Hi, I'm building a 2 way crossover and i'm stuck at the final step. According to the calculation i need a 5.32 Ohms & 4.03 Ohms resistors for the L-pad to give 180 watts to the horn driver. I read that the resistors wattage needs to add up to the wattage that its going to give out, is that correct?
Thanks for the help in advance
Thanks for the help in advance
Its my first time building a 2 way crossover. The horn driver is the RCF N850 crossed at 1400 Hz.
The Low Frequency driver is the RCF MB15H401 pushing 1400 watts program.
What is the application for this system -will you be actually "pushing" that much power through the drivers, or is that simply their maximum rated handling before the appearance of magic smoke?
1400 W of peak / program into a 100dB sensitivity woofer, or max of rated 180W program into the compression driver with sensitivity of 109dB will be PFL
1400 W of peak / program into a 100dB sensitivity woofer, or max of rated 180W program into the compression driver with sensitivity of 109dB will be PFL
Its my first time building a 2 way Xover. The compression driver is the RCF N850 crossed at 1400hz. The driver is rated at 180 watts above 1200 hz.
For the low frequency, im using the RCF MB15H401 rated at 1400 watts program. According to a few forms i read, i need to lower the horn sensitivity to match the woofer sensitivity (109db - 100db = 9db) I entered the information into a L-Pad calculator and i need R1 = 5.16 Ohms & R2 = 4.4 Ohms to reduce 1400 watts to 176 watts. (this is rated at program power) At Aes power power, 700 watts will be reduce to 88.12 watts.
My question is, does the wattage on the resistors needs to add up to the output its going to put out? and do i work with the program wattage or Aes wattage?
this is for a 2 way pro audio cabinet, 1400 is the program power, 2800 will be peak. I'm not going to actually push 1400 wattage to the cabinets, i might push 700 watts, but i just want to build the xover to handle 1400 wattage program power.
Forget about how many wattage i need or will be using, my question is, does the resistors wattage needs to add up to the wattage that it will put out? If yes, do I use the AES power of 700 watts or program power of 1400 watts do determine the wattage of the resistors?
Accordingto an L-Pad calculator i need Resistor1 = 5.16 Ohms & Resistor2 = 4.4 Ohms to reduce 1400 watts to 176 watts. (this is rated at program power) At Aes power power, 700 watts will be reduce to 88.12 watts.
Actually, you can't really "forget about how many wattage ..will be used" (sic), but Cal is more well versed in PA /pro audio systems than the usual denizens of the "Full-Range" driver forum in which this was originally posted, and his answer should be reliable.
Be aware however, that the big systems that he's trucked over to our annual get togethers have almost exclusively used active cross-overs over and bi or tri-amped. There's very good reasons why that approach is used in such set-ups.
Be aware however, that the big systems that he's trucked over to our annual get togethers have almost exclusively used active cross-overs over and bi or tri-amped. There's very good reasons why that approach is used in such set-ups.
If you have any doubts, double the size of the resistor and parallel them. IOW, instead of one 4 ohm, use two 8 ohms. You can also use small light bulbs in series to control the power. The higher the wattage, the more resistance. They are often used in PA gear so when you see them starting to light, you know you are getting close to the max.
"20 watt resistors should be fine. "
Best laugh I have had in quite a while!
100W resistors would be best. I use multiple 25W types from Madisound. A 250W Hafler DH500 slagged the original single 25W resistors so bad there was smoke rolling out of the cabinet, solder ran out of the joints, and the cabinet charred.
"small light bulbs in series ....the higher the wattage, the more resistance. "
Backwards.
A #1156 lamp makes a pretty good protector for most 8Ω compression drivers. The cold resistance is on the order of 0.3Ω, hot it becomes about 6Ω or so. On musical transients it will not heat up enough to change the sound, yet will protect the driver from a blast of feedback.
Best laugh I have had in quite a while!
100W resistors would be best. I use multiple 25W types from Madisound. A 250W Hafler DH500 slagged the original single 25W resistors so bad there was smoke rolling out of the cabinet, solder ran out of the joints, and the cabinet charred.
"small light bulbs in series ....the higher the wattage, the more resistance. "
Backwards.
A #1156 lamp makes a pretty good protector for most 8Ω compression drivers. The cold resistance is on the order of 0.3Ω, hot it becomes about 6Ω or so. On musical transients it will not heat up enough to change the sound, yet will protect the driver from a blast of feedback.
How big an amplifier are you using? Let's do a little math... It you run a 1400W amp, you will be getting about 467 watts of AVERAGE POWER out of it right about the time it's turned up so you can barely tell what song is playing anymore. Equal power split between low and high is usually about 300 Hz, falling at 3dB per octave. At 1200 Hz, that's 1/8 of the power, or 58 watts to the highs, the rest to the woofer. Assume double that because you're clipping badly forcing more power to the highs. So 116 watts. Attenuate by 9 dB and you put 15 watts to the horn driver. And 101 watts into the L-pad. At more sensible volumes those power figures decrease. For typical music you'd be dropping that by a factor of 4. Still, a 25 watt resistor is iffy - at full power they reach a case temperature of 250 C (!). Put a little acoustic damping material next to it, and "How about a little fire, Scarecrow?"
A full power feedback tone could be as high as 176 watts with a 9 dB pad. With the 1156 in series with it, it will drop about 5 dB when hot, putting 50 watts to the driver. Most 1" and all 2" will take that for at least the time it takes to get to the fader and save your ears.
A full power feedback tone could be as high as 176 watts with a 9 dB pad. With the 1156 in series with it, it will drop about 5 dB when hot, putting 50 watts to the driver. Most 1" and all 2" will take that for at least the time it takes to get to the fader and save your ears.
"Put a little acoustic damping material next to it, and "How about a little fire, Scarecrow?"
Seen that in some Yamaha club series, the yellow fiberglass caught on fire.
"Again, if he uses 20 watt resistors, I doubt very much he will ever blow them."
No, the solder just ran out of the joints and the cabinet charred, but the resistors still worked.
Most wirewound resistors are specified for a 250°C rise over ambient, solder melts far lower than that, and wood burns lower too.
Use whatever you like.
I rebuilt the networks in question three times before they would hold up, the first two times the inductors also caught fire, and the solder joints on the BeCu voice-coil leadout wire of the compression driver melted too (I repaired with a higher temp solder).
Seen that in some Yamaha club series, the yellow fiberglass caught on fire.
"Again, if he uses 20 watt resistors, I doubt very much he will ever blow them."
No, the solder just ran out of the joints and the cabinet charred, but the resistors still worked.
Most wirewound resistors are specified for a 250°C rise over ambient, solder melts far lower than that, and wood burns lower too.
Use whatever you like.
I rebuilt the networks in question three times before they would hold up, the first two times the inductors also caught fire, and the solder joints on the BeCu voice-coil leadout wire of the compression driver melted too (I repaired with a higher temp solder).
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