6db doubles voltage and adds one S-unit
12db doubles power

50uV is the commonly-accepted antenna voltage for "S9" (50mV is "S9+60", meaning 60db above S9)
0.2uV is S1.  A good receiver can copy signals at 0.1uV, just slightly under S1

Each 6db is double your voltage.  You have to quadruple your voltage to double your power,
so +12db is double your power.

s1 to s9 is 8 S-units, or 6x8=48db.  (256x the voltage, 16x the power)

?? You have to quadruple your power to double your range so +24db is double your range.

db for the purpose of signal level in volts:
1v is +10db
.3v is +0db
.1v is -10db

50uV (S9) is -86db
.2uV (S1) is -134db
.1uV (S0) is -140db
so a standard receiver has an S-meter reading amplification of +140db over db volts

padding S9 down to S1 requires 48db of attenuation, which is 6.3k series, 50.4ohm ground for T pad

the s1/s9 gen uses a 34db pad between S9 and S1, which would only work for S units of 4.25db per,
instead of 6

50uv to 0.2uv is more reliably agreed upon for S9 and S1.

the above is for HF "standards".  VHF uses a different scale for S meters, with S1=0.4uv, 

dbm is used as a fixed-reference db value, where the reference point is 1mW.
dbm=10 x log10(power in watts/.001) or 30 + 10 x log10(power in watts)

10W would be 40dbm
100W would be 50dbm
1500W would be 61.7dbm
dbm are a log product and are therefore added when amplified, instead of multiplied.

so an amplifier may be said to be +10dbm  (typical 10 amp)
10 watts in, is 40dbm, add 10 to get 50dbm out

so an amplifier may be said to be +12dbm
10 watts in, is 40dbm, add 12 to get 52dbm out

to convert dbm to watts:
power in watts = 10^(dbm/10)/1000

50dbm out is 100W
52dbm out is 159W

simulating the S1/S9 box
50uV into the 1.27k/52.3ohm pad

we need to first see how much current is flowing out of the generator.
calculating the resistance network on the pad:
R=1/(1/52.3R)+(1/(52.3R+1270R)) = 50.3R (a little off 50R, but close enough I suppose)

calculating total current flow on the pad input to ground is unnecessary but a good exercise:
E=0.00005V
R=50.3R
I=0.000000994A (very close to 1uA)

calculate current flow through the series and end resistor (unnecessary:
E=0.00005V
R=(52.3R+1270R) = 1322.3R
I=0.000000037A (37pA)

calculate the voltage on the output of the pad
E=.00005/(1270+52.3)*52.3 = 0.000001977V (1.977uV, very close to 2uV)