In today’s blog we will use the same values as last time , R=10Ω, C=10μF, L=10mH, and as promised we’ll look at the current and voltage in this circuit:
In part 2, we looked at the impedance (magnitude only) of the circuit components, and saw this graph:
From Ohm’s Law, V=I*Z, we would expect to see the current as a mirror image of the total impedance, since current is inversely proportional to impedance, or I=V/Z. So without further adieu:
Shown is the magnitude of the current. The voltage on the resistor, V=I*R, is simply the current times 10, and tracks the current (voltage is directly proportional to current). If the resistor were the load in this circuit (or a loudspeaker) this voltage curve is the “voltage transfer” of the circuit. If this were a loudspeaker (assuming it were perfectly resistive, which it never is) and we convert the voltage to dBV, we can just add it to the measured frequency response of the driver to get the frequency response with the filter in place.
For simplicity I always do my calculations by assuming a voltage source of 1Vrms. This allows me to quickly calculate the effect of the filter in dBV (or dB’s compared to 1 Volt). For example 0.1Vrms is -20dBV. In other words if I put 1V into a circuit and get 0.1V on the output I have 20dB of attenuation. If I look at the 0.1000V line on the graph I can easily see where the response will be -20dB.
What about the phase? In my effort to keep these things in bite size chunks, both for me and you, this sounds like a good topic for next time. But I promised to talk about voltages, so before I go, here’s a fun little cliffhanger that’ll stick in your craw…
In a series circuit, the total voltage dropped across the various circuit elements must be equal to the total source voltage. The source voltage is 1 Volt rms. The current through the circuit at 1kHz is 20.86mA (you can see it for yourself on the graph above). At 1kHz the voltage on the resistor is 20.86mA*10 Ohms or 208.6mV. What about the capacitor and inductor? (The voltages on the capacitor and inductor are 331.8mV and 1.310V respectively, see graph below.) So how do you get all the voltages to add up to 1 Volt. Hint: What happens at ~500Hz when there is 1 Volt across the resistor?