Resistor Voltage Drop Compensation

Discussion in 'General Electronics Chat' started by mossman, Feb 28, 2014.

  1. mossman

    Thread Starter Member

    Aug 26, 2010
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    I need to insert a low-value series resistor in the power trace of a PCB to monitor power consumed by a microprocessor IC. I will be using a 50 ohm 1% resistor. At a current of 5 mA, the drop across the resistor will be 250 mV. The processor requires a supply voltage of 3.3V. Question is, is it ok to increase the power supply voltage to 3.55 V to compensate for the resistor drop in order to maintain the proper supply voltage to the processor? If so, is there a general rule of thumb for how much the supply can be increased (e.g. if a larger value resistor were used)? I'm curious if there are any detrimental effects to the processor circuit by doing so. For instance (an extreme case), could I use a 1k resistor (Vdrop = 5V) and increase the power supply to 8.8V?
     
  2. wayneh

    Expert

    Sep 9, 2010
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    A big problem when using a resistor as a voltage regulator is if the current is not constant. If it drops, all the sudden your micro is seeing the full voltage and will likely be destroyed. Resistors works for LEDs and lightbulbs, but not so well with smart devices.
     
  3. BReeves

    Member

    Nov 24, 2012
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    How about a resistor and a zener diode to clamp anything over 3.3 volts? Problem is it may then go below 3.3 volts depending on how much curent the processor is using.
     
  4. mossman

    Thread Starter Member

    Aug 26, 2010
    131
    3
    So if I understand correctly, the only concern is protecting the processor from over-voltage in case of resistor failure. Is this correct? If so, clamping the voltage using a 3.3V zener sounds like the way to go.

    How is the voltage going to drop based on how much current the processor draws? The zener will be shunted to ground so that the voltage will never drop below 3.3V unless the main supply voltage drops below 3.3V.
     
    Last edited: Feb 28, 2014
  5. BReeves

    Member

    Nov 24, 2012
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    If you draw more than your assumed curent through the resistor, the resistor could drop more voltage than you expect. For example if you are using the micro outputs to drive leds the drop across the resistor will depend on how many leds you try to light.
     
  6. mossman

    Thread Starter Member

    Aug 26, 2010
    131
    3
    Right. I understand. Thanks.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    A 250mV drop to measure current seems a little high. A more typical shunt value of 50mV to 100mV is used, thus having less adverse effect on the circuit operating voltage.

    How are you measuring the shunt voltage?
     
  8. mossman

    Thread Starter Member

    Aug 26, 2010
    131
    3
    Sorry, I'm measuring current fluctuations so the greater the resistor the better.
     
  9. Johann

    Senior Member

    Nov 27, 2006
    190
    30
    Hi,

    A shunt giving a 50mV drop can be used if you amplify this mV drop by means of an Op-amp circuit with pre-set gain to suit your readout requirement.

    Just a thought!
     
  10. crutschow

    Expert

    Mar 14, 2008
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    Not necessarily. What circuitry are you using to measure these "fluctuations".
     
  11. mossman

    Thread Starter Member

    Aug 26, 2010
    131
    3
    5X differential probe with 100k impedance, sense resistor = 51 ohms, current through resistor = 5 mA.
     
    Last edited: Feb 28, 2014
  12. crutschow

    Expert

    Mar 14, 2008
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    What is the best sensitivity of the probe with your scope?
     
  13. mossman

    Thread Starter Member

    Aug 26, 2010
    131
    3
    Not sure.

    The current setup I'm using isn't ideal. The probe is connected to an external power supply, which is connected via USB to a real time spectrum analyzer (the diff probe and power supply came with the RSA). So I'm basically using the RSA to power the probe and feeding the 50 ohm output of the power supply to an o-scope. It is working, but I'm wondering if my results could be better.
     
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