Hi, i’ve got a question about the current that flow in a circuit...
Hypotize you have an IC, for example the l293d motor driver, and you want to power it with a battery, which delivers for example 7.4 volts and 1000mAh. Now, the question is: will the l293d(or another circuit not necessarily integrated) use a small portion of these 1000mAh to power the entire circuit(for example 100mAh leaving the remaining 900mAh) or will it absorb all the 1000mAh damaging itself?
The same question can be asked with a simple circuit on protoboard:
For example if i power a circuit which needs 9v and 200mAh with a power supply that delivers for example 9v and 2000mAh, will the excessive current destroy the circuit or will it absorb only the 200mAh from the 2000mAh leaving the remaining 1800mAh to power another circuit?
Ps: all the values(current and voltages) are totally random only to make you understand the question...thanks
I think you should learn Ohm's Law before you start messing around with electronics. There is a minimum knowledge requirement to work with electronics and you are below that minimum.
Ohm's Law states (among other things) that the voltage V = the current I x the resistance R
("V=I*R")
Thus, the current that flows is dependent on the resistance until it exceeds the current capability of the source, so the LM293 has nothing to do with how much current flows. That is determined by the
device (motor) you try to drive with it. If the motor current rating is within the max current rating of the LM293 then everythings fine. If, on the other hand you try to drive a 6A motor you'll fry the LM293.
You also need to understand that mAh and mA are very different things. You can't just use those terms interchangeably.
Milliamps (mA) or Amps (A) are a measure of current. But the 1000mAh that a battery has specified is NOT current. It is capacity, milliamps x hours. So a 1000mAh battery could deliver 1000mA (1A) for 1 hour or 100mA for 10 hours or potentially 10,000mA (10A) for 6 minutes (1/10th of an hour).
So a circuit will never require 200 mAh but it might require 200 mA. And a power supply will never be rated at 2000 mAh but it may be rated at 2000mA (2A).
The same question can be asked with a simple circuit on protoboard:
The same question can be asked if your house mains wiring.
You plug in a radio that takes say 100mA into a socket that can supply 13A for an electric fire. Is that going to work that you can connect two different things with such vastly current requirements into the same source of power.
Back to the other part of the question.
An L293D is old motor driver technology (darlington transistors instead of mosfets), and 'drops' (looses) about 2.4 volt (current dependent) when driving a motor. The motor will only get about 5volt from that 7.4volt.
You could say that an L293D is about 67% efficient in this case.
1/3 of used battery power is wasted as heat inside the LM293D chip.
Same story for it's bigger brother, the L298.
Leo..
Now, the question is: will the l293d(or another circuit not necessarily integrated) use a small portion of these 1000mAh to power the entire circuit(for example 100mAh leaving the remaining 900mAh) or will it absorb all the 1000mAh damaging itself?
Couldn't say without knowing the load rating. If the load was rated for 3A and you connected that battery it would draw 3A from the battery until the battery cooked or the L293 burned up.