I have a 380V PT. Using a bridge rectifier and a capacitor filter, the voltage jumps to about 530Vdc (loaded).
The original design required a 350V PT and gave about 470Vdc (loaded).
Question, can I use a zener between the bridge rectifier and the first filter cap to reduce the DC voltage?
Thanks.
-D
The original design required a 350V PT and gave about 470Vdc (loaded).
Question, can I use a zener between the bridge rectifier and the first filter cap to reduce the DC voltage?
Thanks.
-D
Hi,
Use a C-R-C filter, that'll take care of it
Since the voltage is 530V loaded, you know your current drain. You need to drop it by 60 volts. Therefore your resistor value will be:
R = E/I
or
R(ohms) = 60V/current draw in amps.
The resistor should be rated at least:
P = E*I
or
P(watts) = 60V*current draw in amps
Cheers!
Use a C-R-C filter, that'll take care of it
Since the voltage is 530V loaded, you know your current drain. You need to drop it by 60 volts. Therefore your resistor value will be:
R = E/I
or
R(ohms) = 60V/current draw in amps.
The resistor should be rated at least:
P = E*I
or
P(watts) = 60V*current draw in amps
Cheers!
diaz028 said:I have a 380V PT. Using a bridge rectifier and a capacitor filter, the voltage jumps to about 530Vdc (loaded).
The original design required a 350V PT and gave about 470Vdc (loaded).
Question, can I use a zener between the bridge rectifier and the first filter cap to reduce the DC voltage?
Thanks.
-D
As the other posters indicated, a Zener diode is a BAD idea.
You can achieve your purpose, without generating lots of waste heat. Use CLC filtration. A 10 H. choke and a substantial reservoir (2nd) cap. are needed. Start with 1.5 muF in the 1st filter position and load the PSU down. Increase the amount of capacitance in the 1st position, until the rail voltage comes in right.
Another technique for shedding a few Volts is the use of a hybrid bridge rectifier. The forward drop in the vacuum half of the bridge helps, in this particular situation.
Geek said:Hi,
Use a C-R-C filter, that'll take care of it
Since the voltage is 530V loaded, you know your current drain. You need to drop it by 60 volts. Therefore your resistor value will be:
R = E/I
or
R(ohms) = 60V/current draw in amps.
The resistor should be rated at least:
P = E*I
or
P(watts) = 60V*current draw in amps
Cheers!
Like the schematic I posted?
Hello
You can use LC filter.
LC filter more putdown the Voltage than CLC
LC have soft sound. (have a soft curent waveform and minimal harmonics) sounds better than CLC or CRC,
but if You make stabilisation it's will be better.
Parralel stab better.But this recomandation if you like clear sound and you closed acoustic system.
Many people like painting sound and stabing is cleared sound that bad for them.
You can use LC filter.
LC filter more putdown the Voltage than CLC
LC have soft sound. (have a soft curent waveform and minimal harmonics) sounds better than CLC or CRC,
but if You make stabilisation it's will be better.
Parralel stab better.But this recomandation if you like clear sound and you closed acoustic system.
Many people like painting sound and stabing is cleared sound that bad for them.
Hi,
We're still missing a number for the equation - what's your total current draw?
For now, place a 10 ohm resistor there and measure the voltage across it. This will tell us how much current the circuit wants. Then we can determine the resistor to drop your voltage
You could also try rearranging the circuit as mentioned before, to a L-C (though that may drop it too much).
Cheers!
diaz028 said:What resistor value should I use in this schem?
-D
We're still missing a number for the equation - what's your total current draw?
For now, place a 10 ohm resistor there and measure the voltage across it. This will tell us how much current the circuit wants. Then we can determine the resistor to drop your voltage
You could also try rearranging the circuit as mentioned before, to a L-C (though that may drop it too much).
Cheers!
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