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current capability of op-amp

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cjron

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Hi, i am working with a simple IC741 and determining its current drive capability (o/p current) with a resistive load..provided (inverting ampr, negativ feedback with a gain of 10).between the op-amp output and the load (resistor) there is a series resistance (Rs). can anybody explain me by what way of altering the load and Rs.i can achieve a higher output current and what is the role of output impedance???


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Current delivered from op-amp output to load depends on voltage at the output, load resistance Rl, output resistance Ro and your series resistance Rs. Current through load can be determined from Ohm's law. In that case Io=Uo/(Rs+Ro+Rl). Considering that Ro is internal resistance of opamp which is integrated circuit, you cannot chamge it. So the only wau to increase current is to lower the load resistance or the series resistance. You must take into consideration that the output current causes power dissipation in internal resistance, which can be determined from the formula P=I^2*Ro. You musn't exceed your opamp's maximum power dissipation or you are risking damage to it.
 
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    cjron

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The output current capability of LM741 and many other OPs is defined by current limiting means rather than an output resistance.

Maximum ratings guarantee continuous output short circuit duration, but output swing with different load is only sparsely specified.
 
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Well thq u all, that made me clear with the whole concept now. but is it possible to find the internal resistance of the ampr? and what will happen if the inductance is varied??

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