Help me understand MC34063ACN boost circuit

Discussion in 'The Projects Forum' started by letsbully, Apr 11, 2012.

  1. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
    1
    [​IMG]

    R2 = 1.1K ohm @ 2W
    R1 = 360 ohm @ 2W
    L = 150 uH @ 0.67A (could not find a 170uH inductor)
    Rsc = 0.5 ohm @ 5W
    IC Datasheet

    Vin = 3.5v
    Vout = 5.15v (wanted 5.07v but maybe the inductor is not the right one or the resistors are a tiny bit off)
    Vout (under load) = 4.37v (just what I am looking for)

    Everything is perfect except the following:

    I(in) = 0.57A
    I(out) = 0.25A

    How do I increase the current the IC is "allowed" to output. I replaced the Rsc resistor from the specified 0.22ohms to 0.5ohms because the circuit was only giving 150mA of current with 0.22ohms. I assumed if I used resistors values of 1.1K and 360ohm instead of say 11K and 3.6K, the output current would be greater. Wrong assumption. I want the current output to not be limited by anything except the IC itself (or 1.5A which is what the IC can output).

    The "power supply" can output 3A of current if something is willing to draw it. The device connected to the circuit can draw more than 0.25A (the device can draw 1A). I know the inductor matters but being rated at 0.67A I am also assuming that it can output more current and the IC is actually limiting the output based on the combination of resistors and/or capacitors in the circuit (or is it consistent with what the inductor is capable of outputting with 0.57A going in to the circuit? So if I use an inductor rated for 2A I can expect 1A out?). I'm not really understanding the datasheet in terms of what components limit the current output. What parts can be added/replaced (and why) to NOT limit the circuit to lower current output?

    Thanks a lot for your help!
     
  2. Eric1180

    New Member

    Apr 10, 2012
    1
    1
    If you don't want current limiting just replace the .22Ω Resistor with a wire. It bypasses the current limiting.
     
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  3. crutschow

    Expert

    Mar 14, 2008
    13,003
    3,232
    If you try to put more current through the inductor then its rating it will saturate and possibly blow the power transistor in the chip. For 12V in and 28V out the peak inductor current will be over twice the output current you draw so you must allow for that. Table 12 in the data sheet has a formula for calculating the peak switch current [IPK(switch)], which is also the peak inductor current.

    Do not reduce Rsc to zero since that provides short-circuit protection to the chip. But you can reduce (not increase) its value to increase the available output current.
     
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  4. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
    1
    I replaced the inductor to 150uH @ 1.5A in the circuit.

    I have already attempted to do that. Curiously, the current output decreases when you remove the resistor. With a 0.5 ohm resistor present, the current output is 250mA. With the resistor replaced with a wire, the output current drops to 170mA. I got the following values using different resistors:

    wire = 170mA
    0.22ohm = 150mA
    0.5 ohm = 250mA
    0.51 ohm = 260mA
    0.75 ohm = 230mA
    1 ohm = 240mA
    180 ohm = 15mA
    360 ohm = 10mA
    So when I use the formula in Table 12 I can expect the circuit to output 1/2 of that amount? How does decreased voltage affect that estimate (3v in 5v out)?

    See my results above when it comes to increasing or decreasing the resistance. What can be the possible explanation for the results?

    I am now starting to assume that this IC can not output even close to 1A in a boost scenario. Am I correct in that assumption?
     
  5. crutschow

    Expert

    Mar 14, 2008
    13,003
    3,232
    That is likely true, depending by what you mean as "close". If you calculate the peak switch currents as shown in Table 12, that will tell you the maximum output current you can get. For 12V in and 28V out it will be less than 1/2 of the allowed peak current.
     
  6. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    Yes, but it is easily overcome by adding just a resistor and an external boost transistor.

    It is purposely connected in Darlington mode to prevent saturation of the external transistor as hard saturation will lengthen the turn off time.

    [​IMG]
     
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  7. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
    1
    Thanks!
    I had that thought after reading replies to this thread and actually contacted the manufacturer last night about it. They got back to me confirming that it would work.
    How do I determine the resistance value? Is it just Rsc x 10,000? ;)
     
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