johne@hpvcla.HP.COM (John Eaton) (06/13/88)
<<< < I have a question about the MC7805 5V 1A regulator. Can you parallel < two of these devices together to derive a higher current? ---------- In a word, NO. Unless they are extremely well matched then one will wind up with most of the load and the other will have very little. The best way is to use a 7805 for the regulation and an external power transistor for the drive. The cheapest way is to use a power resistor to bypass the regulator but this requires a known minimun load current and bad things can happen if you aren't careful. John Eaton !hpvcla!johne
wolfgang@mgm.mit.edu (Wolfgang Rupprecht) (06/15/88)
> I have a question about the MC7805 5V 1A regulator. Can you parallel > two of these devices together to derive a higher current? Parallelling them is not very clean. You have the problem that both regulators will have *slightly* different output voltages. The one with the higher voltage will try to carry the full load, up to the point where it goes into current limiting. At this point the voltage will drop till the second one picks up this excess load. The effective source impedance is quite high during this load pass-off. You can always try to use a simple PNP buffer as such: (Any high-current silicon PNP will do.) ~+8v in +5 out o---+----\ /-------------------+----o | \ / | | v / | | --------- +5v reg. | | | +--------+ | | 2.7 ohm| | | | +--\/\/\-+--------+ in out+------+ | | gnd | | | +---+----+ | --- 0.1 uf // | --- 0.1 uf // 10uf tantalum --- 10 uf tant V --- | | | | v v If you have problems with oscillations you can put a few 10's pf capacitance between the base and collector of the PNP. Invert this circuit for a negative supply booster. --- Wolfgang Rupprecht ARPA: wolfgang@mgm.mit.edu (IP 18.82.0.114) TEL: (617) 267-4365 UUCP: mit-eddie!mgm.mit.edu!wolfgang
markz@ssc.UUCP (Mark Zenier) (06/16/88)
Someone in a posting long expired on this system asked. > < I have a question about the MC7805 5V 1A regulator. Can you parallel > < two of these devices together to derive a higher current? My library says a 7805 is good for 1500 ma. If a TO-3 will fit, a LM323 is good for 3A. There is a National part, the LM396 adjustable good for 10A. Is this the "MOOSE" part? Mark Zenier Holder of the Cliff Claven chair at the school of uunet!pilchuck!ssc!markz Unsubstantiated Opinion
al@cs.strath.ac.uk (Alan Lorimer) (06/16/88)
In article <5770@bloom-beacon.MIT.EDU>, wolfgang@mgm.mit.edu (Wolfgang Rupprecht) writes: > > I have a question about the MC7805 5V 1A regulator. Can you parallel > > two of these devices together to derive a higher current? > > You can always try to use a simple PNP buffer as such: > (Any high-current silicon PNP will do.) > > ~+8v in +5 out > o---+----\ /-------------------+----o > | \ / | > | v / | > | --------- +5v reg. | > | | +--------+ | > | 2.7 ohm| | | | > +--\/\/\-+-------+ in out+-----+ > | | gnd | | > | +---+----+ | > --- 0.1 uf // | --- 0.1 uf // 10uf tantalum > --- 10 uf tant V --- > | | > | | > v v Seems a bit involved - I had to think for a while to figure out how this was meant to work, I'm still not sure that is correctly solves the load sharing problem though, is the load taken by the PNP transistor related to the HFE of that transistor, and the 2.7 ohm resistor? Perhaps you could explain the reasoning? Since you want a higher current, you're probably going to need to use a TO3 output device anyway - I know these are a bit more difficult to mount mechanically, but why not just use an LM317K (3A) or 78H05 (5A?). Alternatively, if you wan't really high output currents use an LM723 voltage regulator IC, with as many output transistors as you need. If you are mounting the regulators on the same board as the components you want to drive, then there is no problem, use the 7805s, and power some of the board from one, and the rest of the board from the other, but don't link the power supply lines - you then have full control over the load taken by each of them. Alan. -- UUCP: ...!seismo!mcvax!ukc!strath-cs!al DARPA: al%cs.strath.ac.uk@ucl-cs JANET: al@uk.ac.strath.cs
dale@amc-vlsi.UUCP (Dale Wlasitz) (06/17/88)
>If you have problems with oscillations you can put a few 10's pf >capacitance between the base and collector of the PNP. Invert this >circuit for a negative supply booster. You WILL have problems with oscillations, add 47 pfd's across input to output of the regulator. As a side note...input filtering capacitors can go as large as you like, however do not increase the output capacitance more than 100 uFd's as the cap looks like a short circuit on power up and will nuke the regulator. 8-) Dale
phil@amdcad.AMD.COM (Phil Ngai) (06/17/88)
In article <5030002@hpvcla.HP.COM> johne@hpvcla.HP.COM (John Eaton) writes: >< I have a question about the MC7805 5V 1A regulator. Can you parallel >< two of these devices together to derive a higher current? >In a word, NO. Unless they are extremely well matched then one will wind up >with most of the load and the other will have very little. The best way is I had a friend in college do this. I was skeptical but it worked fine. What is the problem with one taking most of the load? They are over current and temperature protected anyway. >to use a 7805 for the regulation and an external power transistor for the Now you don't have the protection that the 7805 has. -- I speak for myself, not the company. Phil Ngai, {ucbvax,decwrl,allegra}!amdcad!phil or phil@amd.com
david@sun.uucp (David DiGiacomo) (06/17/88)
In article <22117@amdcad.AMD.COM> phil@amdcad.UUCP (Phil Ngai) writes: >In article <5030002@hpvcla.HP.COM> johne@hpvcla.HP.COM (John Eaton) writes: >>< I have a question about the MC7805 5V 1A regulator. Can you parallel >>< two of these devices together to derive a higher current? >>In a word, NO. Unless they are extremely well matched then one will wind up >>with most of the load and the other will have very little. The best way is > >I had a friend in college do this. I was skeptical but it worked fine. > >What is the problem with one taking most of the load? They are over >current and temperature protected anyway. They are not protected from reverse current into the output. I have personally blown them up by wiring them backward. Even hanging a BFC ("big capacitor") on the output can be bad news; you are supposed to have a reverse diode across the regulator to discharge it. What is the problem with buying a 323K or the new wave 5A/10A versions? P.S. I'm sure there is more than anyone would ever want to know on this subject printed on fine quality newsprint in the National regulator handbook.
straka@ihlpf.ATT.COM (Straka) (06/17/88)
In article <22117@amdcad.AMD.COM| phil@amdcad.UUCP (Phil Ngai) writes: |In article <5030002@hpvcla.HP.COM| johne@hpvcla.HP.COM (John Eaton) writes: ||< I have a question about the MC7805 5V 1A regulator. Can you parallel ||< two of these devices together to derive a higher current? ||In a word, NO. Unless they are extremely well matched then one will wind up ||with most of the load and the other will have very little. The best way is |What is the problem with one taking most of the load? They are over |current and temperature protected anyway. I thought that the problem was that if the output voltage of the multiple regulators were not identical, you end up with the regulator with the highest voltage supplying most/all of the current. If the load is sufficient to go past 1.5A, the one regulator may be in (protected) overload condition all of the time. Although these devices are pretty bulletproof, I wouldn't want to run them like this as a matter of course. -- Rich Straka ihnp4!ihlpf!straka Avoid BrainDamage: MSDOS - just say no!
wolfgang@mgm.mit.edu (Wolfgang Rupprecht) (06/18/88)
Me: >> ~+8v in +5 out >> o---+----\ /-------------------+----o >> | \ / | >> | v / | >> | --------- +5v reg. | >> | | +--------+ | >> | 2.7 ohm| | | | >> +--\/\/\-+-------+ in out+-----+ >> | | gnd | | >> | +---+----+ | >> --- 0.1 uf // | --- 0.1 uf // 10uf tantalum >> --- 10 uf tant V --- >> | | >> | | >> v v >al@cs.strath.ac.uk (Alan Lorimer) writes: >Seems a bit involved - I had to think for a while to figure out how this >was meant to work, I'm still not sure that is correctly solves the load >sharing problem though, is the load taken by the PNP transistor related >to the HFE of that transistor, and the 2.7 ohm resistor? Perhaps you >could explain the reasoning? The transistor is the OUTPUT buffer. It is meant to carry the bulk of the current for the high-current case. For low currents the PNP is turned off (by the 2.7 ohm resistor) and the total output current flows through the regulator. The PNP will start conducting as soon as the voltage drop reaches ~0.6v across the 2.7 ohm resistor. This happens when the output current is (~0.6v/2.7ohm) or about 220 ma. (The resistor will be dissipating a constant 0.13w from this point onward.) The regulator only supplies the first 220 ma. plus whatever base current the transistor needs. Even at a conservative HFE of only 20, that's only an additional 240ma of base current for 5 Amps. output current. This means that at this 5A. point the regulator will only be carrying roughly 10% of the output current. If you can keep the input voltage fairly low, let's say 8-9 volts, you will get a power loss of 3-4 Watts/Amp output current. For 5A thats 15-20 Watts. Of this power, less than 1.3-1.8 Watts will be dissipated by the regulator itself. Wolfgang Rupprecht ARPA: wolfgang@mgm.mit.edu (IP 18.82.0.114) TEL: (617) 267-4365 UUCP: mit-eddie!mgm.mit.edu!wolfgang