Voltage Regulator Clarification

Discussion in 'General Electronics Chat' started by deefactorial, Aug 7, 2008.

  1. deefactorial

    Thread Starter Active Member

    Jun 11, 2008
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    I found a external power supply that I can purchase from digikey. But the only one I could find with the least amount of power was a 10W 5v 2A power supply. The one that is used on the Explorer 16 development board from which I am using the schematics from has a 9v 0.75A power supply.

    I am curious what modifications I need to make to the schematic diagram to make this work. I have attached the schematic diagram that was given to me for the explorer 16 development board. I have looked up the spec sheet for the LM1117 Voltage Regulator. It says that the max current that it can handle is 1500mA and it is rated for 800mA. Do I need to add a resistor before the LM1117 so that I do not exceed the max rated current.

    The specs also mention that it is important that I use 10uF Tantalum Capacitors. before and after the LM1117. How is the performance effected by using one Ceramic 0.1uF and one 47uF Tantalum Capacitor on both sides.

    Any replies would be much appreciate.:)
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    With this configuration you will take about 1.25 Volts for both outputs. You have to use resistors to determine the output voltages to the desired values. Check this datasheet for more information
    http://cache.national.com/ds/LM/LM117.pdf

    Also, i suggest you to replace capacitor 29 with a 1000uF electrolytic capacitor
     
  3. deefactorial

    Thread Starter Active Member

    Jun 11, 2008
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    I believe that with the LM1117MP-5.0CT-ND chip and the LM1117MP-3.3CT-ND chip that I am using for this application have the output voltage fixed at 5.0v and 3.3v respectively. So there is no need for external resistors to adjust the output voltage. I may be wrong though and please correct me if I am.

    So if both LM1117 chips take 1.25v from the source then 2.5v remain unused. Do I need to dissipate this current through a resistor or will the chips just draw as much current as they need.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    If you read the datasheet for the LM1117, you will discover that the regulator has 1.2v dropout at it's rated current. That means for a 5v supply, you need a minimum of 6.2v input. For your supply, you could use a "wall wart" type supply rated for 1.6A (1600mA) @ 6.2V to 10V.

    No, the LM1117 has an internal current limiter. If you added a resistor prior to the LM1117, you would likely get wildly unregulated voltage output.

    Also, your schematic shows the adjustable type LM1117, but used as the fixed-type LM1117. You should have the voltage specified in the part number, as LM1117T-5.0 and LM1117T-3.3. If you planned on using the adjustable type (LM1117T-ADJ) then you will need to use a voltage divider network on the output; a pair of resistors, R1 from OUT to ADJ, R2 from ADJ to GND.
    Vout = Vref(1+ R2/R1)+Iadj(R2)
    1.20 >= Vref >=1.29; 1.25 typ
    Iadj = 60uA = 0.00006A

    The datasheet is a good guide to follow. If you don't follow the datasheet, you may not receive optimal performance from the product.

    As far as the 0.1uF caps - you should have one across Vcc/GND or Vdd/GND for each logic-level or low-power analog IC in your circuit; if a particular IC is a power handling IC, you'll need additional capacitance.
     
    Last edited: Aug 7, 2008
  5. deefactorial

    Thread Starter Active Member

    Jun 11, 2008
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    I found a wall mount 9v 1.66A external power supply. So this is a little closer to the specs of what I'm looking for.

    I changed the 47uF capacitors for the 10uF capacitors the spec sheet recommended. Do I still need the 1000uF capacitor? Since the Voltage and the amps matches closely to the specs that I need that I could go with a smaller capacitor for the external power supply.

    The power supply specs mentions that it has 150mV ripple voltage at full load what is the formula for calculating how to compensate for ripple voltage in a circuit.

    From all the things I have read so far. My understanding of what a capacitor does is that it is a temporary "storage container" for an electrical current such that if the current drops it maintains the continuity of the current for as much as it has stored within its "container". When it does this it has the effect of calming turbulent currents that have ripples in them. The size of the "container" determines the size of the ripples it calms. Thats why you put a 0.1uF capacitor right before the Vcc or Vdd of the IC so that the current that is going into the IC has the small ripple voltages taken care of. Does this make sense or am I way out in left field here. Is there any resources or tutorials I could read that would clarify any ambiguities I have about this ?

    Thank you so much for clarifying things for me. Its so hard to navigate the unknown when I don't even know what I'm looking for. Thats why this forum has been such a great help.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    That should be fine.

    It kind of depends upon how long your power lead is.
    1,000uF caps tend to have a pretty high internal resistance. That's why you need the smaller caps in parallel.

    Take a look in the datasheet in the Applications Notes. You'll find that for increased ripple rejection of a 120Hz signal, with an R1 value of 124 Ohms, you should have a cap from the ADJ terminal to ground that is >11uF. Note that for the LM117/LM317, a typical value for R1 is 120 Ohms. This guarantees that there will be a 10mA load on the output at all times, which is required to achieve the guaranteed regulation specification. Note that if you will have a constant load somewhere on your circuit that you are powering, you can subtract that load from the minimum guaranteed regulation requirement when you are calculating a resistor value to use for R1. For example, if you will always have a 5mA load, you can likely use a 240 Ohm resistor for R1.

    You're basically correct. However, large capacitors generally tend to have a high ESR (equivalent series resistance) and high parasitic inductance, so they are ineffective at removing high frequency noise components. That is why smaller caps are used in parallel; they have relatively low ESR and low parasitic inductance, so they are very effective at removing the high frequency components.
    [/QUOTE]
     
  7. AchMED

    Active Member

    Aug 5, 2008
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    I think your confusing impedance with ESR. Aluminum electrolytic's for a 1mF cap we are talking aluminums the esr decreases with can size. For bypassing yes you want generally small size low capacitance capacitors. This typically means MLCC's 0403,0603 or 0805's.

    For a given ripple you can read up TI' slup 060 this will also show you how to estimate rms ripple current for your bulk cap ,for such a low power this shouldn't be a problem. You can also use spice and use a constant power load to check your ripple over various line and load conditions.

    Code ( (Unknown Language)):
    1. http://www-s.ti.com/sc/techlit/slup060.pdf
    Code ( (Unknown Language)):
    1. http://en.wikipedia.org/wiki/Electrical_impedance
     
  8. deefactorial

    Thread Starter Active Member

    Jun 11, 2008
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    It looks like the cord length is 1000mm, I may be wrong the specs are hard to read. Part # T990-P5P-ND

    I have replaced the 1000uF cap with a 330uF 220uF 100uF and 47uF cap, see schematic for reference.

    the ESR of all the caps are:

    0.1uF - N/A - ceramic
    10uF - 500mOhm - tantalum
    22uF - 400mOhm - tantalum
    47uF - 28mOhm - Electrolytic Alum
    100uF - 80mOhm - Electrolytic Alum
    220uF - 300mOhm - Electrolytic Alum
    330uF - 14mOhm - Electrolytic Alum

    total Resistance = 0.014 + 0.3 + .08 + .028 + 0.5 + 0.4 + 0.5 = 1.822Ohm for each 5.0v line and 3.3v line.

    Now these are some of the most expensive Caps so I figure they are the higher quality ones. but that is not always the case. If I should change the type or the ESR value of any of the caps please let me know.

    I added a 22uF cap on the ADJ pin to help regulate noise on the line. From the information that I gathered from the specs if you add a cap to the ADJ line you get less noise on the line. Its not necessary for for it to function but it helps. if you add a cap larger than 25uF you need to add a protection diode. So I choose to add a 22uF tantalum cap. Thank you SgtWookie.

    On the LM1117 spec sheet it says R1 and R2 are integrated into the device for Fixed voltage ICs. So I don't think I need to calculate R1 and R2.

    I read the articles that AchMED posted, I'm still wrapping my mind around the Electrical Impedance article, when I read it I get a visual of a stone on a calm pond. The other slup060 article, there is a lot of electrical engineering jargon I'm not familiar with to extract the formula I need. So what is RMS and how do I calculate what capacitance I need to compensate for 150mv = Voltage Ripple. That article looks like it is explaining how a power supply creates a ripple voltage.

    Spice is a circuit simulator right. I will have to look into that. looks like Geda has some derivatives of Spice I could check out. What has your experience been with circuit simulation?

    Thank you all for taking the time to simplify the complexities of this into something I can understand.
     
  9. AchMED

    Active Member

    Aug 5, 2008
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    If I understand you correctly you have a wall wart 9VDC @ 1.66A. You don’t need to worry about having that much capacitance at the input to your regulator. The wall wart already has bulk capacitance. When it says 9V @ 1.66A that’s what it means when the load draws 1.66A the voltage is 9V with anywhere between 10 to 20% ripple. If the load draws less current the output voltage increases hence the term unregulated supply. They recommend a 10uf tant at the input of the regulators maybe also put a 0.33 or 0.1 uf ceramic in parallel with that and the input is done. Place the ceramic as close to the input lead as possible.


    You don’t want or need polymer organic output caps or input caps .I’m assuming that is what your Aluminum electrolytics are judging by the low ESR (28 mohms for a 47uf aluminum). These are mainly used for SMPS’s.


    The output esr requirements are pretty lax 0.3 ohms to 22 ohms. Select a tantalum cap with an esr in that range or aluminum electrolytic. Your original schematic shows a 47 uf cap so one somewhere in that range . Dont use ceramics on the output.If you are using tantalum caps you have to derate by a factor of three. In other words for your five volt output you would use at least a tantalum cap of at least 15V.


    Here is a possible 47uf output cap for your five volt regulator.


    Part # 718-1086-1-ND from Digikey 0.47 cents in small quantity.


    http://www.vishay.com/doc?40002
     
    Last edited: Aug 9, 2008
  10. deefactorial

    Thread Starter Active Member

    Jun 11, 2008
    33
    0
    Thank you for your help. I am new to this and any help is much appreciated.

    So now that my power supply has changed and it closely matches the specs of the LM1117 chip I do not need large caps on the input. So a 10uF Tantalum 500mOhm 25v and a 0.1uF Ceramic on the input will be sufficient. A 22uF Tantalum 400mOhm 20v on the ADJ pin of the LM1117 will help to regulate the noise. A 47uF Tantalum 700mOhm 16v on the Output of the LM1117 will be sufficient capacitance for the output.

    Will a 47uF Tantalum on the output of the 3.3v work too?

    I've attached a new schematic diagram.

    Unfortunately I sent in the order for all those caps on Friday, I will have to start a new order for the 47uF Tantalum.

    Low ESR aluminum electrolytic caps are typically for High frequency applications so they work with the high frequency ripple current. To get a really balanced power plane is it generally a good practice to use caps that vary in size and ESR values?
     
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