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Old 6th September 2008, 05:33 PM   #1
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Default Input voltage attenuation

Say I have an amplifier that starts clipping with say .667v input and the source component has a fixed output voltage of 2.5v... could I put a simple voltage divider in between? (R1= 2.75K and R2 = 1K).

The amp already has an input attenuator of 100K and the first tube in circuit has a grid stopper of 100R and a grid leak of 470K.


I was also thinking of putting in a selector switch which would have the appropriate R1's to knock down the voltage for sources of 3, 2.5, 2, 1.5 and 1 volt. Different sources have different output voltages and there doesn't seem to be a "standard" value out there and I'd like to be able to accomadate for that variable if possible.

Thanks!

JD
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Old 6th September 2008, 05:36 PM   #2
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Ditch the grid leak. With a 100k pot, it's simply not needed. The idea of a rotary selector switch to obtain proper input sensitivity is a great idea!

All of the amps I build now get PEC pots on the inputs, as I can never be quite sure once I start changing preamps.
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Old 6th September 2008, 05:50 PM   #3
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oops.

meant to say the pot is a 50K attenuator

JD
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Old 6th September 2008, 06:26 PM   #4
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Quote:
Say I have an amplifier that starts clipping with say .667v input and the source component has a fixed output voltage of 2.5v... could I put a simple voltage divider in between? (R1= 2.75K and R2 = 1K).

No, you can't. Commercial products are capable of driving the 10 KOhm IHF "standard" load. The total resistance of the divider can't go below 10 KOhms. 6.19 KOhm and 3.92 KOhm parts should work for you. You "get lucky", in that 3.92 KOhms is OK working into the 50 KOhm pot. already present.

Tiny 1/8 W. metal film resistors inside the RCA plugs, at the amp end, are a good way to do this job.
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Old 6th September 2008, 07:32 PM   #5
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If I'm doing a "selector switch", as long as the total R is around 10K, I'll be ok?

What is the reason for the "luck" with the 3.92KOhm working into the 50K pot?

Say I have a range (for the selector switch) of values that range from 2.49K to 6.98K as R2... how would that work out with luck?

The values of R1 would range from 8.45K down to 3.48K.

This would keep the total of any of the selections at around the 10KOhm standard while giving me around the .67v output I'm looking for for the input voltage target to the amp.

And If I had another amp with a 100K attenuator.. how would that work out?

Also, should I eliminate the grid leak as was suggested?


Thanks !

JD

Quote:
Originally posted by Eli Duttman



No, you can't. Commercial products are capable of driving the 10 KOhm IHF "standard" load. The total resistance of the divider can't go below 10 KOhms. 6.19 KOhm and 3.92 KOhm parts should work for you. You "get lucky", in that 3.92 KOhms is OK working into the 50 KOhm pot. already present.

Tiny 1/8 W. metal film resistors inside the RCA plugs, at the amp end, are a good way to do this job.
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Old 6th September 2008, 07:53 PM   #6
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Quote:
Originally posted by jeffdavison
Also, should I eliminate the grid leak as was suggested
I'd say no. What happens when the pot eventually fails open? Over the years enough crud will build up and the wiper won't touch the track any more.

Keep the grid leak there. Look at the datasheet for the first tube and decide what's the largest value you can put there. I've often seen 1 meg parts used on the common 12A*7 twin triodes, but some of them can handle up to 10 meg.

You want your grid leak to have a higher resistance than the volume pot. A 10:1 ratio is probably a good goal. At the same time, your input network shouldn't have a total resistance less than the recommended 10K. If the input load is too low, the upstream device (preamp, CD, whatever) may have a hard time sourcing enough current to drive it.

I'd just use a 100K pot in parallel with a 1 meg grid leak and call it done. Don't forget your 100 ohm grid stopper.
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Old 6th September 2008, 08:15 PM   #7
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1st tube is a 6SN7. I already am using a 50K attenuator on this amp and the grid leak is 470KOhm...

So I'll just keep them in place

JD

Quote:
Originally posted by Ty_Bower


I'd say no. What happens when the pot eventually fails open? Over the years enough crud will build up and the wiper won't touch the track any more.

Keep the grid leak there. Look at the datasheet for the first tube and decide what's the largest value you can put there. I've often seen 1 meg parts used on the common 12A*7 twin triodes, but some of them can handle up to 10 meg.

You want your grid leak to have a higher resistance than the volume pot. A 10:1 ratio is probably a good goal. At the same time, your input network shouldn't have a total resistance less than the recommended 10K. If the input load is too low, the upstream device (preamp, CD, whatever) may have a hard time sourcing enough current to drive it.

I'd just use a 100K pot in parallel with a 1 meg grid leak and call it done. Don't forget your 100 ohm grid stopper.
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Old 6th September 2008, 08:35 PM   #8
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Quote:
If I'm doing a "selector switch", as long as the total R is around 10K, I'll be ok?
The load presented to the source may not go below 10 KOhms. If the load presented to the source is above 10 KOhms, all is well.

Quote:
What is the reason for the "luck" with the 3.92KOhm working into the 50K pot?
Ty explained the 1:10 impedance rule. 3.92 KOhms is less than 1/10th of 50 KOhms, which is fine. The "tallest" common resistance value that can be used with a 50 KOhm pot. is 4.7 KOhms.

Quote:
Say I have a range (for the selector switch) of values that range from 2.49K to 6.98K as R2... how would that work out with luck?
As this is yet another stepped attenuator situation, a shorting type switch (make before break) is in order. Mouser stock # 105-13572 is a highly suitable and affordable 2 pole Lorlin switch, with 2 to 6 user selectable positions. Make a permanent connection to the pot. from the "bottom" resistor. As the switch gets moved, the amount of attenuation will increase. That is additional resistance gets switched in.
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Old 7th September 2008, 05:24 AM   #9
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So a table like this is good to go?

Vin...........5..........3.75..........2.5........ ..2..........1.5
R1.......... 8660.....8060.........7150.......6190....5110
R2...........1500.....2000.........3010.......3830 ....4990
Vout........ .74....... .75........... .74 ....... .76 ...... .74
Rtotal..... 10160 ....10060 ... 10160 .... 10020 ..... 10100



JD
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Old 7th September 2008, 06:02 AM   #10
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I still vote that you add a zero to all those resistances and keep things at 100k. What are you using for a preamp btw?
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