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Voltage drop reason?

Discussion in 'General Electronics Discussion' started by vinit2100, Mar 12, 2014.

  1. vinit2100

    vinit2100

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    Oct 21, 2013
    Hi, In the attached circuit Nimh.JPG when the switch is ON the voltage across the LOAD dropped to 700mV, What is the reason for this?

    the same happened in the circuit Opamp.jpg. what is the reason for this voltage drop in two cases? The LOAD is same in both cases..
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    What is the load?
    What OpAmp do you use?

    Too high a current through the load will drop the voltage due to the internal resistance of the battery or the limited current drive capability of the opamp, respectively.
     
    Last edited: Mar 12, 2014
  3. vinit2100

    vinit2100

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    Oct 21, 2013
    The load is actually a circuit. CIRCUIT2.JPG. that works with 1.5V. the opamp i used is TLV2464
     

    Attached Files:

  4. vinit2100

    vinit2100

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    Oct 21, 2013
    Harald Karap,
    Sir when i removed the resistor 22 Ohm. and the switch ON. there is no drop in the voltage..
     
  5. shrtrnd

    shrtrnd

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    Jan 15, 2010
    Harald answered this for you. Quite well, I think.
     
    Last edited: Mar 12, 2014
  6. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Yes Harald said this, in case you missed it first time.

    "Too high a current through the load will drop the voltage due to the internal resistance of the battery or the limited current drive capability of the opamp, respectively."
     
  7. Harald Kapp

    Harald Kapp Moderator Moderator

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    The voltage drop across the 22Ohm resistor is 22Ohm*I(load). You probably measure still 1.5V from BT+ to BT- and you will find 0.8V across the 22Ohm resistor which translates to I(load)=36mA
     
  8. vinit2100

    vinit2100

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    1
    Oct 21, 2013
    Sir Harald,
    so that means the LOAD requires less than 36mA for operation. and too much current will cause the voltage drop right? What is the case in opamp ? If op amp supplies more current what will happen?
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    No, it doesn't. It obviously requires at least 36mA, Otherwise the voltage drop across the 22 Ohm resistor were smaller.

    Any current will casue a voltaeg drop by V=R*I. Higher currrent = higher voiltage drop. You can reduce voltage drop to ~0V only by setting R=0Ohm (short circuit) or removing the load, the latter being impractical.

    The voltage drop is independent of the source of the current. It doesn't care whether teh current comes from a battery or an opamp.
    You'll run into problems only if the opamp can supply less current than the load requires. In that case an additional voltage drop will develop across the output transistor of the opamp.
     
  10. vinit2100

    vinit2100

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    Oct 21, 2013
    What is the similarity in operation both the circuits? I have the two circuits waveform, when the switch is ON. which look similar.. here by attaching the same..
    Wave_opamp.jpg
    wave-nimh.jpg

    in both cases the voltage drops to 700mV from 1.5v when the switch is On. !!!
     

    Attached Files:

  11. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Go back to the top of the thread and re-read carefully. It has all been explained to you, including where to measure to verify the statemens made.
     
  12. vinit2100

    vinit2100

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    Oct 21, 2013
  13. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Read the datasheet completely. It says 250mA in the "features", but it will not supply full output voltage at this current. Look at figure 12 (page 7): with a load curent of 100mA the "output voltage to supply rail" wil be ~1000mV. This means that the actual output voltage is 1V less than supply voltage. Your circuit uses 3.3V supply, so at 100mA the output of the opamp will be ~2.3V.

    What is the currrent that is flowing to the load from a good power supply without series resistor? You need to know that as input for designing an appropriate supply.
     
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