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Old 28th July 2011, 08:05 PM   #21
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Yeah, I've just been looking on the Farnell website, and you can get 20nF film caps with a 2.5% tolerance very cheaply. What kind of power supply is usually used to power this kind of circuit? I'm thinking of a transformer with 2 x 0-12 outputs to give me 12 - 0 - -12 output. Then feed this into 2 x bridge rectifiers and use caps on the output for smoothing. Is this the best way or is there something simpler?
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Old 29th July 2011, 04:02 AM   #22
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There is no simpler way, but you should use a voltage regulator after the smoothing caps.

A transformer with 2x12 VAC will give you ~16-17 VDC, even more, when the load is low. You can also build a split supply with a single rectifier.

It is also possible to add the voltage regulator to the power supply you already have and save the additional transformer, rectifier, smoothing caps, fuses, etc. You need to watch the heat dissipation then, because the voltage difference might be quite high. So you will either need heatsinks for the regulators or add a power resistor in series between the amp power supply and each regulator or cascade several regulators. The resistor is the cheapest and easiest method.
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Old 29th July 2011, 09:57 AM   #23
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Thanks pacificblue, never thought of using resistors and using my existing supply. So I just worked out if I used 120ohm resistors from my 30v rails, thats a max current of .25A and 7.5 watts. 7.5 watts should be easily enough to power the active filter. So 120ohm 10watt resistors and +12 and -12 regulators seems a nice and cheap way to power the filter.
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Old 29th July 2011, 11:11 AM   #24
AndrewT is offline AndrewT  Scotland
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Choose your topology first. Then choose your opamp and if need be adjust the topology.
Then choose your supply voltage to suit the opamp and to suit the maximum signal + dB signal overhead. Finally choose the regulator and transformer/supply.

You may decide to use +-5Vdc to suit a low voltage rail to rail opamp or to use +-21.5Vdc with 44V opamp to maximise the signal overhead.
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Last edited by AndrewT; 29th July 2011 at 11:38 AM.
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Old 29th July 2011, 11:23 AM   #25
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Thanks for the help once again. I'm currently reading about filters, there is a lot to learn. I'm an electrical engineer by trade and also do a lot of digital PIC stuff. Havn't really done much with analogue since college.
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Old 29th July 2011, 07:01 PM   #26
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Quote:
Originally Posted by portreathbeach View Post
So I just worked out if I used 120ohm resistors from my 30v rails, thats a max current of .25A and 7.5 watts. 7.5 watts should be easily enough to power the active filter. So 120ohm 10watt resistors and +12 and -12 regulators seems a nice and cheap way to power the filter.
You will need a few V more before the regulator than you can get out of them. E.g. a 12 V regulator will need more than 15 V at its input to work properly. Assuming your 30 V rails are ideally stable the resistor must drop 15 V which means 120 Ohm will work for up to 125 mA. That leaves you with 1,5 W after the regulators and 1,875 W of power dissipation across the resistor.
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Old 29th July 2011, 08:16 PM   #27
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Thanks for that. I've been looking at the LM388 and the maximum current is 8mA. 5 of these would only draw 40mA, so the 120ohm resistors should work fine.
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Old 30th July 2011, 09:04 AM   #28
AndrewT is offline AndrewT  Scotland
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Hi,
I don't have the datasheet but <=8mA for max current does not sound right.
Is that 8mA the max spec for quiescent current?

Look for the maximum output current and then add the quiescent current to get a rough idea of worst case current draw by the opamp.
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Old 30th July 2011, 09:35 AM   #29
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Don't buy the LM388. It is a power amp.

For the LM833 8 mA is the max supply current at 0 V output. 85 mA appear to be a quite comfortable headroom for the output currents with all op amps working at line voltage levels into impedances of several to many kOhms.
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Old 30th July 2011, 09:39 AM   #30
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What op amp do you suggest for the circuit?
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