Please Check my Pad Math

[toddjal]

Ok, so I'm experimenting with building an L pad on an unbalanced input that the manual says is 10k. I'd like to know what resistors to use in the scenarios of 10db, 15db, and 20db. After all of my reading and researching the numbers just don't look right. These equations came from here.
So....

For 10db
K=10^(db/20)
K=3.162
If Rshunt=10k or 10000 then...
Rseries=(k-1)*Rshunt=(3.162-1)*10000
Rseries=21.62k

Those are some pretty big numbers aren't they? Can this be right? With the same shunt of 10k I ended up with 46.2k and 90k for 15 and 20db respectively. Am I missing something? Maybe I need to get a real reading on the input with a meter but I don't know how to use the meter I was given. Would the unit need to be powered on or off for such a reading? These 2 resistors are causing a lot of problems. Any help would be tremendously appreciated.

Todd

[The Scum]

You math looks good to me.

Keep in mind that resistors are somewhat magical components: they come in a very wide range of values, but the bigger values don't cost any more than the smaller values (whereas caps or inductors get more expensive for larger values.).

A quick check for sensibility: look at your 20 dB pad, with it's 10:1 voltage ratio. The total resistance is 10K + 90K = 100K. So if you think about it, 10K is 1/10th of the way up the total resistance, so it gets 1/10th of the voltage.

But that's assuming it's unloaded. With the 10K input behind it, you effectively have two 10K resistances in parallel. 10K || 10K = 5K. The input is loading the pad, and will skew it's attenuation. To prevent this, try smaller values of Rshunt...say 2.2K, or even 1K.

[toddjal]

Thankyou soo much. That makes sense and helps a lot. Thanks

[brianroth]

Run the math....

20 K in series with a 10K "shunt" means that 1/3 of the original voltage is available across the "shunt". IOW, 1 Volt into the divider results in 0.33V across the shunt. Break up the divider string into "three thirds"; two thirds voltage across the series R, and 1/3 across the shunt.

Then, if you use the dB formula (for voltage):

20 (log Vout/Vin):

20 * log (.333)

you end up at just under 10 dB loss.

Bri

[nclayton]

If you can do it with your switch(es) it's probably better to switch in a series resistor AND an additional shunt resistor in parallel with the input impedance to ground. That way you don't have to use resistors as large, the input impedance (looking into the pad) can stay roughly constant, and the impedance driving the REAL input (going out of the pad) will be lower, which means less noise and better frequency response if there's any stray reactance in the input (like miller capacitance, etc.).

For example, for a 10dB pad you could have a 6.8K ohm series resistor followed by a 4.7K shunt (which makes 3197 when paralleled with 10K). Now your input impedance with the pad in place is 10K (6.8K + 3197), your driving impedance is about 2.8K (6.8K || 4.7K. Highish, but much better than 21K or whatever) and you get a 9.9 dB pad: close enough.



Ned