Lee
07-16-2002, 06:08 AM
5.15 Can I split the single pre-amp output from my head unit to drive two amplifiers with a Y-cable? [IDB][This section was written by someone who wishes to remain anonymous, but I will field any questions on the subject -IDB]
Yes. When two loads are connected in parallel (such as with a Y-cable) they get the same voltage as each other. They do NOT get the same voltage as if only one load was connected because the head-unit has an internal resistance (typically around 600 ohms). So, given that the amp has a typical input impedance of around 10k ohms then we get something like this:
----------------------------- ----------------------------
HEAD UNIT ________ | | AMP |
______| |_________Vamp___________ |
| | R(head)| | | | | _ |
__|__ |________| | | __|___ |__| - _ |
/ \ | | | | | -___|__
| Vi | | | |R(amp)| | _- |
\_____/ | | |______| __| _- |
|_______________________________|________| - |
| | |
----------------------------- ----------------------------
for the single amp situation. Please realize that the R(head) and R(amp) are internal to the head unit and amplifier and in fact are not deliberately added resistors but are characteristic of the real world circuits (non-ideal) in the head-unit and amplifier (and eq's, etc.). These numbers are typical, check your specific equipment for its particular specs. the worst case situation is a high source output impedance and low load input impedance.
So, assuming a typical head unit and single amp the voltage seen at the amp (Vamp) is given by (Ohms law/Kirkov's law/1st year EE/high school electronics technology class/etc.):
R(amp)
Vamp1 = Vi * ------------------
R(amp) + R(head)
Vamp1 = Vi * 0.94
Now, putting two amps in parallel from the original signal, R(amp) is effectively halved while R(head) is unchanged. Using the same voltage divider formula we get:
10000/2
Vamp2 = Vi * ---------------------
10000/2 + 600
Vamp2 = Vi * 0.89
So, for an Alpine 4V preout, Vi in the diagram (the open circuit head unit line level output) is 4V. Thus Vamp1 = 3.76V and Vamp2 = 3.56V. With two amplifiers' inputs connected in parallel, the voltage is reduced from 3.76V to 3.56V or approximately 5%, not a big deal.
If you had a more typical 1V preout you would get Vamp1 = 0.95V and Vamp2 = .89V, also not a noticeable drop.
This is also why this is slightly more susceptible to noise than a direct one-to-one connection. If the noise level inserted due to cabling was 0.1V per cable then the noise level in the signal reaching each of the two amps would be a slightly higher percent of the signal level but not doubled. (this is also why the 4V head unit is favored over the 1V unit for noise immunity: 0.1V noise / 3.76V or 3% is much less than 0.1V noise / 0.95V or 10% even in a one to one connection).
Yes. When two loads are connected in parallel (such as with a Y-cable) they get the same voltage as each other. They do NOT get the same voltage as if only one load was connected because the head-unit has an internal resistance (typically around 600 ohms). So, given that the amp has a typical input impedance of around 10k ohms then we get something like this:
----------------------------- ----------------------------
HEAD UNIT ________ | | AMP |
______| |_________Vamp___________ |
| | R(head)| | | | | _ |
__|__ |________| | | __|___ |__| - _ |
/ \ | | | | | -___|__
| Vi | | | |R(amp)| | _- |
\_____/ | | |______| __| _- |
|_______________________________|________| - |
| | |
----------------------------- ----------------------------
for the single amp situation. Please realize that the R(head) and R(amp) are internal to the head unit and amplifier and in fact are not deliberately added resistors but are characteristic of the real world circuits (non-ideal) in the head-unit and amplifier (and eq's, etc.). These numbers are typical, check your specific equipment for its particular specs. the worst case situation is a high source output impedance and low load input impedance.
So, assuming a typical head unit and single amp the voltage seen at the amp (Vamp) is given by (Ohms law/Kirkov's law/1st year EE/high school electronics technology class/etc.):
R(amp)
Vamp1 = Vi * ------------------
R(amp) + R(head)
Vamp1 = Vi * 0.94
Now, putting two amps in parallel from the original signal, R(amp) is effectively halved while R(head) is unchanged. Using the same voltage divider formula we get:
10000/2
Vamp2 = Vi * ---------------------
10000/2 + 600
Vamp2 = Vi * 0.89
So, for an Alpine 4V preout, Vi in the diagram (the open circuit head unit line level output) is 4V. Thus Vamp1 = 3.76V and Vamp2 = 3.56V. With two amplifiers' inputs connected in parallel, the voltage is reduced from 3.76V to 3.56V or approximately 5%, not a big deal.
If you had a more typical 1V preout you would get Vamp1 = 0.95V and Vamp2 = .89V, also not a noticeable drop.
This is also why this is slightly more susceptible to noise than a direct one-to-one connection. If the noise level inserted due to cabling was 0.1V per cable then the noise level in the signal reaching each of the two amps would be a slightly higher percent of the signal level but not doubled. (this is also why the 4V head unit is favored over the 1V unit for noise immunity: 0.1V noise / 3.76V or 3% is much less than 0.1V noise / 0.95V or 10% even in a one to one connection).