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Simple circuit, say 9V AC to a regulated 6.3 V at at least 2 amps?
Posted by
Chris Wilson
on 19 Jun, 2013 18:12
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Basically I understand rectifiers and just about understand regulators, but can modern regulators handle a current draw of say 2 amps at a regulated 6.3V DC, without also using a power transistor? If so, which regulators? How big a voltage delta between AC out of transformer to required DC out of regulator? Thanks. Follow on from my transformer rating thread (VA rating of a transformer). Thanks.
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How about two lm317 in parallel? You need to add some resistance to the output though
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#2 Reply
Posted by
Alana
on 19 Jun, 2013 18:36
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OK, thanks for the part numbers. I should have asked, this is for a DC supply to valve filaments, two small Russian valves with the filaments in parallel. There will be some inrush current when they switch on. How do these regulators cope with that? I see a data sheet for the LM317 shows a "slow start" circuit, I think it uses a capacitor / resistor charge time "timer". Would that help? Why do I want DC? The circuit is a driver amp for a big tetrode, designed for very low noise, the designer specs DC to reduce chances of hum etcetera. It also allows good voltage control, of course, for better valve life. Thanks. The valve data sheet says 0.55 Amps each filament, so I am playing a bit safe on speccing 2 Amps. Cheers!
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#4 Reply
Posted by
Kremmen
on 19 Jun, 2013 19:17
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As long as you use a linear regulator circuit the voltage you drop will be dissipated in some component, and the magnitude of dissipation will be the same regardless whether you use a power transistor or not. So it might be easier to just use one and have done with it. Any number of circuits out there. Possibly even the 3 component regulator (resistor, zener, transistor) could do but if not, many other simple options exist.
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#5 Reply
Posted by
edavid
on 19 Jun, 2013 19:51
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You shouldn't have to do anything special about inrush current.
If you don't want to find an LM350, you could just use an LM317T for each tube. That's probably cheaper.
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I have bought an LM350 to have a play with, and for stock, as I am in no rush for them, a very cheap pack of 5 LM317T from China, that will take a few weeks to arrive. Sound like useful regulators to have about. If the current draw is more than the spec sheets suggest I'll do something else with a power transistor. All good learning experience! Thanks again.
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#7 Reply
Posted by
Paul Price
on 20 Jun, 2013 00:51
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You should just put two 7805 regulators in parallel on the same heatsink...it will work perfect.
Some people think this idea is poor design and only one of the 7805 will be active...not true. They will quickly balance the load between them and the voltage output will remain constant during this balancing period.
KISS
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#8 Reply
Posted by
Paul Price
on 20 Jun, 2013 00:56
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Someone says, "LM350 would be better, 78xx and LM317 are not ment to be connected in parallel."
Try connecting two or more 78xx in parallel and test this idea! If you can, mount them on the same heatsink.
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#9 Reply
Posted by
edavid
on 20 Jun, 2013 02:35
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You should just put two 7805 regulators in parallel on the same heatsink...it will work perfect.
Some people think this idea is poor design and only one of the 7805 will be active...not true. They will quickly balance the load between them and the voltage output will remain constant during this balancing period.
KISS
No, it will not work perfect(ly), and it really is poor design. The 7805 with the higher voltage output will supply most of the load until/unless it goes into current limit or thermal shutdown.
Now, in this case, the OP needs 6.3V out, so he could trim the voltages on the 2 7805s so they were close enough to share pretty well, especially with a little extra ballast resistance. However, since he has 2 separate loads, he could just supply them from separate regulators and avoid the issue.
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#10 Reply
Posted by
Rufus
on 20 Jun, 2013 02:57
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Some people think this idea is poor design and only one of the 7805 will be active...not true. They will quickly balance the load between them and the voltage output will remain constant during this balancing period.
If it were good design don't you think some of the data sheets might show it as a suggested higher current regulator and sell twice as many 7805s? Two 7805s can have a much as 400mV difference in output and typically only 10mV of load regulation at 1.5A. The chances are one of the regulators will try to do all the work limited by its SAO and thermal protection circuits.
The other will fill in but who knows what the characteristics of a 7805 trying to current limit or shut down are like. The SAO protection possibly has a negative resistance characteristic. The higher the output voltage the less dissipation in the pass transistor and the more current the SAO circuit will allow to pass. A pair of 7805s heavily loaded might just sit there oscillating.
Who knows, the idea is too naff to be worth investigating.
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#11 Reply
Posted by
c4757p
on 20 Jun, 2013 02:58
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No, it will not work perfect(ly), and it really is poor design. The 7805 with the higher voltage output will supply most of the load until/unless it goes into current limit or thermal shutdown.
7805 has a negative temperature coefficient of voltage, so as long as the two are reasonably close to begin with, they should end up sharing.
Two 7805s can have a much as 400mV difference in output and typically only 10mV of load regulation at 1.5A. The chances are one of the regulators will try to do all the work limited by its SAO and thermal protection circuits.
I think it's a reasonable thing to do
for a hobby circuit, if you take the time to match a pair so that they begin close together. As long as the difference isn't so high that one goes into thermal shutdown before they have a chance to balance, they should equalize.
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I'd still stick a 0.1 ohm or something on the output of each one to ensure some current sharing though.
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#13 Reply
Posted by
richard.cs
on 20 Jun, 2013 09:10
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Is there a particular reason for using a 9V transformer and d.c. heating? Would you consider running the heaters in series?
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#14 Reply
Posted by
Paul Price
on 20 Jun, 2013 10:45
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Remember what we are talking about here, heating filaments.
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#15 Reply
Posted by
Paul Price
on 20 Jun, 2013 10:50
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Finding the simplest way to accomplish an electron task can often be the best. All the remarks about output voltage matching an thermal shutdown are immaterial and simply do not happen. They just don't happen with two 7805's in parallel if the output of both total matches the demand.
There is no need to create an electron egalitarian democracy.
The 7805's will always share because one attempting to take more of the load will always drop in output voltage and this drop in output voltage will cause its partner to take up the slack.
It makes no difference for the circuit and it will function perfectly even if one is running a little hotter than the other. The 7805 is designed to work over its range of operating temperature and current sharing will take affect well before either one approaches it's output current or temperature SOA limit.
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#16 Reply
Posted by
Paul Price
on 20 Jun, 2013 10:54
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Edavid says,"No, it will not work perfect(ly), and it really is poor design. The 7805 with the higher voltage output will supply most of the load until/unless it goes into current limit or thermal shutdown."
I say this is not true, for same reasons I have stated before.
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#17 Reply
Posted by
Paul Price
on 20 Jun, 2013 10:56
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Rufus says, "If it were good design don't you think some of the data sheets might show it as a suggested higher current regulator and sell twice as many 7805s? Two 7805s can have a much as 400mV difference in output and typically only 10mV of load regulation at 1.5A. The chances are one of the regulators will try to do all the work limited by its SAO and thermal protection circuits.
The other will fill in but who knows what the characteristics of a 7805 trying to current limit or shut down are like. The SAO protection possibly has a negative resistance characteristic. The higher the output voltage the less dissipation in the pass transistor and the more current the SAO circuit will allow to pass. A pair of 7805s heavily loaded might just sit there oscillating. "
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Does not and cannot happen. Have you ever tried such a circuit?
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#18 Reply
Posted by
Paul Price
on 20 Jun, 2013 10:59
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C4757p says, "I think it's a reasonable thing to do for a hobby circuit"
How can you be sure it not altogether the right thing to do in any reasonable circuit. The characteristics a 7805 under load cause them to share as much as necessary.
A 7805 is not be used to create a laboratory standard voltage reference!
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#19 Reply
Posted by
Paul Price
on 20 Jun, 2013 11:02
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Peter.Michel wrote,"I'd still stick a 0.1 ohm or something on the output of each one to ensure some current sharing though."
Why add components that are unnecessary and make more work for yourself, as I've already said they don't need to exactly match and they will anyway for the reasons I've stated already.
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#20 Reply
Posted by
Paul Price
on 20 Jun, 2013 11:05
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Edavid says,"Now, in this case, the OP needs 6.3V out, so he could trim the voltages on the 2 7805s so they were close enough to share pretty well, especially with a little extra ballast resistance. However, since he has 2 separate loads, he could just supply them from separate regulators and avoid the issue."
Firstly, is there really an issue? The 7805's are already matched close enough and there is again no need for the currents to exactly match.
Why create more work for yourself if there is no proven reason to.
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#21 Reply
Posted by
Paul Price
on 20 Jun, 2013 11:14
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cs.richard says, "Is there a particular reason for using a 9V transformer and d.c. heating? Would you consider running the heaters in series?"
Yes, this tampers with an antique circuit and creates more work soldering than necessary with possible damage to brittle old tube socket pins and wires with rubber or fragile cloth insulation that is ready to drop off at the first toggle or breeze.
Why not try just providing the current you need easily, say with two 7805's in parallel.
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#22 Reply
Posted by
Paul Price
on 20 Jun, 2013 11:17
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Why doesn't anyone try paralleling of 7805's to test their current sharing operation before commenting?
Prove me wrong as well, but Is it good to give advice on how to protect a circuit that doesn't need protecting, to match devices that do not need matching, to offer advice without experimental proof, with just fear and ignorance to guide you?
Has anyone forgotten that we are trying to find a simple solution to this extremely delicate and complex precision design question, heating filaments?
These poor vacuum tubes have managed to cope with unregulated 6.3V AC for enough time for your grandson to enter college, somehow they have survived.
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For a fixed load like a filament, you could just use a resistor - it will get hot, but so will a linear regulator
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Personally I think unregulated 6V AC would be fine, but I am trying to emulate a known REALLY low IMD design, and not experienced enough to want to go off at a tangent from what is known to work. Thanks for the interesting points of view and ideas
