Lightspeed Attenuator a new passive preamp

[AndrewT]

and Bob Pease wrote this

But to get high ZOUT, you usually do have to trim.

For a 1-mA output (all R’s = 10k ±1%), the ZOUT might be as poor as 0.25 MO, +0.25 MO or even –0.25 MO. So, you have to trim. (See the little trim on Figure 2. Even 0.1% resistors would only provide a moderate improvement, to ±2.5 MO.) That App Note also shows how to get high ZOUT without any pots.

[wapo54001]

It appears that I misspoke in my remarks about the AD8277 -- at 5V rail it is current limited to 10ma, not 15ma as I said earlier. This I am gathering from a re-read of the spec sheet.

However, this is still fine for my purposes. For LDRs simulating a 10K pot driving a 10K or larger load, the series LDR doesn't have to go below 200 ohms, and for the shunt LDRs, two LDRs in series/parallel at 10ma or less will deliver 25 ohms. When I tested a dozen LDRs some time ago, ten out of twelve reached 50 ohms at less than 10ma; only two required more current for 50 ohms. Those two can be used as series LDRs. This means that almost all LDRs should be usable in my circuit design, but a little selection will be necessary to insure that all shunt LDRs will achieve 50 ohms or less at 10ma or less.

I really like it that the entire circuit can be operated at 10ma or less because that means the LDRs will never come even close to being stressed enough to change performance characteristics due to heavy current passing through the LEDs.

[AndrewT]

Change R1 to set the new maximum output current.

Io(max) = Vref(max) / R1

[wapo54001]

Originally Posted by AndrewT

Change R1 to set the new maximum output current.

Io(max) = Vref(max) / R1

Not sure if you're saying the current can be higher than 10ma -- page 4 of spec sheet: Short-circuit current limit of +/-10ma for Vs = +2.7V to <+/-5V. Previously, I had been looking at the spec for Vs above +/-5V.

I think that means that in order to use this chip to drive an LDR "pot," you have to use two LDRs in the shunt leg in order to achieve any reasonable minimum resistance. Easily done with Vs > 4V and a current-controlled circuit.

[AndrewT]

Use the discrete version with a soic opamp and 4 precision smd 805 resistors.
or
Use the AD chip and add the discrete transistor (Fig1 or Fig5) to increase the maximum output current.
Go and read the AD paper you linked.

[costis_n]

I am planning to make a passive pre with 100K input resistance.
It will be driven by normal, mpdern sources (DAC, Tuner & Phono pre)
It will drive a valve input stage with an east 220K input resistance.
The source to in to out cables will be less than 1 meter , with around 140pf capacitance each (shielded twisted pair).

Am I okay regarding high freq. response?
Should I modify the input resistance of the active strage?

Thanks to all

[AndrewT]

the 220k input impedance will work very well with a Source impedance of <= 220k/20, i.e. <=11k.
A 100k vol pot has a range of output impedance of [100k +Rs]/4 to zero ohms.
if your Source has an output impedance of 550r then the vol pot presents a maximum source impedance to the valve amp of [100k+500r]/4 = 25k125.

That exceeds 11k by quite a margin.
You will get good signal transfer if the ratio of input impedance:source impedance is ~10:1 and gets pooer by the time you are down to 5:1.
You 100k with the valve amp gives 220k:25k ~ 9:1
Usable, but not ideal.

[costis_n]

Thank you for the number crunching!

So I can up the input stage to, say, 390K, or make the pre to be 50K (it is a TubeCad Stepped series/ladder)

I believe lowering the pre's impedance would be preferable, am I correct?

[AndrewT]

Where is the relevance to Lightspeed?

[costis_n]

None.