the LM338, with lower current output capacity.

    D99 and R99 are not necessary for the "real thing" they are just in this simulation so that you can see the rippled DC output of the bridge (A), the blue trace in the simulated O-scope on the bottom. With 60Hz AC sine wave into the bridge, you get 120Hz rippled DC out of the bridge.

    C1 is a 3,300uF capacitor, which is adequate for loads up to about 1A. Notice the ripple on the DC voltage (B), yellow trace.

    This particular simulated regulator has a Vref of about 1.29v; output voltage is about 11v (C), green trace.

    Note that if you are planning on using a 555 timer to generate your pwm, you will need to regulate the voltage down to at most 16v.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    SgtWookie, doesn't Circuitmaker have a .step directive? Most (all) spice-based simulators do. This would allow you to change the cap (call it Cfilt) value from , say, 1pF to 3300uF, eliminating the need for the extra diode.
    In LTspice, I place this directive on the schematic:

    .step param Cfilt list 1pF 3300uF

    Cheers-
    Ron
     
  10. K-Young

    Thread Starter Member

    Aug 22, 2009
    27
    1
    Hmm, from what you've shown me so far, I can save myself a lot of work if I just use a PC ATX power supply. My motor claims to run at around 3-4 amps when at 12VDC, but I did not measure that kind of current at all when I just connected it right after the bridge. I measured around 500mA:confused:. I guess it's due to the varying voltage.

    I think I'll go through the route of the PC ATX power supply, it'll save me a lot of trouble in the long run and it should be more than able to supply the amount of current I need on the 12VDC line.

    As for the PWM, I'm looking at a Pixaxe right now for the PWM.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Ron_H,
    Nope, not in the free "student version" - it has a parameter sweep function, but I haven't fiddled with it much. Just tried to sweep the cap value from 1m to 10m step 1m; didn't work.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    It will be more efficient, too. Google "ATX bench supply" for lots of ideas. However, don't snip off all but one of the ground wires like some some suggest; one ground wire isn't enough if you're drawing several amps.
    Might also be that your motor is loading down your transformer's output.

    Besides that, you'll already have a regulated 5v supply for your Picaxe.

    OK. You'll need to use a logic-level MOSFET like an IRLZ24 or the like to sink current from your motor, or a power Darlington transistor with a resistor on the base.

    Don't forget that you will need to use a diode across the motor (cathode towards +V) to take care of the reverse EMF when the MOSFET turns off. Something like a 1N5401 (3A 50PIV rectifier diode) will do the trick for you.
     
  13. K-Young

    Thread Starter Member

    Aug 22, 2009
    27
    1
    Thanks SgtWookie, you've saved me a lot of time and trouble. I'll let you guys know if I have any more trouble. You've been a lot of help! :D
     
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