You are correctThe original crossover circuit also has 13 components. There is a resistor placed vertically in the bottom corner. I think you missed that.
Yes, original iron core inductor's DCR is 0.56 ohm at 100 Hz and 1.37 ohm at 1kHz. Air core inductor's DCR is 0.77 ohm at 100 Hz and 1.35 ohm at 1 kHz.Is the difference in dcr of old iron core vs. new air core inductors known?
4,6 mHWhat is the value of inductance for the inductor you are talking about?
I made these measurements with an LCR meter. When I first disassembled the crossover, I didn't have an LCR meter. When I measured it with a multimeter at that time, it showed 0.8 ohms. The air core coil's data sheet says 0.73 ohms.Ok, never mind measuring resistance at 100Hz and 1kHz, just measure it at DC.
I apologise if you already did, but did you subtract the meter lead resistance?When I measured it with a multimeter at that time, it showed 0.8 ohms.
No. I don't know what the resistance of the multimeter wire is. But I don't think there would be an astronomical difference.I apologise if you already did, but did you subtract the meter lead resistance?
I never thought it would have such a high resistance. In this case, the resistance of the iron inductor is very low. But it is very difficult to find an air core inductor with such low resistance.Lead resistance is typically 0.2 to 0.5 ohms.
Now, for low frequency I am using the original inductor and the original capacitor, so for low frequencies the crossover remains the same as it was originally. Here it is not just the resistance of the inductor but also the capacitor that I think is affecting the low frequencies.
About that for a simple crossover. Assuming lowest impedance is near 100Hz that amount would be minimally audible.perhaps about 0.5 dB
Its the sum of the contact resistance of the selector switch, lead plugs and the lead wires themselves. Contact resistances can be surprizingly high - read a few switch datasheets... For low resistances a 4-wire multimeter would be used to get accurate measurements down to milliohms, or you can do the same with a bench supply (with current meter) and a multmeter set to milli-volts - divide multimeter voltage by supply current to read resistance. This takes the leads carrying current out of the measurement, you probe the device-under-test directly for voltage drop.I never thought it would have such a high resistance.
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