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I am designing high pass RL filter. R is 550 [Ohm], L is 220 [uH]. I calculated that time constant is therefore 0.0000004 [s] and -3 dB pulsation 2 500 000 and frequency 397 887 [Hz]. I put this on the breadboard and connected JDS 6600 signal generator and Rigol DS1054Z oscilloscope to test it. For testing I apply sine signal of 5 [V] (Vpp, amplitude) and varying frequency. As expected I observe decreased amplitudes of the signal at the output for low frequencies and increasing amplitudes for signals of increasing frequency. Using transfer function

|H(j omega)| = 1 / sqrt(1 + (1 / (omega tau)^2))

I calculated that 0.95 signal attenuation should happen at 1 210 781 Hz (|H(f=1210781)|=0.95) and that would mean that I should see about 4.75 V at this freq, but in practice I get this value at about 1 MHz.

Why do I see such a big difference?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ A couple of points to note: 1.your text says R is 550 ohms but your schematic shows 6 ohms. 2. Have you accounted for the output impedance of your generator? 3. What is the tolerance of your components (an inductor could easily be 20%)? 4. The slope of the filter response (one pole) is not very rapid. It can be difficult to accurately determine the frequency where the attenuation is exactly 0.95 especially using an oscilloscope. You should plot the response over a fairly wide range of frequencies and then compare it to the theoretical. One point is not enough. \$\endgroup\$ – Barry Jun 25 '18 at 0:42
  • \$\begingroup\$ More points to note: Since you only have an RL filter, the value of R factors heavily into the roll-off point. Please check your math as 6 ohms has almost no effect in an RL filter. Also, you could cut your scope photo's in half. Only the initial readings help us. \$\endgroup\$ – Sparky256 Jun 25 '18 at 1:00
  • \$\begingroup\$ For circuits as simple as this a simulation will almost always serve as a basis for comparison, provided you account for the possible parasitics that interfere with your setup. \$\endgroup\$ – a concerned citizen Jun 25 '18 at 6:53
  • \$\begingroup\$ Thanks, it is 550 - corrected. I input signal with different frequency and check whats the output, therefore we have couple of points (images) of impulse response, 40kHz - 520 mV, 100kHz - 1 V, 390kHz - 3.04 V, 700kHz - 4.24 V, 1000kHz - 4.8 V, 1200kHz - 5.04 V, 1500kHz - 5.16 V \$\endgroup\$ – 4pie0 Jun 25 '18 at 13:05
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What you may actually be testing

schematic

simulate this circuit – Schematic created using CircuitLab

Assuming 1:1 probe

You should see a low Q <0.3 breakpoint around 1~2MHz with 50% attenuation when coax load pF impedances falls to equal rising ZL(f). enter image description here

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  • \$\begingroup\$ Probe is set to 10x. So basically you are saying that I need to tak into account impedance of the JDS signal generator and impedance, capacitance of the oscilloscope. Is that correct? \$\endgroup\$ – 4pie0 Jun 25 '18 at 13:11
  • \$\begingroup\$ Yes notice I have red lines for probe range then you add your probe values in this range and choke capacitance is also significant lowering self resonant f. This is a good way to measure C total with accurate L. But naive to expect a flat response. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 25 '18 at 13:23
  • \$\begingroup\$ Could you provide some more info (links) how to read this chart? \$\endgroup\$ – 4pie0 Jun 29 '18 at 14:12
  • \$\begingroup\$ Search for RLC monograph this is 50 yrs old info. This is impedance from which you can estimate any single unknown. By the intersection. Like a slide rule before your time? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 '18 at 14:17

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