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Old 2nd February 2013, 07:04 AM   #21
Mooly is online now Mooly  United Kingdom
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Here are two ways to do it. Switching grounds is never a good idea and the opamp needs a constant ground referenced bias to maintain the DC conditions.

The first one attenuates the input and the second the output. Both have there merits and disadvantages.
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Old 2nd February 2013, 09:57 AM   #22
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I would tend towards the circuit that attenuates the output, since none of
the orig. signal is lost. Wouldn't that one be more accurate?

What are the two resistors, R5 and R7 all about?

Here's an updated schematic w/o the coupling cap on the input.

Thank you Mooly
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Old 2nd February 2013, 11:12 AM   #23
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The inputs are AC coupled for both circuits. That is good practice for most circuits because you just don't know what might get applied. The 470K defines the DC conditions for pin 3. It could be 1ohm or 100Meg ohm as far as the DC conditions are concerned. Make it to low though, and it will lower the input impedance. So 470K and 0.47uf is a good compromise. Together they form a high pass filter with a -3db point of 0.72Hz. The attenuator values (the two 10K's) are outside this filter and so can be altered at will.

The second circuit is similar at the opamp input side. The disadvantage of this circuit is that if the input is so large (such that it cause the opamp to clip) then no amount of attenuation on the output can correct that because the signal has already been distorted. The first one can attenuate any level on input and provide a distortion free output.

The 68 ohm isolates the opamp output from any capacitive loading. Some opamps can become unstable with light capacitive loads (sometimes oscillating at many 10's of Mhz). Its just good practice to include it unless you are 100% sure no issue exists by leaving it out.
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Old 2nd February 2013, 12:47 PM   #24
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What kind of an input impedance are we talking about here?
Ignoring the switch/divider network stuff, i calc. the impedance of your C1/R3 filter to be about 470k over the audio range which seems sufficiently high.

On of the reasons for using the op amp, besides adjustable gain, was as an impedance matcher for the drv134.

Ok next try..... This time i favor the 1st circuit after your functional explanation. After all, its all about attenuating an input that is too high w/o distortion/clipping! With the RC network on the input, guess i wouldn't need any additional capacitor here.

Because the op amp will be driving a drv134, which is close to the op amp, maybe i don't have to worry about capacitive loads, e.g. could omit the R4?

Btw. when the switch is closed, e.g. R1 is bypassed, don't I lose part of the signal over R2? But i guess even with R3 tied to ground, switching R2 away from its ground would also change the gnd. potential a bit, which as you pointed it is not a good idea.

Again many thanx Mooly!
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Old 2nd February 2013, 06:12 PM   #25
Mooly is online now Mooly  United Kingdom
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Quote:
Originally Posted by chris2 View Post
What kind of an input impedance are we talking about here?
Ignoring the switch/divider network stuff, i calc. the impedance of your C1/R3 filter to be about 470k over the audio range which seems sufficiently high.
Yes, 470k is the input impedance of the opamp part of things.

The first circuit adds either 20k (10k +10 k) or just 10k (depending on switch position) across the 470k. In practice that difference is negligable.

So there's a "downside" to the first approach. The input impedance is non constant and "lowish". If the attenuator is fed from normal modern line sources such as CD players/DVD etc then its not a problem because all these can drive loads of that value easily.

Quote:
Originally Posted by chris2 View Post
Ok next try..... This time i favor the 1st circuit after your functional explanation. After all, its all about attenuating an input that is too high w/o distortion/clipping! With the RC network on the input, guess i wouldn't need any additional capacitor here.
You should always have the capacitor Suppose the source had a significant DC offset. You don't want that passed down the chain.

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Originally Posted by chris2 View Post
Because the op amp will be driving a drv134, which is close to the op amp, maybe i don't have to worry about capacitive loads, e.g. could omit the R4?
Maybe The IC itself will present a capacitive load due to the internal semiconductor junctions... it may be a few 10's of pf's or several hundred pf. I have no idea unless the data sheet mentions it specifically. So good practice... assume there could be a problem.

Quote:
Originally Posted by chris2 View Post
Btw. when the switch is closed, e.g. R1 is bypassed, don't I lose part of the signal over R2? But i guess even with R3 tied to ground, switching R2 away from its ground would also change the gnd. potential a bit, which as you pointed it is not a good idea.
With the switch closed all the voltage appears across R2. It doesn't lose any in that sense. If the source is capable of driving a 10k (such as the CD/DVD players etc) then all the output voltage of the player appears across R2
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Old 2nd February 2013, 07:55 PM   #26
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Quote:
Originally Posted by Mooly View Post
So there's a "downside" to the first approach. The input impedance is non constant and "lowish". If the attenuator is fed from normal modern line sources such as CD players/DVD etc then its not a problem because all these can drive loads of that value easily.
I will be driving this mainly from a preamp with output impedance of 47 ohms. I read somewhere that a good matching of input to output impedance is 100:1, so if that's true, in my case its 10,000:1


Quote:
Maybe The IC itself will present a capacitive load due to the internal semiconductor junctions... it may be a few 10's of pf's or several hundred pf. I have no idea unless the data sheet mentions it specifically. So good practice... assume there could be a problem.
The data sheet for the drv134 states the following:
Input Impedance: 10k
Should be driven by a low impedance source such as an op amp or buffer
Output Impedance: 50 ohms
Load Capacitance, Stable Operation, CL Tied to Ground (each output): 1uF
Not sure if that's "exactly" what u meant?
Short-Circuit Current: +- 85mA

Does the Short-Circuit Current spec. mean that both output signals, when shorted to gnd. pull 170mA max? In other words the absolute max current draw of the driver. Or do i got that all wrong?

Cheers,
Chris
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Old 2nd February 2013, 08:25 PM   #27
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Chris, please be careful with the quote function. I have cleaned up a few of your posts.
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Old 2nd February 2013, 08:47 PM   #28
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Sure Pano,

pls. tell me what it is I'm doing wrong?
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Old 3rd February 2013, 06:37 AM   #29
Mooly is online now Mooly  United Kingdom
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If the data sheet suggests it should be driven by a buffer then the first version is the one to go for. Keep the 68 ohm though.

I would take the short circuit current to be as you say in that each output can source or sink around 85ma maximum.

Your preamp sounds fine to drive any version. What really matters is whether the preamp can swing its full output voltage across the connected attenuator. For example a TL071 type opamp would allow a minimum load of around 2K and still be able to deliver full output voltage. Anything below that and the limited current drive ability of that opamp would limit the total outout available. Something like the OPA134 would happily drive 600 ohm load to full output.
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Old 3rd February 2013, 10:00 AM   #30
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Hi Mooly,

here are the lineout-analog specs of my preamp, from the manual:
output imp.: 47 ohms
output range, e.g. what i see on the display when turned down/up all the way: -90db/+15db
max analog out voltage: > 3.5 vrms or (9.0 vrms on subwoofer out)
analog freq. range: 2Hz - 20kHz.

I also took some measurements on the line out, playing a whole "track" from itunes on my computer.
itunes was turned up to about half way and my preamp was turned up to the max. volume setting.

Here's what i got:
Min voltage: 0.4
Max voltage: 620 mV
Avg. voltage: 260 mV

Seems to be a big diff., e.g. approx. 5 times, between whats specified as the max voltage and what i measured on some arbitrary music...

Are those numbers to be understood as peak to peak values, e.g. 3.5 => -1.75 - +1.75?

Cheers,
Chris
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