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 Post subject: CLC filter questions.
PostPosted: Oct Fri 11, 2019 7:25 pm 
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Joined: Jan Fri 04, 2019 11:32 pm
Posts: 34
Location: Sheridan, Wyoming
I have an isolation transformer, 115/115 @35VA. I am using it as a power transformer, feeding a 4 diode bridge rectifier, and am getting a 2 amp,118 volt output from the bridge rectifier{ without a load on it}.I then have this current going through a 47mf electrolytic cap, a choke coil of 5 henry, 150 milliamps, another 47mf electrolytic cap, then to the load. I am having troubles. It looks to me like the 150 milliamp choke cannot handle the 2 amps. I have another choke of 225 milliamps. I could use this one if needed. My question is this: Would a simple resistor be more useful here, rather than the coil? If so, what value resistor would work?Any help appreciated........William


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Fri 11, 2019 7:37 pm 
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Location: Livermore, CA
If you want a choke to handle 2 amps it should be rated 2000 ma (2 amps) or more. A lower rated choke will have too high of resistance and heat in use.

Could use a resistor instead of choke. Every ohm will drop 2 volts at 2 amps. With that kind of current filter caps should be much higher in value.

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 Post subject: Re: CLC filter questions.
PostPosted: Oct Fri 11, 2019 7:43 pm 
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Joined: Jan Fri 04, 2019 11:32 pm
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Location: Sheridan, Wyoming
Thanks Norm.


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Fri 11, 2019 8:12 pm 
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williampage wrote:
I have an isolation transformer, 115/115 @35VA. I am using it as a power transformer, feeding a 4 diode bridge rectifier, and am getting a 2 amp,118 volt output from the bridge rectifier{ without a load on it}.


Maybe my dyslexia is messing with me, but I don't understand how you can be getting 2 amps out of this transformer. 35VA/115V= .304 A. Maybe I can learn something here if anyone cares to explain.


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Fri 11, 2019 9:05 pm 
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Joined: Jun Sat 15, 2019 7:43 pm
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You cannot measure amperage without a load, there is no current flow. Your transformer is only rated for 300ma. Something doesn't make sense here

DM


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Fri 11, 2019 10:04 pm 
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Joined: Jan Fri 04, 2019 11:32 pm
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Location: Sheridan, Wyoming
I will try measuring the current with a load, and then take it from there. Thanks, guys! William


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Sat 12, 2019 3:56 am 
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Location: Perrysburg, OH, 43551 U.S.A.
Of course you may be able to get 2A out of a transformer rated for 300mA. The transformer just won't last very long.
John

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 Post subject: Re: CLC filter questions.
PostPosted: Oct Sat 12, 2019 4:10 pm 
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Location: Lexington, KY USA
Yes, the filter choke had best be rated for more than your actual load current.

If you are transformer shopping, be sure to realize that for a filtered 2A DC output, you will need a VA rating far higher than 2A X 120V.

I think Hammond have a nice set of suggestions about power transformer ratings for DC rectifier supplies.

And the caps need to be closer to 4700uF or 10000uF.

If you don't actually need the 2A DC output current, smaller filter components can be used, depending on the maximum current requirement.

Don't forget your bleeder resistor.

Ted


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Sat 12, 2019 6:36 pm 
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williampage wrote:
I will try measuring the current with a load, and then take it from there. Thanks, guys! William

You can't measure power supply current any other way!! With no load, there is no current----Or maybe I'm missing something??

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"Even if you don't understand Ohm's Law, you are still required to obey it."


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Mon 14, 2019 8:04 pm 
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Location: Texas, U.S.A.
As Usually Lurking indicates, the transformer volt-ampere (VA) rating will need to be higher than the direct current (dc) load power needed. Using a full wave bridge rectifier, the VA rating will need to be 123% or 1.23 times the maximum dc load power needed [1]. With a 35 VA transformer, the maximum dc output power available (including power used in the resistance of filter choke or resistor) will be 35 VA / 1.23 = 28.4 watts. For 2 amps at 118 volts, the load power is 236 watts. Commonly, instead of VA, the voltage and current will be given for the secondary. The product of the voltage and current is the VA. For example, a transformer secondary labeled 150 volts and 300 mA would be 45 VA.

In the application you describe, use of a resistance instead of a choke may be practical if the power loss and voltage drop it causes are acceptable. The resistance would need to be 1.3 ohms if input and output capacitors of 10,000 microfarads (mF or uF) are used, and it would give 2.6 volts drop across it at the 2 amperes load current. The power loss would be 5.2 watts. Conservative practice would require the resistor power rating to be 10 watts. If a choke is used, much smaller capacitors are needed (1000 uF) and power loss in the dc resistance of the required 180 millihenry choke inductance will be very small.

The following graph can be used to determine the filter input capacitance and output voltage for full wave rectifiers [2]:
Attachment:
FULL_WAVE_RECTIFIER_GRAPH.jpg
FULL_WAVE_RECTIFIER_GRAPH.jpg [ 303.11 KiB | Viewed 1795 times ]


1. In this case, where the load resistance is not known, the stated 2 amp load current can be used in approximating the load resistance. The dc voltage from the rectifier cannot be higher than the peak secondary voltage, so given the secondary RMS voltage of 115, the peak would be 1.4 x 115 = 161 volts dc. Subtracting 1 volt for the rectifier forward voltage gives 160 volts. 160 V / 2 A = 80 ohms approximate RL.

2. Assuming, for illustration, total source resistance Rs of 2 ohms gives Rs / RL = 2 ohms / 80 ohms = 0.025. (the actual dc resistance of the secondary should be measured, and solid state rectifiers will present small enough resistance to be ignored in this application). Entering the graph on the right side Y axis at 0.025 and going horizontally across to the left side Y scale, Edc / Eac peak, it is seen that the maximum dc voltage from the rectifier under load will be about 88% of the peak ac secondary voltage. So 0.88 x 160 V = 141 volts dc.

3. Now, a more accurate load resistance RL can be calculated. 141 V / 2 A = 70.5 ohm RL. 2 ohms Rs / 70.5 RL = 0.028. Entering the right side Y axis at 0.028 it can be seen that the maximum output voltage will be about 85% of peak and that the minimum (omega x C x RL) for this percentage is about 40.

4. Solving for C (the capacitor connected directly to the bridge); C = 40 / (omega x RL). Omega is 2 x Pi x ripple frequency. So, 40 / (6.28 x 120 Hz x 70.5 ohms) = 753 microfarads (uF) minimum capacitance, unless I've made a blunder of some sort. The closest standard capacitance can be used, which is 1000 uF. The ripple voltage across the capacitor will be about 2% of 141 volts or 2.8 volts.

5. The additional choke and capacitor form a low pass filter. The cutoff frequency can be chosen to be one tenth of the ripple frequency. So 120 Hz / 10 = 12 Hz cutoff. For convenience, the output capacitor can be the same as the input capacitor, 1000 uF (0.001 farad).
Choke inductance L = 0.0253 / (Fsquared x C) [3]. So L = 0.0253 / ((12 x 12) x 0.001) = 176 millihenrys. The reactance of 176 mH is 132 ohms at 120 Hz. The closest standard inductance is 180 mH. The ripple voltage at the filter output should be about 2.8 V / 110 = 25 millivolts [2, fig. 15.18] (if I haven't made a blunder of some sort).

Greater capacitance and/or inductance can be used if the calculated ones are not on hand.

[1] R. Visintini, "Rectifiers", Power Converters for Particle Accelerators, CERN course book, 2004, p. 141

[2] Landee, Davis, Albrecht, Electronic Designers' Handbook, 1957, pp. 15-10 to 15-18

[3] V.F.C. Veley, The Benchtop Electronics Reference Manual, 4th ed., 1994, p. 204

-----------------
Eric LaGess
WB5HDF


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 Post subject: Re: CLC filter questions.
PostPosted: Oct Mon 21, 2019 2:35 am 
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Joined: Nov Mon 06, 2017 2:35 pm
Posts: 120
Location: Texas, U.S.A.
williampage wrote:
...try to figure out how to get the load for all 4 of the tubes... At this point, I feel totally lost, and cannot go forward until I get this figured out. If you know how to do this, how about figuring the amps needed for two 12sk7gt tubes, one 12sj7gt tube, and one 12v6gt tube? Or. Could you direct me to a site which lists, in order, the steps to accomplish this? I have looked for info on building a 240 volt, diode bridge rectified, CLC power supply out of my new EDCOR xpwr119-120 power transformer. The transformer is 120 volt, 60 hz primary, and the secondary is 240 volt@ 200 ma. , and has 12.6 volt, 3 amp filament winding. I need info on the microfarad rating for my electrolytic capacitors and need to know if a choke of 30 H would be ok. I would like to know what value resistor would be best for the bleeder resistor. What I am asking is for someone to direct me to a site with an existing 240 or 250 volt power supply schematic is shown. I can just copy that schematic and use the values given there. Thanks.... I think I am dealing with a total of 69 milliamps current at the plates of the 4 tubes totaled... William

William,
Known:
1. Desired output voltage 250.
2. Transformer secondary rated 240 V ac at 200 milliamps = 48 VA.
3. Full wave solid state bridge rectification.
4. Circuit to be supplied uses 70 mA plate current.
4. Dc output voltage from full wave rectifier with no load and no bleeder = 240 x 1.4 = Eacp = 339 volts.
5. 300 V chosen as maximum permissible output voltage with bleeder but no load on supply.
4. Capacitor input filter connected to rectifier.
5. An additional LC (inductance / capacitance) filter is desired to follow the rectifier capacitor.
6. A 5 henry, 150 mA choke is on hand. Measure resistance of choke. 105 ohms will be used here.

To be determined:
1. Resistance of transformer 240 V secondary, Rs. Use ohmmeter to measure. 50 ohms will be used in this example.
2. Resistance of choke, Rchoke. Use ohmmeter.
3. Filter capacitance.
4. Bleeder resistance and power rating.

Method:
1. Look up data on plate current for each tube, add up all. Result = 70 mA (0.07 amp) needed by tubes.
2. Calculate equivalent load resistance of tubes. Rl = 250 / 0.07 = 3571 ohms
3. Determine C1 capacitance needed on rectifier positive terminal for omega x C1 x Rl = 30. C1 = 30 / (6.28 x 120 x 3571)
Result: C1 = 10 microfarads (uF or mF). The 47 uF you already have is okay.
3. Calculate /Rs / Rl. For example 50 ohms / 3571 ohms = 0.014
4. Use Fig. 15.9 graph, above, to determine Edc / Eacp with /Rs / Rl = 0.014 and omega x C1 x Rl = 30. Result: Edc / Eacp = 0.93.
5. 339 x .93 = around 315 V dc output from rectifier at the 70 mA load.
6. 315 V dc out is too high, need to drop 315 - 250 = 65 V.
7. A series resistance, R1, can be used to drop the 65 volts.
8. Resistance to drop 65 V at 70 mA = 65 V / 0.07 A = 928 ohms. 928 - Rchoke = 928 - 105 = 823 ohms. 820 ohms will be suitable for R1.
9. 820 ohms x 0.07 amps squared = 4 watts. Will need 5 to 10 watt rating for R1. The dropping resistor is placed between the rectifier positive and the choke.
10. Reactance of 5 H choke at 120 Hz = 6.28 x 120 x 5 = 3768 ohms.
11. The reactance of the output capacitor can be one tenth of 3768 ohms. 3768 / 10 = 377 ohms. C2 = 1 / (6.28 x 120 x 377) = 3.5 uF. For convenience, C2 can be 10 uF or 47 uF instead.
12. Bleeder resistance, R2, calculation. Roughly 315 V - 300 V = 15 V to be dropped across the 925 ohm (R1 + Rchoke). 15 V / 925 ohms = 16 mA. 300 V / 0.016 A = 18,750 ohm bleeder, R2. I would use 20 k ohms. The additional current due to the bleeder will drop the output voltage to around 230 V when the 70 mA load is on the supply.
13. 300 V x 0.015 = 4.5 watts for the bleeder. I'd use a 10 watt rated resistor.
14. Approximate total dc power: 4.5 watts bleeder + 4.5 watts dropping resistor + (.07 x 230) = 25 watts.
15. 48 VA / 1.23 = 39 watts. 25 watts is less than 39 watts, so sufficient power is available from the transformer.
Attachment:
RECT_FILTER_01.jpg
RECT_FILTER_01.jpg [ 70.18 KiB | Viewed 1677 times ]

----------------
Eric LaGess
WB5HDF


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