I have an audio circuit that I need to find the internal impedance of the circuit. How do I do this by only changing the output impedance of the circuit and taking measurements? I know I did this back in school but forgot how to solve. Basically, I need to find internal resistance of a black box without knowing the total voltage being supplied. Thanks!
Black box impedance…how to test
audiocircuit analysisimpedance
Related Solutions
Simple amplification of audio is the most trivial demand for a BJT and virtually any transistor (including germanium) will do the job you want. Maybe look for BJTs that are referenced as good audio amplifiers but, given that the marketing boys would see some folk being put off buying a particular transistor because it is pigeon-holed by the label.
The level of audio does not need to be greater than 0.6 volts because the right BJT amplifier will set its bias point to accommodate all the + and - swings of the audio signal it wishes to amplify.
This sounds very much like a homework question to me but as its Xmas...
Barry is quite correct if we were to consider every possibility but let us assume for this particular case that the question is actually solvable or at least approachable. i.e. the box contains only a few passive components arranged in a simple circuit. How could we approach it?
Consider a generalised black box - terminals A and B are the inputs and terminals C and D are the outputs.
Use the information you have - You have been told that it is a (simple?) demodulator - so there's a good chance it will have a diode somewhere in the circuit and possibly a capacitor.
A first approach would be with resistance (continuity) measurements. If you are lucky B and C are connected together to form a common connection which simplifies things a bit.
Measuring resistance/continuity between two points and reversing the measurement voltage ( e.g A -> B and B -> A) should pick up a diode (or rectifier type junction) connected between those terminals. A bi-directional low resistance may indicate a wire or an inductor. A changing resistance value may indicate a capacitor as it charges up - short it out and see if it charges up again.
The next step would be to try putting some form of signal into the box and see what comes out (into a known load such as a resistor), measuring input voltage, current and phase (useful for capacitance/inductance type circuits). You could also try using a variable DC input of some sort as well as a variable AC source. An oscilloscope would be an invaluable tool to see the waveforms as well as measure inputs and outputs.
This is where the possibilities for the unknown circuit multiply but by using logic and reason (based on the characteristics of different devices) to suggest the most probable solution you can take the analysis a bit further. For example, if there was no continuity between input and output terminals for a DC voltage BUT an AC signal could get through then you could be dealing with either a capacitive (series) circuit OR a transformer. Think about how you would eliminate one choice or the other.
As a training exercise use this approach to analyse the circuit below and see if you can answer the following questions -
(1) How would you know there was a diode between A and C?
(2) How would you know it was a Germanium diode and NOT a Silicon one?
(3) How would you know there was a 10k resistor across the inputs?
(4) How could you determine the value of the 0.1uF capacitor from outside the box?
Related Topic
- Electronic – Finding the schematic of a black box with unknown impedances
- Electronic – Audio – How does I/O impedance affect the device
- Effective Source Impedance – How to Calculate Effective Source Impedance in an Audio Mixing Circuit
- Electronic – How should I probe when measuring output impedance? Why do I get inconsistent readings
Best Answer
If you model the black box as a voltage source in series with an impedance, then you can obtain enough data to calculate the values of these parameters by measuring the output voltage open circuit and with a resistive load at 2 different frequencies. You will have to solve the equations for a voltage divider.