It was a strange description.Thanks Mike, but very strange!! specially the NO 4!
Why the HALF OUTPUT is the base parameter and is when the resistor is equal to the input/output impedance?!!
A very stange method to determine the impedances to me!
It doesn't have to be half, half just makes the maths very easy that's all.Thanks audioguru,
But Why the HALF is the key of the input/output impedance????
I am not able to see main relationship between the input impedance and the half output!? Is it due to something like 3dB freq or half output power which is used to describe F3dB?
Is this the reason of what does happen regarding the equivalence resistor values and the input impedance?-3dB is half the power but is 0.707 times the voltage.
-6dB is half the voltage.
Yeah but yet nobody told the reason?When the output drops to half the level with your added resistor then the input impedance of the amplifier is the same as the value of the resistor.
What you're doing here is making a potential divider - two resistors, one known, one unknown.Is this the reason of what does happen regarding the equivalence resistor values and the input impedance?
Yeah but yet nobody told the reason?
Nigel told that the half value is not critical which this makes the understanding the reason much harder!
Which math are you talking about?
Thanks a lot Nigel, It clearly makes sense now. I took it just when you mentioned it as a potential divider. It is a very good trick to measure the input impedance.What you're doing here is making a potential divider - two resistors, one known, one unknown.
By measuring either the voltage, or the current, you can easily calculate the value if the unknown resistor - simple ohms law.
It's simple enough maths - BUT if you make the known resistor the same value as the uknown one, you don't have to calculate anything at all, just read it off.
Easiest way is to use a potentiometer, adjust it for half the original signal, then remove it and measure it's resistance with a meter,
Like I said, the simple maths - work it out, it's only ohms law.Thanks a lot Nigel, It clearly makes sense now. I took it just when you mentioned it as a potential divider. It is a very good trick to measure the input impedance.
But If my brain serves, I remember that there was another method to measure the impedances by measuring Vopen then Vfull load by a known resistor then use a formula to calculte the impedances?
I think what have been explained here is how to measure Output impedance. Input impedance has different procedure, may be thats what you remember.But If my brain serves, I remember that there was another method to measure the impedances by measuring Vopen then Vfull load by a known resistor then use a formula to calculte the impedances?
Measuring power amplifiers is a VERY, VERY different matter!Using the resistor method to reduce the output voltage to 1/2 for a circuit with a lot of negative feedback, such as many op amp circuits and power amplifiers, will give an interesting result. The output voltage will initially have little variation with load change, since the closed loop output impedance is very low. Then the voltage will suddenly drop when you reach the current limit of the output. Thus the measurement to 1/2 output voltage basically gives you the open loop output impedance of the circuit, not the closed loop impedance during normal operation.
To measure the closed-loop output impedance you would just vary the resistance to a value above the current limit point, which will typically give only a small voltage change. Then use ohm's law to calculate the output impedance. This can be in the milliohm region for a circuit with significant feedback.
Or you could measure the voltage drop at the output when you connect a 4 ohm load, but the voltage drop will be VERY small.I think you measure the output impedance of a power amplifier by injecting a signal into its output from another amplifier that has an extremely low measured output impedance, in series with a very low value resistor.