Andy's Blog

September 2009 - Posts

• Analog Filters 101 – Part 3

  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? Posted Sep 24 2009, 08:52 PM by Andy W with 3 comment(s)

• Analog Filters 101 – Part 2

  Now with pictures! I have been neglecting my blog as of late, so I decided to write at home, and take smaller chunks.  Hopefully the pictures will help to visualize the concepts – it always helps me, I’m a visual type. In the previous installment we talked about inductors and capacitors, their reactive nature, and how to calculate their impedances.  Next time we’ll talk about the voltage and current and show how the impedances affect the current that flows through a circuit.  After all it’s the current through the voice coil that determines the output level of a speaker. To start our conversation today we will use a simple series RLC circuit. The total impedance seen by the voltage source is Z=Z(R)+Z(C)+Z(L). I’ve picked values for the R, C, and L that will do something interesting in the area of interest – the audio band:  R=10Ω, C=10μF, L=10mH.  So let’s get started by doing an example at 1,000 Hz. The impedance of a resistor is constant with frequency (at least we are going to assume it is for the time being) so Z(R1k)=10Ω. The impedance of the capacitor is Z(C1k)=1/(2πfC)=1/(2*3.14159*1000*0.000010)=15.92Ω The impedance of the inductor is Z(L1k)=2πfL=2*3.14159*1000*0.010=62.83Ω We could repeat this calculation ad nauseum for every frequency, but since I have a computer… I went ahead and used a logarithmic scale for frequency since it impossible to see much of anything on a linear frequency scale.  The graph above shows the capacitive and inductive impedance for 10μF and 10mH in the audio band (plus an octave on either end).  Whenever we see a curve in the graph it’s always interesting to see what would happen if we plotted the curve on a logarithmic scale. And as expected we get straight lines.  Since decibels are logarithmic we can deduce that a capacitor or inductor in the signal path will induce a linear-in- dB change in the current flowing through the circuit.  But that’s a blog for another day.  Just for fun, I added a curve showing the total impedance of the circuit as seen from the voltage source. One more thing for today… Notice that the impedance of the capacitor falls exactly tenfold for every tenfold increase in frequency, and vice versa for the inductor.  If all you had was a piece of log-log graph paper and knowledge of only one data point, you could draw the entire curve for any value of capacitor or inductor. Posted Sep 17 2009, 11:31 PM by Andy W with 7 comment(s)