He needs 8.2 Volts final. Add the 1.5V burden of the regulator and that's 9.7 Volts.
Now... use a 1 Farad cap, silicon bridge, and a 9 V secondary, the Vmin is 9.15 Volts. Most of your analysis is correct; but ripple must be calculated using the current draw of 2.55 Amps; an effective resistance of 3.22 Ohms at the tube; or 3.8 at the tube/reg combo...
Download PSUDII... I don't use sims much either... but this one is worth it.
Now... use a 1 Farad cap, silicon bridge, and a 9 V secondary, the Vmin is 9.15 Volts. Most of your analysis is correct; but ripple must be calculated using the current draw of 2.55 Amps; an effective resistance of 3.22 Ohms at the tube; or 3.8 at the tube/reg combo...
Download PSUDII... I don't use sims much either... but this one is worth it.
Hmmm. OK, let's do Vmin on paper.
Once again, your ideal capacitor voltage with an RMS voltage of 9V will be 12.73V, supposing no diode drops. Add two diode drops at 0.7V each for a bridge rectifier and you're down to 11.33V.
Now let's draw 2.55A. A 1F capacitor (which is what you just said- note that I believe this is a typo) charged to 11.33V is holding 11.33 coulombs of charge; that's the definition of a farad, it holds 1 coulomb at a differential of one volt, two at two, and so forth. At 2.55A, you'll draw 2.55 coulombs per second, the charge transferred by a current of one ampere flowing for one second being the definition of a coulomb. Since the recharge happens every 1/120 of a second, that means that the discharge will occupy 1/240 of a second, because after that it's going back up again, and charging, and the rate is highest at the beginning (as you should know from the shape of the wave). During 1/240 of a second, at 2.55 coulombs per second, you'll draw 10.6 millicoulombs from the cap. This will reduce its charge to 11.32 coulombs, which in a 1F cap will reduce its voltage to 11.32V. OK, let's make it take a whole 1/120 of a second; during that time it will discharge to 11.31V.
Oops.
OK, let's scale back to a 10mF cap; I think maybe 1F was a typo, since you were talking about 10000uF earlier in the thread. This one holds 113.3mC of charge at the stated voltage of 11.33V, and when we take our 10.6mC out, that leaves us with 102.7mC of charge, which in that cap is 10.27V.
Oops.
OK, now we know that the cap won't start charging again until the voltage reaches the voltage across the cap, which is falling. And that's not going to be until a ways through the cycle- certainly less than 1/120 of a second, but more than 1/240. So can we guesstimate that? Well, twice 10.6 is 21.2- and 113.3-21.2=92.1mC. And that brings our voltage down to 9.21V. So it must be more than 9.21V, but less than 10.27, and probably much closer to 9.21 than 10.27. But it sure as heck ain't 9.15. It's more like 9.4V- if you wanna do the math to see where those two intersect, go ahead, this is just a forum and it just ain't that important to me.
Oops.
But fine, we'll call it 9.15.
Ever do the calculations for Vav of a rippling circuit? What you do, you calculate the ripple, then subtract half of it from the peak. Standard stuff, right out of Malvino (I liked Malvino; been using it all my life. YMMV.). So here we have 11.33 and 9.15V, and I get 10.24V(avg). Put it another way, if I put a resistor out there, and it draws 2.55A(avg) from this thing (without the regulator), and I put a multimeter with a magnetic movement on it, it will tell me 10.24V. For that matter, so will my Fluke- or perhaps just a bit more, since the Fluke is a true RMS reader, whereas magnetic meter movements are by their very nature average readers, and root-mean-square is a bit higher than average.
I think that Vav is what you need to use here; in fact, it's what I've been using my whole life, and friend, I've designed and built a boatload of power supplies. And I mean professionally. Like for money and stuff, you know, as my job and whatnot. The reasons you want to use Vav are first, because that's the voltage you use to calculate the power dissipation in the device, and second, because that's the voltage that the data sheet is specifying against, because that's the voltage that the techs can measure when they're checking a lot (and I mean a manufacturing lot, not "a bunch") of chips to see if they meet spec or not, because that's the value that a meter can measure.
I also have to point out that most single-unit bridges drop 1.2V, not 1.4, and that's according to their spec sheets, which agree with my measurements over the years; and if you're anal about it, you can go get Schottky diodes that only drop 0.2V each for a total of 0.4 across the bridge. In this case, if you're spending $50 on a great big capacitor, an extra $3 for fast-recovery Schottky diodes don't seem like a lot. So there's one source of a few extra tenths of a volt, if you need it. Another source of extra volts is the transformer; that's why I recommended a toroid. Add a couple windings and Bob's your uncle.
And the final consideration is power dissipation. You want this thing to dissipate as little as possible; that's why you're getting an LDO regulator. The more volts it drops, the more power it dissipates for a given current; and with a nice steady current demand, and particularly for a tube heater supply, which isn't exactly the most highly spec'ed item in the whole entire world, you want to shave it about as close as you can to reduce heating inside the chassis, since you already have a heat problem before you even start. You start using Vmin to calculate your regulator voltage instead of Vaverage, you're wasting power, and adding headroom you'll never use, at the expense of extra cooling and perhaps a failed device. Hey man, people use UNRECTIFIED CURRENT STRAIGHT OUT OF A TRANSFORMER to run heater supplies, and the results aren't all that bad- dude, it's a HEATER. With a regulator and a rectifier and a filter cap, this guy is building a gold-plated heater supply already. It don't have to be platinum. It'll work FINE, trust me.
I'll leave it at that. It seems like enough to establish my point.
Once again, your ideal capacitor voltage with an RMS voltage of 9V will be 12.73V, supposing no diode drops. Add two diode drops at 0.7V each for a bridge rectifier and you're down to 11.33V.
Now let's draw 2.55A. A 1F capacitor (which is what you just said- note that I believe this is a typo) charged to 11.33V is holding 11.33 coulombs of charge; that's the definition of a farad, it holds 1 coulomb at a differential of one volt, two at two, and so forth. At 2.55A, you'll draw 2.55 coulombs per second, the charge transferred by a current of one ampere flowing for one second being the definition of a coulomb. Since the recharge happens every 1/120 of a second, that means that the discharge will occupy 1/240 of a second, because after that it's going back up again, and charging, and the rate is highest at the beginning (as you should know from the shape of the wave). During 1/240 of a second, at 2.55 coulombs per second, you'll draw 10.6 millicoulombs from the cap. This will reduce its charge to 11.32 coulombs, which in a 1F cap will reduce its voltage to 11.32V. OK, let's make it take a whole 1/120 of a second; during that time it will discharge to 11.31V.
Oops.
OK, let's scale back to a 10mF cap; I think maybe 1F was a typo, since you were talking about 10000uF earlier in the thread. This one holds 113.3mC of charge at the stated voltage of 11.33V, and when we take our 10.6mC out, that leaves us with 102.7mC of charge, which in that cap is 10.27V.
Oops.
OK, now we know that the cap won't start charging again until the voltage reaches the voltage across the cap, which is falling. And that's not going to be until a ways through the cycle- certainly less than 1/120 of a second, but more than 1/240. So can we guesstimate that? Well, twice 10.6 is 21.2- and 113.3-21.2=92.1mC. And that brings our voltage down to 9.21V. So it must be more than 9.21V, but less than 10.27, and probably much closer to 9.21 than 10.27. But it sure as heck ain't 9.15. It's more like 9.4V- if you wanna do the math to see where those two intersect, go ahead, this is just a forum and it just ain't that important to me.
Oops.
But fine, we'll call it 9.15.
Ever do the calculations for Vav of a rippling circuit? What you do, you calculate the ripple, then subtract half of it from the peak. Standard stuff, right out of Malvino (I liked Malvino; been using it all my life. YMMV.). So here we have 11.33 and 9.15V, and I get 10.24V(avg). Put it another way, if I put a resistor out there, and it draws 2.55A(avg) from this thing (without the regulator), and I put a multimeter with a magnetic movement on it, it will tell me 10.24V. For that matter, so will my Fluke- or perhaps just a bit more, since the Fluke is a true RMS reader, whereas magnetic meter movements are by their very nature average readers, and root-mean-square is a bit higher than average.
I think that Vav is what you need to use here; in fact, it's what I've been using my whole life, and friend, I've designed and built a boatload of power supplies. And I mean professionally. Like for money and stuff, you know, as my job and whatnot. The reasons you want to use Vav are first, because that's the voltage you use to calculate the power dissipation in the device, and second, because that's the voltage that the data sheet is specifying against, because that's the voltage that the techs can measure when they're checking a lot (and I mean a manufacturing lot, not "a bunch") of chips to see if they meet spec or not, because that's the value that a meter can measure.
I also have to point out that most single-unit bridges drop 1.2V, not 1.4, and that's according to their spec sheets, which agree with my measurements over the years; and if you're anal about it, you can go get Schottky diodes that only drop 0.2V each for a total of 0.4 across the bridge. In this case, if you're spending $50 on a great big capacitor, an extra $3 for fast-recovery Schottky diodes don't seem like a lot. So there's one source of a few extra tenths of a volt, if you need it. Another source of extra volts is the transformer; that's why I recommended a toroid. Add a couple windings and Bob's your uncle.
And the final consideration is power dissipation. You want this thing to dissipate as little as possible; that's why you're getting an LDO regulator. The more volts it drops, the more power it dissipates for a given current; and with a nice steady current demand, and particularly for a tube heater supply, which isn't exactly the most highly spec'ed item in the whole entire world, you want to shave it about as close as you can to reduce heating inside the chassis, since you already have a heat problem before you even start. You start using Vmin to calculate your regulator voltage instead of Vaverage, you're wasting power, and adding headroom you'll never use, at the expense of extra cooling and perhaps a failed device. Hey man, people use UNRECTIFIED CURRENT STRAIGHT OUT OF A TRANSFORMER to run heater supplies, and the results aren't all that bad- dude, it's a HEATER. With a regulator and a rectifier and a filter cap, this guy is building a gold-plated heater supply already. It don't have to be platinum. It'll work FINE, trust me.
I'll leave it at that. It seems like enough to establish my point.
In the 1 Farad sim I neglected to adjust the value of ESR to an appropriately low value. This caused an error. Since those specs aren't readily available; let's start with a published value of 100mOhm for 10,000uF and then assume that ESR is inversely proportional to capacitance from that point.
Also, a 1n4007 is inappropriate here... a 1n5400 has a suffucient current rating and will help because of reduced losses in bulk resistivity.
Using 10,000uF with 100mOhm:
Vmax = 11.4V & Vmin = 9.1V (below the target minimum of 9.7V)
Using 100,000uF with 10mOhm:
Vmax = 10.8V & Vmin = 10.5V (meets target except for low-line)
In the first case, apparent Vmax is very near the napkin value; as long as the cap is undersized and the Vmin req. is ignored. I say apparent because during the moment of peak charge, the 9A charge current is dropping about 0.9V across the ESR of the cap; the true Vmax of the cap is really about 10.7V. ESR, secondary, and bulk diode resistance seem to make a difference when a larger (still not large enough) capacitance is used.
You will be happy to know that the sim does indeed agree with your assertions that:
Vcap = Qcap / C
&,
dV = (I *dt) / C
Yes indeed, Malvino provides a handy-dandy graph (fig. 5-12) to muck through all this. Malvino states, "...this applies when the primary and secondary winding resistances are negligable, and when the diode bulk resistances are small enough to ignore." Not so good in this case. Malvino's formulas and factors are all well and good; when used strictly within Malvino's framework... perhaps why his text was more popular in tech schools rather than universities.
Hmmm... I don't need a copy of Malvino to know that Vmin equals Vmax minus Vripple. Because we are driving a regulator here, we are not really interested in Vavg. We are interested in Vmin.
I could go on, I suspect you will. It is not about worthless banter trying to show the engineers what you know (or don't).
The purpose here is to help jarthel design a regulated supply for a heater. That is want he wants and the reasons why are not germane to the discussion. We do lack a final ripple spec. But we do know that to avoid saturation within the regulator's integrater we should stay above the drop-out voltage. Keep in mind, I have not studied the drop-out recovery characteristics of his chosen reg.
With a 9Vrms secondary, he will need huge a capacitance (for ripple) and huge diodes (for inrush).
This can be done a a bit more elegantly, albiet with some power loss. I have only done a couple of DC heater supplies, certainly not enough to fill a boat, but I did find that the low voltages and high currents challenge a under-kill napkin design severely.
jarthel send me a PM.
Also, a 1n4007 is inappropriate here... a 1n5400 has a suffucient current rating and will help because of reduced losses in bulk resistivity.
Using 10,000uF with 100mOhm:
Vmax = 11.4V & Vmin = 9.1V (below the target minimum of 9.7V)
Using 100,000uF with 10mOhm:
Vmax = 10.8V & Vmin = 10.5V (meets target except for low-line)
In the first case, apparent Vmax is very near the napkin value; as long as the cap is undersized and the Vmin req. is ignored. I say apparent because during the moment of peak charge, the 9A charge current is dropping about 0.9V across the ESR of the cap; the true Vmax of the cap is really about 10.7V. ESR, secondary, and bulk diode resistance seem to make a difference when a larger (still not large enough) capacitance is used.
You will be happy to know that the sim does indeed agree with your assertions that:
Vcap = Qcap / C
&,
dV = (I *dt) / C
Yes indeed, Malvino provides a handy-dandy graph (fig. 5-12) to muck through all this. Malvino states, "...this applies when the primary and secondary winding resistances are negligable, and when the diode bulk resistances are small enough to ignore." Not so good in this case. Malvino's formulas and factors are all well and good; when used strictly within Malvino's framework... perhaps why his text was more popular in tech schools rather than universities.
Hmmm... I don't need a copy of Malvino to know that Vmin equals Vmax minus Vripple. Because we are driving a regulator here, we are not really interested in Vavg. We are interested in Vmin.
I could go on, I suspect you will. It is not about worthless banter trying to show the engineers what you know (or don't).
The purpose here is to help jarthel design a regulated supply for a heater. That is want he wants and the reasons why are not germane to the discussion. We do lack a final ripple spec. But we do know that to avoid saturation within the regulator's integrater we should stay above the drop-out voltage. Keep in mind, I have not studied the drop-out recovery characteristics of his chosen reg.
With a 9Vrms secondary, he will need huge a capacitance (for ripple) and huge diodes (for inrush).
This can be done a a bit more elegantly, albiet with some power loss. I have only done a couple of DC heater supplies, certainly not enough to fill a boat, but I did find that the low voltages and high currents challenge a under-kill napkin design severely.
jarthel send me a PM.
Dude,
Seriously... down load PSUDII from Duncan's amp pages.
The only compaints I've read about it pertain to slight inaccuracies in some of the tube rectifier models.
I totally share your attitude towards sims... too many lazy people use them and somehow skip the fundamentals. I do alot of stuff on spreadsheets... so I can "watch" and trust the math.
This little program is breezey to use though. You can't screw up a key value like I did and get a good result... garbage-in garbage-out.
Seriously... down load PSUDII from Duncan's amp pages.
The only compaints I've read about it pertain to slight inaccuracies in some of the tube rectifier models.
I totally share your attitude towards sims... too many lazy people use them and somehow skip the fundamentals. I do alot of stuff on spreadsheets... so I can "watch" and trust the math.
This little program is breezey to use though. You can't screw up a key value like I did and get a good result... garbage-in garbage-out.
I'm sorry, I was unclear- ripple is a major concern here. My estimate is that the caps have to handle upwards of 8A of ripple current. That's gonna be expensive- REAL expensive.
BTW, since you do seem to know your way around, and since it's on-topic- my general procedure is to estimate the equivalent square wave of the ripple current, then use the duty cycle equation:
Irms = Ip*(D^0.5)
Do you know a better way?
ETA: And while I'm at it, some good construction advice:
Don't put the regulator or the rectifier inside the same enclosure with the tubes, particularly not above them. If you're using one of those metal boxes with the tubes sticking up from sockets and all the wiring underneath, underneath is where to put them.
BTW, since you do seem to know your way around, and since it's on-topic- my general procedure is to estimate the equivalent square wave of the ripple current, then use the duty cycle equation:
Irms = Ip*(D^0.5)
Do you know a better way?
ETA: And while I'm at it, some good construction advice:
Don't put the regulator or the rectifier inside the same enclosure with the tubes, particularly not above them. If you're using one of those metal boxes with the tubes sticking up from sockets and all the wiring underneath, underneath is where to put them.
If you are using the sim look at the last column in the output table... your rms cap current is there.
Otherwise I know no handy formulas... I have a cheat sheet for RMS currents... you can break your waveforms into "slices" that fit the different shapes, use the equations and then sum... it's a PITA.
And yes, I am seeing the huge rms currents in the first cap here as well, ~4 Amps and that is asking for a huge cap anyway.
Check the boxes to the left of the data output table and te waveform will show on the graph area.
Part of the DC heater problem is that we can't find 10 & 11 volts trans.
Otherwise I know no handy formulas... I have a cheat sheet for RMS currents... you can break your waveforms into "slices" that fit the different shapes, use the equations and then sum... it's a PITA.
And yes, I am seeing the huge rms currents in the first cap here as well, ~4 Amps and that is asking for a huge cap anyway.
Check the boxes to the left of the data output table and te waveform will show on the graph area.
Part of the DC heater problem is that we can't find 10 & 11 volts trans.
How nice. It might almost be worthwhile just for that.poobah said:
In my experience, yes it is.poobah said:
I find that Cornell Dubilier Electronics products from Mouser are pretty much the only way to go. The Panasonic TS-EDs are nice, but low capacitance, and the ripple ratings of the TS-HC aren't quite all they might be.poobah said:
Yep, figured that one.poobah said:
I have good news. Hammond makes what you need; check out the 167 series. Digikey has them, pretty much all of them. The 167P10 or 167P11 is probably the ideal transformer for this purpose. $25 and $30 a pop. I had to special order a Hammond 182 series toroid for my amplifier, and it came the next week, so they don't fool around (at least in my experience). The 167 series, BTW, is the shielded version, with a full faraday cage. I still say that a toroid is the way to go, though, particularly for a static load like this one is.poobah said:
Perhaps chapter 8 of this document might also be of some use
http://www.onsemi.com/pub/Collateral/HB206-D.PDF
http://www.onsemi.com/pub/Collateral/HB206-D.PDF
Ha! Got it in one.
Use the 167P10 transformer, and the 103M063A052 cap, Mouser part number 5985-63V10000 on page 540 of the current catalog. You'll need two of them, but the major advantage is they're only $8 a pop, so you get out of it for $16. BTW, the ESR on those is 0.027, so two in parallel is 0.0135Ohm. Current ripple handling is 8.8Arms, and according to the sim, you get only about 6A, so that's fine.
Use GBU402, Mouser p/n 821-GBU402, rectifier- 150A surge, 100PIV, 4A, $0.69 a pop- or get a little more ambitious and spend a whole simolean on theg 200/200/6 one.
Get the transformer from digikey, unless you can get Mouser to let you special order it and they don't charge you more than $25 or so for it.
Let us know if you need help with the decoupling and massaging and bypass caps for the LM1085. The data sheet doesn't tell you everything you need to know, though it comes close. I know a coupla tricks.
ETA: Sorry about that folks, I shoulda mentioned the caps are 10mF 63V.
EATA: ...and it looks like you're gettin out the door for under 50 simoleans... not bad.
Use the 167P10 transformer, and the 103M063A052 cap, Mouser part number 5985-63V10000 on page 540 of the current catalog. You'll need two of them, but the major advantage is they're only $8 a pop, so you get out of it for $16. BTW, the ESR on those is 0.027, so two in parallel is 0.0135Ohm. Current ripple handling is 8.8Arms, and according to the sim, you get only about 6A, so that's fine.
Use GBU402, Mouser p/n 821-GBU402, rectifier- 150A surge, 100PIV, 4A, $0.69 a pop- or get a little more ambitious and spend a whole simolean on theg 200/200/6 one.
Get the transformer from digikey, unless you can get Mouser to let you special order it and they don't charge you more than $25 or so for it.
Let us know if you need help with the decoupling and massaging and bypass caps for the LM1085. The data sheet doesn't tell you everything you need to know, though it comes close. I know a coupla tricks.
ETA: Sorry about that folks, I shoulda mentioned the caps are 10mF 63V.
EATA: ...and it looks like you're gettin out the door for under 50 simoleans... not bad.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.