How to simulate Power Effeciency of series parallel Charge Pump?

Discussion in 'The Projects Forum' started by ICD_life, Nov 14, 2014.

  1. ICD_life

    Thread Starter New Member

    Nov 12, 2014
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    0
    Hello All,

    I have designed Series Parallel Charge Pump that is used in Power Management circuit to charge the battery.
    Details:
    It has variable input from < 1V
    Output voltage must be > 1.8V
    Input Side using variable Voltage Source [in cadence environment]
    Question:
    I don't have load current fixed(assuming must be around 500uA). How do I calculate the power efficiency?
    1. Do I fix a constant VOLTAGE source at the output of CP as load and check for input and output current
    2. Do I fix a constant CURRENT source at the output of CP as load and check if output voltage reaches 1.8V

    Kindly help me understand the methodology to check the power efficiency of Charge Pump
     
  2. DickCappels

    Moderator

    Aug 21, 2008
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    632
    Assuming that the input and output are DC, calculate:
    (output voltage x output power) / (input voltage x input current)

    A good way to load the output with a resistor. If you need to test at a fixed current, then yes, use a current source.
     
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  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    LTSpice is an enhanced version of Spice especially tailored to make these calculations trivial. For example, it has a one-keystroke method of plotting the power in any component on the schematic, including the source, the load, the package, etc. Once plotted, it has waveform arithmetic which can integrate any plotted voltage, current, or power waveform to find average and RMS. It even has some macros that will tabulate efficiency....

    A stepped current source can be used as a variable load to find efficiencies at various output power levels. A stepped voltage source can be used to vary the input voltage while looking at the output voltage, power level, and efficiency.
     
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  4. ICD_life

    Thread Starter New Member

    Nov 12, 2014
    9
    0
    Thank you for the details. Yes input and output is DC. I am actually following this method since the current is not fixed.
    Initially:
    1) The Charge Pump conversion ratio is 3[at the output charge pump I have fixed a capacitive load]
    2) Input DC voltage is VIN = 1V
    3) IDEALLY: Output must be 3 V

    Efficiency Simulation:
    1) I fix a voltage source parallel to the capacitive load, vary the voltage from 1V to 3V step by step and check for input and output currents for each step in voltage variation.
    for ex Vload(@ output) = 1V; I_VIN = xx mA, I_VOut = yy mA.
    2) so I calculate Power Efficiency = ( Vload * I_Vout ) / (VIN *I_VIN) = (1 *xx)/(1*yy)

    Is this right method ?

    or do I fix a current source and then check for the voltage copnversion?
     
  5. DickCappels

    Moderator

    Aug 21, 2008
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    I think so (both statements 1 and 2 appear to be correct).
     
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  6. ICD_life

    Thread Starter New Member

    Nov 12, 2014
    9
    0
    Hello Sir,

    I think I found my answer. Yes ideally both must work but the CADENCE simulaions gives the maximum flexibility to ideal sources which can push the schematic circuits to extreme levels to give that particular current or Voltage. So I have decided to use transistor Model of Current Source like the Current Mirrors at the load which makes the efficiency calculation more efficient. Hopefully this approach is by far good.

    Thank you for your time and inputs.
     
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