ECL: Why'd they use negative voltages?
in [Design]
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From: Joerg on 8 Jun 2010 17:31 Jim Thompson wrote: > On Mon, 07 Jun 2010 17:01:03 -0700, Jim Thompson > <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: > >> On Mon, 7 Jun 2010 16:56:24 -0700, "Joel Koltner" >> <zapwireDASHgroups(a)yahoo.com> wrote: >> >>> I realize it was the early '60s and all, but why does ECL generally use 0V for >>> VCC and -5.2V for VEE, rather than, oh, say... 5V for VCC and 0V for VEE? >>> Something related to how things were done when toobs ruled? (I realize that >>> you can almost always run ECL off of 5V/0V -- and apparently this was popular >>> practice at one time?) >>> >>> And why 5.2V anyway? (Granted, 5.2V is no stranger than 6.3V filament >>> transformers, I suppose...) >>> >>> ---Joel >> Noise immunity is better with 0/-5.2V >> >> I was there when they (Narud, Seelbach, Philips, et al) did that. >> >> ...Jim Thompson > > See.... > > http://www.computerhistory.org/semiconductor/timeline/1965-Custom.html > Are you the guy on the right, with the wide-rimmed glasses? <ducking for cover> -- SCNR, Joerg http://www.analogconsultants.com/ "gmail" domain blocked because of excessive spam. Use another domain or send PM.
From: Vladimir Vassilevsky on 8 Jun 2010 17:36 John Larkin wrote: > On Tue, 08 Jun 2010 15:22:15 -0500, Vladimir Vassilevsky > <nospam(a)nowhere.com> wrote: > > >> >>John Larkin wrote: >> >> >>>On Tue, 08 Jun 2010 14:42:13 -0500, Vladimir Vassilevsky >>><nospam(a)nowhere.com> wrote: >>> >>> >>> >>>>John Larkin wrote: >>>> >>>> >>>> >>>> >>>>>One of the old Motorola books has a class-D amp that uses bipolar >>>>>supplies, half-bridge mosfets, LC filter, DC coupled to a grounded >>>>>loudspeaker. The problem is that, if you're swinging the load, say, >>>>>positive, you take power out of the + supply and pump power *into* the >>>>>- supply, and blow up its filter caps. >>>> >>>>The problem is known as "rail pumping". >>>> >>>> >>>> >>>>>Their fix was cute: an idler >>>>>circuit off to the side, a pair of mosfets switching at 50% duty >>>>>cycle, pumping V+ and V- into a dummy grounded inductor. That >>>>>automagically equalized the supply voltages. >>>> >>>>There is one small problem with this solution: it doesn't work. If >>>>anything is slightly off balance, that creates virtually unlimited >>>>current. >>> >>> >>>Nothing is off balance, because the idler forces things to balance. >>>The "virtually unlimited current" is what pulls down the higher supply >>>and boosts the lower one. In fact, it transfers energy from the >>>unused, higher voltage side of the supply to the actively-loaded, >>>lower-voltage side. Sorta cute. >>> >> >> >>Now think what happens if the duty cycle is not exactly 50/50. Or if the >> /+/ and /-/ supplies are slightly different. > > > If the power supplies are slightly different, the inductor sees a net > average DC. So the inductor current rises. The inductor then extracts > energy from one supply (the one with the higher magnitude) and pumps > energy into the other one. That tends to equalize the supply voltage > magnitudes. > These are soft supplies, especially in the absorbing-energy direction. > They aren't stiff, absolute voltages. Which is why it works. So the lower rail is getting powered from the higher rail through the idler circuit. With the corresponding circulating high currents, losses and asymmetry. It looks worse if you consider the waveform of the idler current. BTW, once I made similar circuit. FETs were controlled by the CPU; that allowed to optimize the operation (For any considerable currents and voltages, don't think of running this in uncontrollable mode). It worked reasonably well, however I didn't like it. > > Of course, a true h-bridge solves the same problem more efficiently. H-bridge needs twice as many FETs, drivers, protections and feedbacks; that is pricey. Also, it is impossible to connect two channels in a bridge configuration to double the output power. For those reasons, half bridge is usually preferred in audio. Synchronous rectifiers in the power supplies are good solution for this problem and for the efficiency reasons, but they come with a price also. Bottom line is everybody just uses big capacitor tanks; this is dumb, cheap and good enough for the job. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
From: John Larkin on 8 Jun 2010 17:53 On Tue, 08 Jun 2010 16:36:39 -0500, Vladimir Vassilevsky <nospam(a)nowhere.com> wrote: > > >John Larkin wrote: > >> On Tue, 08 Jun 2010 15:22:15 -0500, Vladimir Vassilevsky >> <nospam(a)nowhere.com> wrote: >> >> >>> >>>John Larkin wrote: >>> >>> >>>>On Tue, 08 Jun 2010 14:42:13 -0500, Vladimir Vassilevsky >>>><nospam(a)nowhere.com> wrote: >>>> >>>> >>>> >>>>>John Larkin wrote: >>>>> >>>>> >>>>> >>>>> >>>>>>One of the old Motorola books has a class-D amp that uses bipolar >>>>>>supplies, half-bridge mosfets, LC filter, DC coupled to a grounded >>>>>>loudspeaker. The problem is that, if you're swinging the load, say, >>>>>>positive, you take power out of the + supply and pump power *into* the >>>>>>- supply, and blow up its filter caps. >>>>> >>>>>The problem is known as "rail pumping". >>>>> >>>>> >>>>> >>>>>>Their fix was cute: an idler >>>>>>circuit off to the side, a pair of mosfets switching at 50% duty >>>>>>cycle, pumping V+ and V- into a dummy grounded inductor. That >>>>>>automagically equalized the supply voltages. >>>>> >>>>>There is one small problem with this solution: it doesn't work. If >>>>>anything is slightly off balance, that creates virtually unlimited >>>>>current. >>>> >>>> >>>>Nothing is off balance, because the idler forces things to balance. >>>>The "virtually unlimited current" is what pulls down the higher supply >>>>and boosts the lower one. In fact, it transfers energy from the >>>>unused, higher voltage side of the supply to the actively-loaded, >>>>lower-voltage side. Sorta cute. >>>> >>> >>> >>>Now think what happens if the duty cycle is not exactly 50/50. Or if the >>> /+/ and /-/ supplies are slightly different. >> >> >> If the power supplies are slightly different, the inductor sees a net >> average DC. So the inductor current rises. The inductor then extracts >> energy from one supply (the one with the higher magnitude) and pumps >> energy into the other one. That tends to equalize the supply voltage >> magnitudes. >> These are soft supplies, especially in the absorbing-energy direction. >> They aren't stiff, absolute voltages. Which is why it works. > >So the lower rail is getting powered from the higher rail through the >idler circuit. With the corresponding circulating high currents, Only as much current as needed to equalize the supply voltages. That turns out to be less current than the half-bridge is pumping into the speaker. losses >and asymmetry. It looks worse if you consider the waveform of the idler >current. It's a triangle. The more the L, the smaller the triangle. > >BTW, once I made similar circuit. FETs were controlled by the CPU; that >allowed to optimize the operation (For any considerable currents and >voltages, don't think of running this in uncontrollable mode). It worked >reasonably well, however I didn't like it. > >> >> Of course, a true h-bridge solves the same problem more efficiently. > >H-bridge needs twice as many FETs, drivers, protections and feedbacks; >that is pricey. Also, it is impossible to connect two channels in a >bridge configuration to double the output power. For those reasons, half >bridge is usually preferred in audio. > Synchronous rectifiers in the power supplies are good solution for >this problem and for the efficiency reasons, but they come with a price >also. A synchronous rectifier in the supplies doesn't help. If you push power back into a supply, as the half-bridge likes to do, then energy has got to go somewhere. A synchronous rectifier just pushes it uphill into the caps of the bulk supply. > >Bottom line is everybody just uses big capacitor tanks; this is dumb, >cheap and good enough for the job. For AC loads, yes. John
From: John Larkin on 8 Jun 2010 17:56 On Tue, 08 Jun 2010 14:29:38 -0700, Joerg <invalid(a)invalid.invalid> wrote: >John Larkin wrote: >> On Tue, 08 Jun 2010 09:05:42 -0700, Joerg <invalid(a)invalid.invalid> >> wrote: >> >>> John Larkin wrote: >>>> On Tue, 08 Jun 2010 09:12:24 +0200, Jeroen Belleman >>>> <jeroen(a)nospam.please> wrote: >>>> >>>>> John Larkin wrote: >>>>>> 10KH type ECL does work fine at Vcc=5, Vee=0. That's called "PECL" >>>>>> mode, originally "pseudo ECL" and lately "positive ECL". One generally >>>>>> references all the signals to a nice 5-volt copper pour. >>>>> I once designed a VME module with a lot of PECL with signals terminated >>>>> into +3V, and which also had some circuitry running between +3V and >>>>> GND. The +3V net was shared. Since the combined current of all the PECL >>>>> terminators largely exceeded the consumption of the stuff between +3V >>>>> and GND, I used a negative regulator with its input connected to GND to >>>>> make the +3V. >>>>> >>>>> That raised some eyebrows, but it worked fine. >>>>> >>>>> Jeroen Belleman >>>> >>>> It's fun to use regulators "upside down." >>>> >>>> We need a good bipolar-drive regulator. I use LM8261s for small stuff, >>>> and occasionaly LT1010s for bigger stuff. >>>> >>> Check out the big fat audio amp hybrids. Of course, one has to be >>> careful and pick one that isn't going obsolete next year because a >>> particular car stereo was discontinued. >>> >>> Heck, if you want to go green on this you might even consider class-D. >>> Then claim your carbon offset :-) >> >> A non-H-bridge class D amp can have circulating-current problems... it >> takes current out of one supply rail and dumps it into the other. >> Conservation of energy. >> > >Thou shalt use an inductor towards the load :-) Of course. The inductor causes the problem. If a bipolar half-bridge drove a resistive load, there would be no circulating current. > >Essentially this is how synchronous buck converters work, They use unipolar bulk supplies, and can dump their circulating current into ground. John
From: Joerg on 8 Jun 2010 18:01
Vladimir Vassilevsky wrote: > > > John Larkin wrote: > >> On Tue, 08 Jun 2010 15:22:15 -0500, Vladimir Vassilevsky >> <nospam(a)nowhere.com> wrote: >> >> >>> >>> John Larkin wrote: >>> >>> >>>> On Tue, 08 Jun 2010 14:42:13 -0500, Vladimir Vassilevsky >>>> <nospam(a)nowhere.com> wrote: >>>> >>>> >>>> >>>>> John Larkin wrote: >>>>> >>>>> >>>>> >>>>> >>>>>> One of the old Motorola books has a class-D amp that uses bipolar >>>>>> supplies, half-bridge mosfets, LC filter, DC coupled to a grounded >>>>>> loudspeaker. The problem is that, if you're swinging the load, say, >>>>>> positive, you take power out of the + supply and pump power *into* >>>>>> the >>>>>> - supply, and blow up its filter caps. >>>>> >>>>> The problem is known as "rail pumping". >>>>> >>>>> >>>>> >>>>>> Their fix was cute: an idler >>>>>> circuit off to the side, a pair of mosfets switching at 50% duty >>>>>> cycle, pumping V+ and V- into a dummy grounded inductor. That >>>>>> automagically equalized the supply voltages. >>>>> >>>>> There is one small problem with this solution: it doesn't work. If >>>>> anything is slightly off balance, that creates virtually unlimited >>>>> current. >>>> >>>> >>>> Nothing is off balance, because the idler forces things to balance. >>>> The "virtually unlimited current" is what pulls down the higher supply >>>> and boosts the lower one. In fact, it transfers energy from the >>>> unused, higher voltage side of the supply to the actively-loaded, >>>> lower-voltage side. Sorta cute. >>>> >>> >>> >>> Now think what happens if the duty cycle is not exactly 50/50. Or if >>> the /+/ and /-/ supplies are slightly different. >> >> >> If the power supplies are slightly different, the inductor sees a net >> average DC. So the inductor current rises. The inductor then extracts >> energy from one supply (the one with the higher magnitude) and pumps >> energy into the other one. That tends to equalize the supply voltage >> magnitudes. >> These are soft supplies, especially in the absorbing-energy direction. >> They aren't stiff, absolute voltages. Which is why it works. > > So the lower rail is getting powered from the higher rail through the > idler circuit. With the corresponding circulating high currents, losses > and asymmetry. It looks worse if you consider the waveform of the idler > current. > > BTW, once I made similar circuit. FETs were controlled by the CPU; that > allowed to optimize the operation (For any considerable currents and > voltages, don't think of running this in uncontrollable mode). It worked > reasonably well, however I didn't like it. > >> >> Of course, a true h-bridge solves the same problem more efficiently. > > H-bridge needs twice as many FETs, drivers, protections and feedbacks; > that is pricey. Also, it is impossible to connect two channels in a > bridge configuration to double the output power. For those reasons, half > bridge is usually preferred in audio. > Synchronous rectifiers in the power supplies are good solution for this > problem and for the efficiency reasons, but they come with a price also. > > Bottom line is everybody just uses big capacitor tanks; this is dumb, > cheap and good enough for the job. > Pumping is normally only an issue with low frequency AC drive. There is another trick: If you have two amps (usually the case with audio) operate them 180 degree out of phase down there. -- Regards, Joerg http://www.analogconsultants.com/ "gmail" domain blocked because of excessive spam. Use another domain or send PM. |