I have a BIG problem understanding ohm's law for a crystal earphone and hope to get a complete response from you guys to solve my problem.
Maybe you guys know that a crystal earphone has very large impedance compared to common ones. That means you will have a sound even if you rub the connected wires too. When the wires are rubbed together there is a very low current in the wires which causes the crystal earphone to vibrate. According to ohm's law I=V/R We know that the above formula is correct for impedance too. So we have a very high impedance say 2Mohm, so can we say that we have to have a very low voltage at the ends of the wires to have a vibration? That means when we rub the wires there is a very low voltage (caused by rubbing the wires together) which causes a very low current in the ends of the wires? If so, how those very low current and voltage are able to cause the HIGH IMPEDANCE crystal earphone to vibrate? Even if we replace the crystal earphone with a very high resistor and feed the resistor with a low voltage source then nearly no or too low current will pass through the resistor and it is too small to do any work.
I am getting confused with this issue!
I'm not sure even what your question is. It is good to try and really come to terms with Ohms Law rather then just memorizing the formulas. When you have a true understanding of it then further concepts and laws of electronics start to make sense.
I'm not sure even what your question is. It is good to try and really come to terms with Ohms Law rather then just memorizing the formulas. When you have a true understanding of it then further concepts and laws of electronics start to make sense.
I know what ohm's law is saying but it seems I was not able to say my meaning.
When you have a high resistance resistor and put it on say 12V it will draw a very low current and vice versa.
Now you suppose that you have a high impedance device(crystal earphone) which is drived by a too low voltage( which means that a very low current will pass through it).My question is why we have the voice in the earphone while it is drived by too low voltage and current(we will have a too low current going through the earphone and should not do any work due to high impedance earphone).
We do not have that voice with a 8ohms earphone?!
I like to know the main reason/ I hoped someone could help me and explain the reason by ohm's law too.
Why they are very sensitive? Do you want to say that sensityity comes from other thing rather than the earphone impedance? I must say why they are very efficient?
we can see a such thing at the input of an OPAMP to.
Why they are very sensitive? Do you want to say that sensityity comes from other thing rather than the earphone impedance? I must say why they are very efficient?
Why they are very sensitive? Do you want to say that sensityity comes from other thing rather than the earphone impedance? I must say why they are very efficient?
we can see a such thing at the input of an OPAMP to.
efficiency is independent of impedance, it's another attribute entirely. Two loudspeakers with the same impedance and power ratings can have different efficiency ratings as that is the efficiency of converting electrical input power to mechanical (acoustic) output power.
One way to look at the thing is that the crystal earphone, which its really high impedance (in the megohms vs. that of say a 32 ohm Walkman earphone), doesn't hardly load down a circuit at all. So when you rub the connecting wires together or scratch them across another metal surface, there will be noise in the earphone because that teensy galvanic voltage generated between dissimilar metals is not loaded down hardly at all and can have an effect on the earphone. Nearly any other standard earphone -- they're usually electromagnetic -- has a really low impedance and will load that voltage source down so that it will have no effect on the earphone.