Maintaining a Vbe Multiplier's bias value
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From: Robert Baer on 9 Feb 2010 02:33 Jamie wrote: > John Larkin wrote: > >> On Mon, 08 Feb 2010 17:39:49 -0800, Jon Kirwan >> <jonk(a)infinitefactors.org> wrote: >> >> >>> I think this fits in sci.electronics.design, not .basics. >>> >>> I'd like to consider the Vbe multiplier often used in audio >>> amplifiers to maintain a bias voltage for the output stage. >>> The purpose is to better mitigate against ripple in the >>> unregulated power supply rails and against the the VAS >>> voltage output resulting from amplified signal voltages. >>> >>> (The only active device under consideration is a BJT, though. >>> No JFETs or MOSFETS or opamps or other ICs.) >>> >>> The basic starting form for a Vbe multiplier is shown in Fig. >>> 1 and the bias voltage output is indicated there. Assume Q1 >>> is thermally coupled in some magic way, for now, in just the >>> right way so that if the current through the Vbe multiplier >>> were perfectly stable, that the bias voltage would track just >>> as needed (The 'Eg' of Q1 is exactly what's needed for the >>> output stage's temperature tracking in some nice way and the >>> values of R1 and R2 are set correctly and the thermal >>> coupling and location is somehow where it needs to be.) The >>> focus is on the Vbe multiplier's variation of bias in the >>> face of changes in sourcing current at the top of Fig. 1. >>> >>> >>>> : +V >>>> : | >>>> : resistor or >>>> : current source >>>> : | >>>> : ,---+---, >>>> : | | >>>> : \ | >>>> : / R2 | >>>> : \ +-----> upper quadrant >>>> : / | ^ >>>> : | | | >>>> : | |/c Q1 BIAS >>>> : +-----| VOLTAGE >>>> : | |>e | >>>> : \ | | >>>> : / R1 | v >>>> : \ +-----> lower quadrant >>>> : / | >>>> : | | >>>> : '---+---' >>>> : | >>>> : VAS ---' >>>> : >>>> : FIGURE 1 >>> >>> If I use a resistor as the load for the VAS, it's obvious to >>> me that the Vbe multiplier will need to cope with varying >>> currents. But even if I use a BJT (or two) to make a current >>> source sitting above the Vbe multiplier, it's still not going >>> to hold entirely still with +V ripple and with varying VAS >>> drive voltages. That variation will ultimately manifest >>> itself in a varying Vbe bias voltage. That will change the >>> operating point for the output stage. >>> >>> If it is class-A, I suppose it doesn't matter that much. But >>> I don't want to be forced into class-A operation. Nor do I >>> want to be forced into regulated rails. So it becomes a >>> little more important, I think, to get this nailed down >>> better. >>> >>> There's the problem, anyway. >>> >>> To quantify how bad all this really is, I tried my hand at >>> figuring out the small signal analysis of the Vbe multiplier. >>> If I got a first approximation about right, it is based >>> squarely upon the small-re of the BJT. The very familiar >>> value for (kT/q)/Ic. >>> >>> There is also the value of R2 shown in Fig. 1, but since its >>> effect is only affected by the change in base current, I >>> believe it's contribution is divided by Q1's beta. So the >>> actual equation is something like: >>> >>> R_ac = (1/Ic)*(kT/q)*(1+R2/R1) + R2/beta >>> >>> For a 2X multiplier where R2 is about R1, this is: >>> >>> R_ac = (2/Ic)*(kT/q) + R2/beta >>> >>> The Vbe multipler value is: >>> >>> V_bias = Vbe*(1+(R2/R1)) + R2*Ic/beta >>> >>> (The latter term being a correction for base current.) >>> >>> Ignoring base current for now and assuming I had Ic set >>> around 5mA and placed R1=R2=1k for the 2X factor, this R_ac >>> value works out to about 15.4 ohms. >>> >>> A variation of half an mA in Ic yields about 7.7mV change in >>> the bias point. >>> >>> I decided to see if the Early effect made much of a >>> difference. The adjustment appears to be something like >>> this: >>> >>> R_early = dV/dI = -Ic/VA*R^2 >>> >>> If I'm interpreting it right, it really does show as negative >>> resistance added to R_ac. The fuller equation, then, >>> including the Early effect, would be: >>> >>> R_ac = (2/Ic)*(kT/q) + R2/beta - Ic/VA*R^2 >>> >>> (Which requires a quadratic solution to solve for R.) >>> >>> If R_ac is 15.4 ohms and Ic is around 5mA, a VA of 100V would >>> suggest about R_early=-10mOhms. Which is roughly a factor of >>> 1500 less than 15.4 Ohms. Since it now appears to be on the >>> order of 0.1% or so for typical Ic, VA, and, R_ac values, I >>> think I can ignore it for these considerations. >>> >>> So drop it, I will. >>> >>> I had scouted around a few weeks back (not for this reason) >>> and found what is shown in Fig. 2. I remembered it, but >>> didn't understand it then. >>> >>> >>>> : +V >>>> : | >>>> : resistor or >>>> : current source >>>> : | >>>> : ,---+---, <-- node A >>>> : | | >>>> : | \ >>>> : | / R3 >>>> : \ \ >>>> : / R2 / >>>> : \ | >>>> : / +-----> upper quadrant >>>> : | | ^ >>>> : | |/c Q1 | >>>> : +-----| BIAS >>>> : | |>e VOLTAGE >>>> : \ | | >>>> : / R1 | v >>>> : \ +-----> lower quadrant >>>> : / | >>>> : | | >>>> : '---+---' >>>> : | >>>> : VAS ---' >>>> : >>>> : FIGURE 2 >>> >>> I think I now understand why R3 was there. Changes in Ic >>> create changes in Q1's collector voltage, per Ic*R3. The >>> result is that dV=dI*R3. If R3 is on the order of the above >>> computed R_ac, then variations at node A caused by changing >>> currents through the Vbe multipler (most of which are seen as >>> Ic changes) will be neatly compensated for the change in the >>> voltage drop caused by R3. >>> >>> However, that can only be set for some assumed Ic. Nearby >>> changes will work pretty well. But further deviations will >>> start to show problems again. Also, the Fig. 2 version will >>> use a slightly higher multiplier value to get node A up high >>> enough for the R3 drop to hit the right place required to >>> bias the output stage. That higher multiplier means that >>> while, let's say, the two (or four, if that's it) output >>> BJT's Vbe values vary over temp and the thermally coupled Q1 >>> above also varies it's own Vbe value, the multiplier other >>> than 2 (or 4) will mean the variation of the bias will match >>> at only one place -- if it ever did more than one spot. How >>> important that is, I've not considered yet. >>> >>> I'm wondering about additional topology changes to improve >>> the performance still more. Obviously, if they are crazy and >>> wild, I'm probably going to live with the above and be done >>> with it. But I think there's got to be something still >>> better. Another BJT as a bypass route across Q1 and R3? >> >> >> Hang a big capacitor across it. >> >> John >> > actually, I was going to suggest a diode in the base circuit to VAS to > help with thermo issues with that type of circuit.. > > oh well. > > A DCT would be better..and use of a matched pair better..and on the same die better still..National makes them.
From: Robert Baer on 9 Feb 2010 02:34 Jim Thompson wrote: > On Mon, 08 Feb 2010 17:39:49 -0800, Jon Kirwan > <jonk(a)infinitefactors.org> wrote: > >> I think this fits in sci.electronics.design, not .basics. >> >> I'd like to consider the Vbe multiplier often used in audio >> amplifiers to maintain a bias voltage for the output stage. >> The purpose is to better mitigate against ripple in the >> unregulated power supply rails and against the the VAS >> voltage output resulting from amplified signal voltages. >> >> (The only active device under consideration is a BJT, though. >> No JFETs or MOSFETS or opamps or other ICs.) >> >> The basic starting form for a Vbe multiplier is shown in Fig. >> 1 and the bias voltage output is indicated there. Assume Q1 >> is thermally coupled in some magic way, for now, in just the >> right way so that if the current through the Vbe multiplier >> were perfectly stable, that the bias voltage would track just >> as needed (The 'Eg' of Q1 is exactly what's needed for the >> output stage's temperature tracking in some nice way and the >> values of R1 and R2 are set correctly and the thermal >> coupling and location is somehow where it needs to be.) The >> focus is on the Vbe multiplier's variation of bias in the >> face of changes in sourcing current at the top of Fig. 1. >> >>> : +V >>> : | >>> : resistor or >>> : current source >>> : | >>> : ,---+---, >>> : | | >>> : \ | >>> : / R2 | >>> : \ +-----> upper quadrant >>> : / | ^ >>> : | | | >>> : | |/c Q1 BIAS >>> : +-----| VOLTAGE >>> : | |>e | >>> : \ | | >>> : / R1 | v >>> : \ +-----> lower quadrant >>> : / | >>> : | | >>> : '---+---' >>> : | >>> : VAS ---' >>> : >>> : FIGURE 1 >> If I use a resistor as the load for the VAS, it's obvious to >> me that the Vbe multiplier will need to cope with varying >> currents. But even if I use a BJT (or two) to make a current >> source sitting above the Vbe multiplier, it's still not going >> to hold entirely still with +V ripple and with varying VAS >> drive voltages. That variation will ultimately manifest >> itself in a varying Vbe bias voltage. That will change the >> operating point for the output stage. >> >> If it is class-A, I suppose it doesn't matter that much. But >> I don't want to be forced into class-A operation. Nor do I >> want to be forced into regulated rails. So it becomes a >> little more important, I think, to get this nailed down >> better. >> >> There's the problem, anyway. >> >> To quantify how bad all this really is, I tried my hand at >> figuring out the small signal analysis of the Vbe multiplier. >> If I got a first approximation about right, it is based >> squarely upon the small-re of the BJT. The very familiar >> value for (kT/q)/Ic. >> >> There is also the value of R2 shown in Fig. 1, but since its >> effect is only affected by the change in base current, I >> believe it's contribution is divided by Q1's beta. So the >> actual equation is something like: >> >> R_ac = (1/Ic)*(kT/q)*(1+R2/R1) + R2/beta >> >> For a 2X multiplier where R2 is about R1, this is: >> >> R_ac = (2/Ic)*(kT/q) + R2/beta >> >> The Vbe multipler value is: >> >> V_bias = Vbe*(1+(R2/R1)) + R2*Ic/beta >> >> (The latter term being a correction for base current.) >> >> Ignoring base current for now and assuming I had Ic set >> around 5mA and placed R1=R2=1k for the 2X factor, this R_ac >> value works out to about 15.4 ohms. >> >> A variation of half an mA in Ic yields about 7.7mV change in >> the bias point. >> >> I decided to see if the Early effect made much of a >> difference. The adjustment appears to be something like >> this: >> >> R_early = dV/dI = -Ic/VA*R^2 >> >> If I'm interpreting it right, it really does show as negative >> resistance added to R_ac. The fuller equation, then, >> including the Early effect, would be: >> >> R_ac = (2/Ic)*(kT/q) + R2/beta - Ic/VA*R^2 >> >> (Which requires a quadratic solution to solve for R.) >> >> If R_ac is 15.4 ohms and Ic is around 5mA, a VA of 100V would >> suggest about R_early=-10mOhms. Which is roughly a factor of >> 1500 less than 15.4 Ohms. Since it now appears to be on the >> order of 0.1% or so for typical Ic, VA, and, R_ac values, I >> think I can ignore it for these considerations. >> >> So drop it, I will. >> >> I had scouted around a few weeks back (not for this reason) >> and found what is shown in Fig. 2. I remembered it, but >> didn't understand it then. >> >>> : +V >>> : | >>> : resistor or >>> : current source >>> : | >>> : ,---+---, <-- node A >>> : | | >>> : | \ >>> : | / R3 >>> : \ \ >>> : / R2 / >>> : \ | >>> : / +-----> upper quadrant >>> : | | ^ >>> : | |/c Q1 | >>> : +-----| BIAS >>> : | |>e VOLTAGE >>> : \ | | >>> : / R1 | v >>> : \ +-----> lower quadrant >>> : / | >>> : | | >>> : '---+---' >>> : | >>> : VAS ---' >>> : >>> : FIGURE 2 >> I think I now understand why R3 was there. Changes in Ic >> create changes in Q1's collector voltage, per Ic*R3. The >> result is that dV=dI*R3. If R3 is on the order of the above >> computed R_ac, then variations at node A caused by changing >> currents through the Vbe multipler (most of which are seen as >> Ic changes) will be neatly compensated for the change in the >> voltage drop caused by R3. >> >> However, that can only be set for some assumed Ic. Nearby >> changes will work pretty well. But further deviations will >> start to show problems again. Also, the Fig. 2 version will >> use a slightly higher multiplier value to get node A up high >> enough for the R3 drop to hit the right place required to >> bias the output stage. That higher multiplier means that >> while, let's say, the two (or four, if that's it) output >> BJT's Vbe values vary over temp and the thermally coupled Q1 >> above also varies it's own Vbe value, the multiplier other >> than 2 (or 4) will mean the variation of the bias will match >> at only one place -- if it ever did more than one spot. How >> important that is, I've not considered yet. >> >> I'm wondering about additional topology changes to improve >> the performance still more. Obviously, if they are crazy and >> wild, I'm probably going to live with the above and be done >> with it. But I think there's got to be something still >> better. Another BJT as a bypass route across Q1 and R3? >> >> Getting this nailed down should help mitigate against both >> unreg supply ripple (on one side, anyway) putting hum into >> the output and also against large scale changes in the VAS >> amplified signal voltage (which means distortion.) >> >> Jon > > What's a "VAS"? > > What exactly are you trying to do? > > My nickname, as a kid engineer at Motorola (48 years ago), was "Vbe" > Thompson, because I could pull so much magic with Vbe compensation > methods ;-) > > (Vbe multipliers generally are used just to create a smaller dead-band > that is temperature stable. Class AB bias is an art form of which I > am expert, but cannot divulge publicly at this time :-) > > ...Jim Thompson That makes YOU biased...
From: Ban on 9 Feb 2010 02:36 "Jon Kirwan" <jonk(a)infinitefactors.org> schrieb im Newsbeitrag news:jg91n5d684ru5imsq1cfcjpjd1vddg2b2l(a)4ax.com... >I think this fits in sci.electronics.design, not .basics. > > I'd like to consider the Vbe multiplier often used in audio > amplifiers to maintain a bias voltage for the output stage. > The purpose is to better mitigate against ripple in the > unregulated power supply rails and against the the VAS > voltage output resulting from amplified signal voltages. > No this is not the purpose of this stage. It is used as an adjustable Zener and is used to create and temperature compensate the bias voltage of the output stage. A pur DC function. Since the power Transistors draw quite a bit of quiescent current the ripple of the power supply will be higher then without. The supply ripple is reduced mainly by negative feedback from output to the input stage. snip> >>: +V >>: | >>: resistor or >>: current source >>: | >>: ,---+---, >>: | | >>: \ | >>: / R2 | >>: \ +-----> upper quadrant >>: / | ^ >>: | | | >>: | |/c Q1 BIAS >>: +-----| VOLTAGE >>: | |>e | >>: \ | | >>: / R1 | v >>: \ +-----> lower quadrant >>: / | >>: | | >>: '---+---' >>: | >>: VAS ---' >>: >>: FIGURE 1 > > If I use a resistor as the load for the VAS, it's obvious to > me that the Vbe multiplier will need to cope with varying > currents. But even if I use a BJT (or two) to make a current > source sitting above the Vbe multiplier, it's still not going > to hold entirely still with +V ripple and with varying VAS > drive voltages. That variation will ultimately manifest > itself in a varying Vbe bias voltage. That will change the > operating point for the output stage. > > If it is class-A, I suppose it doesn't matter that much. But > I don't want to be forced into class-A operation. Nor do I > want to be forced into regulated rails. So it becomes a > little more important, I think, to get this nailed down > better. > In class A you do not use this kind of bias generator snip I think you should understand before calculating. It is not much of a use to view this stage in isolation without the output stage and the associated NFB paths. ciao Ban
From: Robert Baer on 9 Feb 2010 02:46 Tim Williams wrote: > "Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message > news:okr1n55h5dvjjklg760dllkqq50v7s38ib(a)4ax.com... >> Part of the function of the Vbe multiplier is to also track >> the Vbe requirements for the output stage as it heats up and >> cools down. > > The general idea is to put the Vbe transistor on the same heatsink as the > outputs, if not glued to a transistor directly. > > Unfortunately, for widely mismatched current densities, this doesn't work. > http://webpages.charter.net/dawill/Images/Ampere.gif > In this boringly typical circuit, the 2N3904 Vbe mult. doesn't have enough > tempco to compensate the far beefier (= lower current density??) output > darlingtons. > > I was thinking of adding another CCS so a constant voltage drop appears on > the Vbe's base divider resistor. Algebraically subtracting a fairly stable > voltage results in the effective tempco (percentwise) increasing. The base > divider ratio has to be changed to compensate. > >> In this case, I want it to track the output stage so I'm >> going to have to couple it thermally in some useful way. What >> I'm considering, right now, is how to make it immune to >> unregulated supply variations and VAS output voltage swings. > > Don't worry about stability -- as John said, bypass and forget about it. > Most of the dynamic VAS/CCS current flows into the output stage, since > that's what it's there for anyway. The capacitor helps turn on the N side / > turn off the P side for rising edges and vice versa. > > As for PSRR, the CCS's and gobs of feedback keep that in check. Of course, > in principle you need something to start the CCS's. ICs do this with a JFET > (i.e. current regulating diode) or bandgap reference (e.g., TL431), or > sometimes both, to set a master current, from which everything else is > mirrored. Most discrete circuits just use a resistor, which is "0%" PSRR, > but it's not all that bad because the currents are balanced (*on average*, > which means you'll see IMD products when it's moving). > > Tim > It is not so much as mismatched current densities, it is mismatched processes that greatly alter the pseudo-log I/V relationship in the power device. for small signal transistors, one can get over 10 decades matching to ideal; with some power devices one can be lucky to get that over 3-5 decades.
From: Robert Baer on 9 Feb 2010 02:48
John Larkin wrote: > On Tue, 9 Feb 2010 00:28:27 -0600, "Tim Williams" > <tmoranwms(a)charter.net> wrote: > >> "Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message >> news:okr1n55h5dvjjklg760dllkqq50v7s38ib(a)4ax.com... >>> Part of the function of the Vbe multiplier is to also track >>> the Vbe requirements for the output stage as it heats up and >>> cools down. >> The general idea is to put the Vbe transistor on the same heatsink as the >> outputs, if not glued to a transistor directly. >> >> Unfortunately, for widely mismatched current densities, this doesn't work. >> http://webpages.charter.net/dawill/Images/Ampere.gif >> In this boringly typical circuit, the 2N3904 Vbe mult. doesn't have enough >> tempco to compensate the far beefier (= lower current density??) output >> darlingtons. >> >> I was thinking of adding another CCS so a constant voltage drop appears on >> the Vbe's base divider resistor. Algebraically subtracting a fairly stable >> voltage results in the effective tempco (percentwise) increasing. The base >> divider ratio has to be changed to compensate. >> >>> In this case, I want it to track the output stage so I'm >>> going to have to couple it thermally in some useful way. What >>> I'm considering, right now, is how to make it immune to >>> unregulated supply variations and VAS output voltage swings. >> Don't worry about stability -- as John said, bypass and forget about it. >> Most of the dynamic VAS/CCS current flows into the output stage, since >> that's what it's there for anyway. The capacitor helps turn on the N side / >> turn off the P side for rising edges and vice versa. >> >> As for PSRR, the CCS's and gobs of feedback keep that in check. Of course, >> in principle you need something to start the CCS's. ICs do this with a JFET >> (i.e. current regulating diode) or bandgap reference (e.g., TL431), or >> sometimes both, to set a master current, from which everything else is >> mirrored. Most discrete circuits just use a resistor, which is "0%" PSRR, >> but it's not all that bad because the currents are balanced (*on average*, >> which means you'll see IMD products when it's moving). >> >> Tim > > This topology, thermally coupled Vbe multiplier, was mediocre 50 years > ago. And still is. > > John > > > So,use multiple DCTs in series if the desired multiple is near an integer (2?). |