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what does hard to drive mean?

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cafeblackforest

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Hi!

I wanted to know what these phrases mean in the context of voltage/current-


A) 'hard to drive' - does it mean it requires more current or more voltage (they are proportional acc. to Ohm's law but what does it usually refer to) ?


B) 'loading a circuit'- what does this mean.. when one circuit is connected to another, and if the second circuit loads the first one, does it mean it causes a high voltage drop across itself, and when this happens does it mean the second circuit has a high impedance/resistance?

Thanks in advance.
 

A) 'hard to drive' - does it mean it requires more current or more voltage (they are proportional acc. to Ohm's law but what does it usually refer to) ?

I would normally associate it with complex loads e.g. highly inductive or capacitive or resonant so they might cause instability in the driving circuit.

Keith
 

A) 'hard to drive' - does it mean it requires more current or more voltage (they are proportional acc. to Ohm's law but what does it usually refer to) ?
Ohm's law only applies to resistors. Many electrical loads do not obey Ohm's law. For example, current and voltage are not even nearly proportional in a diode.

Consider also a reactive load such as a capacitor:
If it is driven with a simple sine wave voltage source, then the current will also be a sine wave of the same frequency, however the phase will be different. In fact half the time the current through the capacitor will be flowing in the opposite direction to what it would through a resistor.

If the voltage across the capacitor is anything other than a sine wave, then the current waveform will be completely different to the voltage waveform.
 

B) 'loading a circuit'- what does this mean.. when one circuit is connected to another, and if the second circuit loads the first one, does it mean it causes a high voltage drop across itself, and when this happens does it mean the second circuit has a high impedance/resistance?

For my opinion, the meaning is as follows:
When you have two parts of a circuit (two stages) it is possible to calculate the transfer functions of both parts separately only if the second stage does not "load" the first stage, which means: ...does not draw any current from the first stage. Such a simplification (H=H1*H2) is allowed only in case both stages are decoupled (without any "loading" effect) using a buffer.
Of course - if the 2nd stage is a high-impedance network (if compared with the first stage) and if a certain error is allowed, you can neglect the "loading error". But the final result will be an approximation only.
 

"Hard to drive" has a hidden qualifier behind it, dropped for
convenience or ignorance of details. Could be something
like "without damaging overshoots" or "given that I insist
on using on-chip CMOS low voltage low current output
buffers" or "while maintaining required edge rate and eye
opening" or whatever.

This is all contextual and you have to pry the details from
their shell, because nobody advertises their difficult
nature on the front page.
 

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