I am thinking of getting a 9VAC 4.17A transformer (no CT) to supply a basic bridged rectified LM1085 circuit (similar to LM317 but lower voltage drop-out). The intent was to have an output voltage of 8.2V.
The hammond website (http://www.hammondmfg.com/pdf/5c007.pdf) says:
average voltage after the 1st cap: .9 * Vsecondary
peak voltage after the 1st cap: 1.41 * Vsecondary.
if the average is .9 * Vsecondary, a 9VAC transformer won't be enough. plus you have the voltage drop-out associated with the LM1085.
I intend to draw 2.6A from the transformer. just wondering what are people practical experience with the topic.
Thanks for the help.
The hammond website (http://www.hammondmfg.com/pdf/5c007.pdf) says:
average voltage after the 1st cap: .9 * Vsecondary
peak voltage after the 1st cap: 1.41 * Vsecondary.
if the average is .9 * Vsecondary, a 9VAC transformer won't be enough. plus you have the voltage drop-out associated with the LM1085.
I intend to draw 2.6A from the transformer. just wondering what are people practical experience with the topic.
Thanks for the help.
I don't know about your specifics, but, in general, you'd like to have some head room so that you regulator can do its job correctly. Often the datasheets for the regulator will suggest a minimum input % over your expected output. Just shooting from the hip here, I'd say you might want a Vsec rms that is 25% higher than your expected Vp(rect).
poobah said:You would be better off to use a 12 VAC secondary...
Go to Duncan's Amp Pages and download PSUDII. Nifty program... FREE!
I have used it and am using it.
the problem is, that is software. and sometimes reality is different
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I just want to avoid a situation where I need to drop alot of voltage on the LM1085 when I do not need to
You can change the resistive load to a "current sink"... this will help your sim. Right click on the resistor and you can change it.
Seems like your cap is too small because the Vmin & Vmax are so different.
OR,
Be sure to set the simulation to 5 seconds or so to evaluate... you could be looking at the first start up voltages rather than the steady state conditions after the supply has charged up.
Seems like your cap is too small because the Vmin & Vmax are so different.
OR,
Be sure to set the simulation to 5 seconds or so to evaluate... you could be looking at the first start up voltages rather than the steady state conditions after the supply has charged up.
poobah said:You can change the resistive load to a "current sink"... this will help your sim. Right click on the resistor and you can change it.
Seems like your cap is too small because the Vmin & Vmax are so different.
OR,
Be sure to set the simulation to 5 seconds or so to evaluate... you could be looking at the first start up voltages rather than the steady state conditions after the supply has charged up.
why is a current load be different from a resistive load? I have the voltage that it requires and I have the current requirements. it's easy enough to derive the resistance using those 2.
ps. the psu will be supplying tube heaters.
4700u is too small?
I've set it to simulate for 30secs. (or 30000 ms in PSUD2)
From a calculational point of view, you should allow for +/- 10% on the line. For 120VAC in the US, this is +/- 12V; for 240 in Britain, it's 24, and so forth. You need to make sure that the highest won't blow things up, and the lowest won't brown your regulator out.
The RMS rating on the secondary of a transformer (and Hammond always give RMS) is Vpeak/2^0.5 (the peak voltage divided by the square root of two). Vpeak is half of Vp-p. Yes, your 120VAC (in the US) is actually 340 volts peak-to-peak. Full-wave rectification with first-approximation diodes with an ideal capacitor across the output and no load gives you Vpeak-(the number of diodes * 0.7V) on the capacitor once it's charged up; real capacitors leak a bit, so there'll be a little ripple, and if you add a load, the equivalent resistance to the load will cause that ripple to grow based on the time constant of that resistance and the capacitance. To calculate ripple, you must know the load.
Your "9V RMS" transformer will therefore yield 9*2^0.5=12.73V at an unloaded ideal capacitor, and very nearly that for a real one, perhaps 12.6V. Since you don't have a center tap, you're going with a bridge rectifier, which means two 0.7V drops, or 1.4V; now you're down to 11.2V. With an LDO regulator, you'll have a little extra left over, and in fact the LM1085 has a maximum dropout of 1.5V at 3A, which is its maximum current capability. That means that if everything is perfect, you could get 9.7V out of it; allowing 10% for line drops, you can probably count on being able to get 8.8V. This allows 9% margin for voltage drops across the transformer due to internal resistance and for ESR in your real-world capacitors; in addition, the real drop for your power diodes is generally less than the 0.7V first approximation, some go as low as 0.2V. Check the data sheet for your rectifier.
Having bought a Hammond toroidal transformer just recently, that was rated at 30VRMS CT, and found it putting out 32V RMS according to my Fluke 23, and having gotten 43V from it measured across a cap after full-wave rectification, I'd say their formulae are extremely conservative. My peak measurements with my o'scope were in line with this, too- 45Vpeak- so I seriously doubt my meter is off by any significant amount.
Off-hand, I'd say you're OK for minimum. At maximum, that is, +10% on the line voltage, the voltage coming out of the transformer will be 9.9VRMS, which gives you 14Vpeak; that's 13.6V if the diodes only drop 0.2V each, and now you're dropping 5.4V across your regulator. At 2.6A, you're dissipating 17W of power. As long as the ambient temperature stays below 100C, which is the boiling point of water, you should be fine; note however that a TO-220 heat sink is a good idea, and ventilation is mandatory, although convection should be sufficient and I don't think a fan is required (unless you already need one for the amplifier). And your 3A dissipation if the input is the ideal 120VAC will be less than this; more like 8 or 10W, which a TO-220 should handle OK if perhaps being a little dicey, and a heatsinked one should have no trouble with at all. So it looks like you're OK at maximum, too.
And the last part is, if you get a toroid, you can always add a couple windings to it if it's a little bit low. So I recommend it. Just make sure you know precisely what the input is; if you're only getting 112V tonight, make sure you don't add windings to bring that to 120V and boost it up beyond the power handling of your regulator.
Measure twice, cut once.
The RMS rating on the secondary of a transformer (and Hammond always give RMS) is Vpeak/2^0.5 (the peak voltage divided by the square root of two). Vpeak is half of Vp-p. Yes, your 120VAC (in the US) is actually 340 volts peak-to-peak. Full-wave rectification with first-approximation diodes with an ideal capacitor across the output and no load gives you Vpeak-(the number of diodes * 0.7V) on the capacitor once it's charged up; real capacitors leak a bit, so there'll be a little ripple, and if you add a load, the equivalent resistance to the load will cause that ripple to grow based on the time constant of that resistance and the capacitance. To calculate ripple, you must know the load.
Your "9V RMS" transformer will therefore yield 9*2^0.5=12.73V at an unloaded ideal capacitor, and very nearly that for a real one, perhaps 12.6V. Since you don't have a center tap, you're going with a bridge rectifier, which means two 0.7V drops, or 1.4V; now you're down to 11.2V. With an LDO regulator, you'll have a little extra left over, and in fact the LM1085 has a maximum dropout of 1.5V at 3A, which is its maximum current capability. That means that if everything is perfect, you could get 9.7V out of it; allowing 10% for line drops, you can probably count on being able to get 8.8V. This allows 9% margin for voltage drops across the transformer due to internal resistance and for ESR in your real-world capacitors; in addition, the real drop for your power diodes is generally less than the 0.7V first approximation, some go as low as 0.2V. Check the data sheet for your rectifier.
Having bought a Hammond toroidal transformer just recently, that was rated at 30VRMS CT, and found it putting out 32V RMS according to my Fluke 23, and having gotten 43V from it measured across a cap after full-wave rectification, I'd say their formulae are extremely conservative. My peak measurements with my o'scope were in line with this, too- 45Vpeak- so I seriously doubt my meter is off by any significant amount.
Off-hand, I'd say you're OK for minimum. At maximum, that is, +10% on the line voltage, the voltage coming out of the transformer will be 9.9VRMS, which gives you 14Vpeak; that's 13.6V if the diodes only drop 0.2V each, and now you're dropping 5.4V across your regulator. At 2.6A, you're dissipating 17W of power. As long as the ambient temperature stays below 100C, which is the boiling point of water, you should be fine; note however that a TO-220 heat sink is a good idea, and ventilation is mandatory, although convection should be sufficient and I don't think a fan is required (unless you already need one for the amplifier). And your 3A dissipation if the input is the ideal 120VAC will be less than this; more like 8 or 10W, which a TO-220 should handle OK if perhaps being a little dicey, and a heatsinked one should have no trouble with at all. So it looks like you're OK at maximum, too.
And the last part is, if you get a toroid, you can always add a couple windings to it if it's a little bit low. So I recommend it. Just make sure you know precisely what the input is; if you're only getting 112V tonight, make sure you don't add windings to bring that to 120V and boost it up beyond the power handling of your regulator.
Measure twice, cut once.
ok...
You model the transformer at 9 Vrms, 4.17 amos with 5% reg.
Your schematic should show 9.45 volts & 108 mOhm resistance.
4007 bridge...
22,000 uF cap with 0.050 Ohm internal res.
A 3.22 Ohm resistor for tube load. (it does work better... for another reason)
Do you see Vc1 min as 8.97V & max is 10.15V ???
What is the drop out voltage of your reg?
You model the transformer at 9 Vrms, 4.17 amos with 5% reg.
Your schematic should show 9.45 volts & 108 mOhm resistance.
4007 bridge...
22,000 uF cap with 0.050 Ohm internal res.
A 3.22 Ohm resistor for tube load. (it does work better... for another reason)
Do you see Vc1 min as 8.97V & max is 10.15V ???
What is the drop out voltage of your reg?
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