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Old 30th October 2014, 07:09 AM   #1
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Default lme49810 input impedance

How much is the input impedance of the lme49810 chip? Could it be drive directly with a pot volume ?
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Old 30th October 2014, 08:05 PM   #2
AndrewT is offline AndrewT  Scotland
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the input impedance should be in the datasheet.
But that is not what the vol pot has to drive.
The vol pot sees the Cable and the RF filter and the Rin all in parallel.

The cable could be 20pF (no cable) to 1000pF (7m of 150pF/m cable)
The RF filter could be 100pF to 1500pF
Rin could be from 22k to 1M0.
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Old 3rd November 2014, 09:46 AM   #3
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The LME49810 is not an op amp, it's an amplifier chip. Note that it operates at dangerous voltage levels.

A number such as 30k might be a usable guess for the input impedance (Zin) of the chip itself. Feedback makes the effective circuit impedance several times that. But lacking information from the manufacturer there's no way to know the real number without experimenting--and it seems there's not a lot of wiggle room with this chip.

So hanging a potentiometer (5k? 10k?) on the input seems problematic. It would work, sure, but audio quality is hard to predict.

Note that the manufacturer's suggested input circuit has a Zin of around 7k, which is on the low side. It appears this chip wants to be driven by a solid state preamp, mixer, or similar, these can have output impedances (Zout) in single digits.

Just to mention it, the required heat sink for 300 watts is simply huge, i.e. expensive. Or fan cooled, of course. Ummm...do you really need 300 watts?

If you're in a group bear in mind that 300 watts is not double the sound level of 150 watts, which is not double the sound level of 75 watts. The log law applies to amps, which means there's a diminishing return, at some point the more you pay the less you get. More smaller amps is better than one big amp.

You wouldn't be interested in a nice LM1875? Or for higher output wattage an LM3875, or LM3886?

Data sheet for the LME49810: http://www.ti.com/lit/gpn/lme49810
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Last edited by bentsnake; 3rd November 2014 at 10:09 AM.
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Old 3rd November 2014, 10:47 AM   #4
JMFahey is offline JMFahey  Argentina
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Quote:
Originally Posted by greierasul View Post
How much is the input impedance of the lme49810 chip? Could it be drive directly with a pot volume ?
Input resistor is 33K to ground so a reasonable assumption would be 20/25K.

Yes, you can connect a pot at the input, which will then be in parallel with the input impedance, so, say, a 25K to 50K one will result in the driving signal needing to easily drive, say, at least 10K.

That's not the main problem, but you will need above 1V RMS to fully drive it.

Not a standard signal source level, except some which has built in preamplification, so for general purpose use you will also need a preamp.
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Old 10th December 2014, 04:37 AM   #5
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Originally Posted by JMFahey View Post
Input resistor is 33K to ground so a reasonable assumption would be 20/25K.

Yes, you can connect a pot at the input, which will then be in parallel with the input impedance, so, say, a 25K to 50K one will result in the driving signal needing to easily drive, say, at least 10K.

That's not the main problem, but you will need above 1V RMS to fully drive it.

Not a standard signal source level, except some which has built in preamplification, so for general purpose use you will also need a preamp.

If the output impedance of pre is 100 ohms then will it be able to drive good fidelity even at low volumes?
In the above car what is 20 in 20/25k?
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Old 11th December 2014, 06:52 PM   #6
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Originally Posted by rhythmsandy View Post
If the output impedance of pre is 100 ohms then will it be able to drive good fidelity even at low volumes?
In the above car what is 20 in 20/25k?
The rule of thumb is that input impedance should be ten times the source's output impedance. So 100 ohms output impedance would be OK to drive any input impedance from 1k on up. The volume setting doesn't enter in.

20/25k in this context means "around 20k to 25k."
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Old 11th December 2014, 07:09 PM   #7
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here is a thread in DIY that explains the relationship between input bias current and input impedance. You will note that the input bias current of the opamp in question is 200na which is similar to the LME part according to the data sheet.

Input bias current and impedance of an op amp?
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Old 11th December 2014, 07:36 PM   #8
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I'll ask TI about it. It's odd that this and input bias offset current is missing.
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Old 15th December 2014, 11:26 AM   #9
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The input bias current is stated to be 200nA max and 100nA typical. This tells me that the input structure is bipolar. This also tells me that the input bias current offset will be on the order of nA; most likely in the 10's of nA.


As as SWAG, the input impedance will be close to 100kΩ||5pF
Quote:
Hi Peranders,

Here are some comments I got for you:
"a) Regarding the input impedance, it depends on the operating condition.
i) If it is open loop condition, the input impedance between In+ and In- will be about 5.12k to 10.24k ohms.
ii) If it is closed loop condition, it will be multiply about feedback amount. For example, if the feedback amount is 40dB, the input impedance will be about 100 times of the open loop input impedance.

b) Regarding input bias offset current, it will be about 2% of input bias current. In this case 100uA of input bias current in typical value, so the input bias current offset is about 2nA typ and 4nA max. But this is not guaranteed."

Andy
Here are the answers.... from TI... and 2 to 4 uA. Not nA.
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Old 15th December 2014, 05:15 PM   #10
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Correction: 2-4 nA....
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