If you want to have a lower the min volume by quite a margin, and have to buy less units to get a quad matched set, then I would go for NSL32SR2S.
Cheers George
Actually, I'm trying to source 25 pcs (most economical qty) of sr2s from Crest Components in Australia. But it's been two days since I ordered my requirements, and they're still "checking their stock" for the item. I reckon they don't have it and also trying to source from 3rd parties.
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Looks like that amp has a pcb designed for a hybrid, with a pair of transistors in place of a dual triode.
If your in Australia you can get the sr2 from Element 14 seeing your buying 25, you'll get a quad match, maybe two.
http://au.element14.com/advanced-ph...hLookAhead&searchView=table&iscrfnonsku=false
Cheers George
I don't live in Australia George, but I have a cousin who lives in Melbourne who can ship the parts to me. Actually, Crest has the sr2 per their website (like the sr2s) and which is about $3 cheaper than Farnell/Element14. I can only hope Crest would have the sr2, i.e. if no sr2s.
But a question though, if connecting in parallel the SR3 may solve the impedance issue, would that remove the matching requirement as well?
Sorry, am not an electronics guy so I'm kinda slow with these stuff.
if you use multiple devices rather than the usual two, then you will find that the shunt pair can use the LDR in parallel and the series pair can use the LDR in series.
The LED parts are still in series, i.e. the same current passes through the LEDs.
The reason I recommend this is based on the effects of unwanted resistance in the series side or shunt side when at the extremes of the vol pot rotation.
Take maximum volume first.
The two series LDRs are at minimum resistance, maybe 100r each.
The two shunt LDRs are at maximum resistance, maybe 30k each.
The series pair in the series position add up to 200r.
The parallel pair in the shunt position add up to 15k.
The attenuation is 15k/{15k+200+Rs} if Rs is low say 200ohms, then the attenuation comes out at ~ -0.23dB
Now I'll look at the minimum volume:
The two series LDRs are at maximum resistance, maybe 30k each.
The two shunt LDRs are at minimum resistance, maybe 40r each.
The series pair in the series position add up to 60k.
The parallel pair in the shunt position add up to 20r.
The attenuation is 20r/{20r+60k+Rs} ~-69.6dB
The max is hardly affected by the two 100r LDR being in series
and the min is well down @ nearly -70dB without stressing the LDRs with excessive LED light.
Based on the above, select your matched pairs and find which gives the lowest LDR for the least LED current. Use these low LED current in the shunt side. Use the two that require the slightly higher current in the series position. That is why I show slightly different LDR values of 40r and 100r. I assumed the 40r LDR were the selected pair that required the least LED current.
The LED parts are still in series, i.e. the same current passes through the LEDs.
The reason I recommend this is based on the effects of unwanted resistance in the series side or shunt side when at the extremes of the vol pot rotation.
Take maximum volume first.
The two series LDRs are at minimum resistance, maybe 100r each.
The two shunt LDRs are at maximum resistance, maybe 30k each.
The series pair in the series position add up to 200r.
The parallel pair in the shunt position add up to 15k.
The attenuation is 15k/{15k+200+Rs} if Rs is low say 200ohms, then the attenuation comes out at ~ -0.23dB
Now I'll look at the minimum volume:
The two series LDRs are at maximum resistance, maybe 30k each.
The two shunt LDRs are at minimum resistance, maybe 40r each.
The series pair in the series position add up to 60k.
The parallel pair in the shunt position add up to 20r.
The attenuation is 20r/{20r+60k+Rs} ~-69.6dB
The max is hardly affected by the two 100r LDR being in series
and the min is well down @ nearly -70dB without stressing the LDRs with excessive LED light.
Based on the above, select your matched pairs and find which gives the lowest LDR for the least LED current. Use these low LED current in the shunt side. Use the two that require the slightly higher current in the series position. That is why I show slightly different LDR values of 40r and 100r. I assumed the 40r LDR were the selected pair that required the least LED current.
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I finally got my 25 pcs of the NSL32SR2S from Crest Australia. And I have my Kemo tester ready (new, yellow version). From the first post, I gather the testing should be up to 20mA only (?) as the new Kemo tester has 35mA and 50mA.
Since it is a range of mA, to do the matching do you just look for the least average of the measured differences in resistance across the mAs for the 25 pcs or just average the resistances measured and compare each? Thanks.
These are resistance values measured at the LDR side at various milliamps slots available in the tester. Ignore the first column, it is just the identifier for each LDR.
Using an LDR attenuator you will find that most of your resistances are between 500ohms and 20k ohms.
You have only looked at resistance values below 200ohms.
That covers a range of attenuation you will hardly ever use.
As I said earlier try checking tracking between your devices of a matched pair for currents from 10mA down to 0.01mA
eg
assume you have 4k of series resistance and 4k of shunt resistance feeding into an Rin of 47k.
That gives you an attenuation of ~6.4dB
Move to 10k for series and 600ohms for shunt and the attenuation becomes -25dB
You are down to -25dB on your volume attenuator and your lowest (required) resistance is still three times higher than any of your measurements.
You have only looked at resistance values below 200ohms.
That covers a range of attenuation you will hardly ever use.
As I said earlier try checking tracking between your devices of a matched pair for currents from 10mA down to 0.01mA
eg
assume you have 4k of series resistance and 4k of shunt resistance feeding into an Rin of 47k.
That gives you an attenuation of ~6.4dB
Move to 10k for series and 600ohms for shunt and the attenuation becomes -25dB
You are down to -25dB on your volume attenuator and your lowest (required) resistance is still three times higher than any of your measurements.
if you choose to have an LDR as the series element and an LDR as a the shunt element.
Then these two parts get you some attenuation.
If both are ~100ohms the the combination gives you a Rin of 200ohms. They will blow up when they receive a loud signal.
I have seen mention of 7k for Rin.
That means the shunt plus the series elements add up to ~7000ohms.
For half volume (-6.02dB) the upper (series) and lower (shunt) need to both be about 3500ohms.
I gave examples in post5596.
Do you know how a vol pot works?
Then these two parts get you some attenuation.
If both are ~100ohms the the combination gives you a Rin of 200ohms. They will blow up when they receive a loud signal.
I have seen mention of 7k for Rin.
That means the shunt plus the series elements add up to ~7000ohms.
For half volume (-6.02dB) the upper (series) and lower (shunt) need to both be about 3500ohms.
I gave examples in post5596.
Do you know how a vol pot works?
I don't have anything readily available for measuring down to .01mA. I only have the Kemo tester for now. It might take a while.
Yes, I think I know how a volume pot works. But am not really an experienced electronics DIY guy. I just thought this project is easy enough for my skill level. Thanks.
Yes, I think I know how a volume pot works. But am not really an experienced electronics DIY guy. I just thought this project is easy enough for my skill level. Thanks.