Maintaining a Vbe Multiplier's bias value
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From: Jim Thompson on 12 Feb 2010 15:30 On Fri, 12 Feb 2010 12:18:33 -0800, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Fri, 12 Feb 2010 08:16:34 -0800 (PST), George Herold ><ggherold(a)gmail.com> wrote: > >>On Feb 12, 9:52�am, Jim Thompson <To-Email-Use-The-Envelope-I...(a)My- >>Web-Site.com> wrote: >>> On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs >>> >>> >>> >>> >>> >>> <pcdhSpamMeSensel...(a)electrooptical.net> wrote: >>> >On 2/11/2010 10:35 PM, George Herold wrote: >>> >> On Feb 11, 1:14 pm, John Larkin >>> >> <jjlar...(a)highNOTlandTHIStechnologyPART.com> �wrote: >>> >>> On Thu, 11 Feb 2010 10:08:38 -0600, "Tim Williams" >>> >>> >>> <tmoran...(a)charter.net> �wrote: >>> >>>> "Jim Thompson"<To-Email-Use-The-Envelope-I...(a)My-Web-Site.com> �wrote in >>> >>>> messagenews:r358n59g5vkv4brn2vc795lhoineb2jvhd(a)4ax.com... >>> >>>>> And a 2.5V "dead-band", but it _is_ precisely known, and temperature >>> >>>>> stable. �Interesting thought if you have high enough power supplies. >>> >>> >>>> Bonus: the dead band allows you to use that TL431 "Vbe" mentinoed earlier. >>> >>> >>>> Too bad they're so slow (hardly capable for audio). �Does anyone make "fast" >>> >>>> regulators (without being stupid LDOs)? >>> >>> >>>> Tim >>> >>> >>> If you drive both adjust pins with the signal input, the 317 output is >>> >>> Vin+1.25 and the 337 output is Vin-1.25. Connect them to the output >>> >>> through a couple of resistors, valued to set the idle current. Where's >>> >>> the deadband? >>> >>> >>> Or you can take the output from the 317 output pin, with the 337 now >>> >>> acting like a constant-current sink to the 317. >>> >>> >>> I like to use LM1117s as power emitter followers, inside the loop of >>> >>> an opamp. That makes a cheap, well protected power driver, for load >>> >>> cell excitation and such. I did a bunch of tests to see whether >>> >>> flailing the adj pin can damage the regulator, and never managed to >>> >>> break one. >>> >>> >>> John >>> >>> >> Cool! �I think I got it... though if I try it in the future and let >>> >> the smoke out of something... then I might have questions. >>> >>> >> George H. >>> >>> >IIRC the LM395 is basically an LM309 with the voltage reference removed. >>> > �Emitter-follower regulators are nearly bulletproof unless you >>> >discharge a cap into the output. >>> >>> >Cheers >>> >>> >Phil Hobbs >>> >>> Close, but no cigar, LM395 = LM317 with some metal rearrangements. >>> >>> I did this analysis for ICE back in 1980: >>> >>> � � � �http://analog-innovations.com/SED/ICE-LM195-LM117.pdf >>> >>> Additionally: Amusing myself with the thoughts of complementary- >>> follower-style power amplifiers made from LM317/LM337 pairs, it fails >>> because both, internally, are NPN's pass devices, so the LM337 has GBW >>> and stability issues plus it needs substantial idle load to stay >>> _vaguely_ stable. >>> >>> � � � � � � � � � � � � � � � � � � � � ...Jim Thompson [snip] >>> - Show quoted text - >> >>Thanks Jim, I always wondered why there wasn't a compliment to the >>LM395. (And why the negative voltage regulators would 'wig out' if >>you didn't follow the cap reccomendations.) >> >>George H. > >There are lots of positive LDOs around these days that have PNP >(namely high impedance) outputs. They tend to be very picky about >output capacitance ESR and such, with appropriately evasive >datasheets. I had one powering an FPGA core, putting a 0.2 volt p-p, >50 KHz triangle into the poor chip. That caused a lot of jitter. > >John It would have increased the die cost by about 1� to compensate it properly on chip ;-) I do it successfully all the time using PMOS pass devices. ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
From: bg on 12 Feb 2010 15:43 big snip - This is an example of a common emitter voltage amplifier. It might be one of the easiest stages to design. Normaly, I would start off by knowing what I need for the stage, for example I need a voltage gain of 20, an input impedance of 10K, Z out of ?? etc ---- An important thing to notice is that I do not labor over any tedious calculations. All of the part values are based on simple resistor ratios. Knowing how the ratios work, and how they affect each other, allows me to adust just about any parameter with a quick mental estimate. Example - - - - The circuit is based on the 2n3904 again, and there are probably hundreds if not thousands of other transistors that could fit in this circuit and work just as well. The voltage gain of a common emitter stage is reduced to RL/ RE. The emitter resistor, RE , has to provide enough feedback for this to be true. I can't Make RE = 1 ohm and Rl = 1megohm and have a gain of 1 million. There are limits. If I intended to run a CE amp with a 1 ma collector Q point current, and with a 15 volt supply, I might choose 7.5K for RL which would put Vc at Vcc/2. I know there will be an emitter resistor for feedback, and because there is a voltage drop across RE, I will want to raise the collector Q point voltage a bit higher to achieve the largest signal swing possible. For a starting point, I'll make RL = 6.8k and RE 680 ohms. When the transistor is cut off Vc = 15 volts. When the transistor is saturated RL and RE form a voltage divider across the 15 volt supply, therefore Vc = 1.4 volts. The midpoint between 15v and 1.4v comes out to 8.2 volts. Notice I neglected Vce when the transistor is saturated. The overall picture is more important at this point than details. I now can expect a voltage gain of 10, a collector Q point of 8.2v at 1ma, and an output resistance of about 6.8k. Up to this point, here is the sum total of calculations - The 1ma collector current was a choice because I know than a 2n3904 is intended to operate at a collector current in that range. I want a voltage gain of ten, no particluar reason, again it was a choice. Re = RL/10 I do that in my head. I did use a calculator to arrive at the desired Q point voltage. The next step - The resistor RB which connects between the base and ground needs to be small compared to the input impedance of the transistor. I know from experience , and also because I read the book "transistor circuit design" , that RB should typically be about 10 to 20 times larger than RE. So I'll go with 6.8K. I can do this because I know that the input of the transistor will appear as an open circuit compared to RB. It isn't any different than a simple voltage divider. For example if I had a divider using two 1K resistors, I know that any load of 10K or higher isn't going to significantly change the divider ratio. I'll still get a division of about 1/2 right? As the load dips below 10K, the output of the divider will be less. Gut feel, and how much accuracy I'm willing to loose, determines if I need to crunch some numbers. The only calculation I made for RB is - - RB equals 10 to 20 RE The next step, is to select the forward bias resistor RA that connects between the base and Vcc. For this resistor I made an educated guess and hit the exact value on the very first shot. I knew that the ratio of Rb to Ra would be close to 10 to 1. 68k is a standard value so I chose that and expected to tweak it later. But in all fairness, this is how I arrived at that guess. The voltage across RB should be Vbe + Vre. We know Vre is 1ma through 680 ohms, and notice that I neglected the base current flowing through RE becuase IE probably equals Ic + 1%Ic, the one percent doesn't matter. Vbe is a tough one to call. Typically it ranges from .5 to .7 volts. Seeing as I will probably miss the target, and I will want to use a standard value, I'll do the math, see where I come out and tweak values if neccessary. The math tells me that the current through RA is the same current as through RB + Ib. The current through RB is Vbe + Vre / Rb = about 1.6 volts/6.8k = 235 micro amp. Notice that I neglected Ib, and I can because I expect it to insignificant. Therefore RA drops 15 volts - 1.6 volts at 235 micro amp which works out to 57k. The next step If you put all of this into circuitmaker , the collector Q point comes in at 6.2 volts if Ra = 57k. An obvious tweak is to raise Ra to the next standard value, but for the model I used, Ra works out to be 68k for a collector Q point of 8.37v which was my first guess. The beauty of this is that I can scale Ra and Rb up or down so that standard values can be used. As long as I keep Rb around 10RE. Circuitmaker shows that the input impedance is 5.8k up until the input capacitance becomes an issue. My 6.8k Rb is being shunted by Ra and the transistor input impedance so I would expect the input Z to be less than 6.8K. The parallel equivalent of Ra and Rb is a hair under 6.8k, therefore the majority of the shunting effect is due to the transistor. I could quite easily lower Ra and Rb to make the transistor input Z even less noticeable. Circuit maker also confirms that the output resistance is 6.8k and the voltage gain is 19.571db (10x would be 20db). The peak to peak output signal clips around 13 volts. You might have noticed that I never mentioned Beta. Over the range of collector current that this stage operates , I can expect beta to be alot higher than my target voltage gain. I have an open loop current gain probably 10 times higher than my voltage gain with feedback. If I consider the equation A' = A/ 1+AB, I could solve for B or just plug numbers in to see that the stage gain is controlled by the feedback, and as long as Beta remains fairly high, it is not a factor. From experience, I know that my stage voltage gain will be much lower (due to feedback), than the transistor current gain. Therefore I can ignore the transistor current gain. As a voltage amplifier for low frequencies, I would expect that my preceeding stage will need an output resistance about 10 times lower than this stage's input resistance. The preceeding collector load resistor would need to be 680 ohms or less. The following stage would need an input impedance of 10 times this stage's output impedance or 68k. Those values are what I would be aiming for if I was putting this stage between two voltage amplifiers. Summary - #1 Select a transistor that is meant to operate at the collector current for which you intend to operate. The collector load resistor IS the output impedance. That value combined with the collector Q point current determines the Q point voltage. Any value can be varied to zero in on the best compramise of Zout, Ic, RL, Vc etc. #2 Select your voltage gain with, Gain = RL/RE Again vary any value to achieve whatever compramise best suits your situation. #3 Consider what the peak to peak output signal needs to be. Make sure that your Vcc and Q point can accomodate that signal swing. Your peak to peak output signal is found by considering the collector voltage when the transistor is open, and then when it is shorted. To avoid clipping at the signal peaks, Vcc should be high enough so that the signal does not need to swing to those extremes. In other words , leave some headroom. More headroom will avoid distorting at the signal peaks. It is not uncommon for Vcc to be 50% or higher than the peak to peak signal swing. The only limit is that the transistor's maximum power dissapation or voltage rating might be exceeded. #4 Revisit steps one through three. These steps determine the voltage gain and the maximum output signal that can be ahieved #5 Select a value for Rb. Rb should be low compared to Re. Rb will be close to the input impedance of your amp so consider that too. As you raise the value of Rb, the transistor shunting effect becomes more significant and the input Z of the amp becomes less than Rb. Again it is a compramise. There is no limit for how low you go with Rb, but there would be benifits to using something like a common base amplifier instead of a common emitter amplifier if Rb needs to be very low. #6 Select a value for Ra. This is best done by considering the bias voltage needed across Rb. If an exact figure for VBE is not known, and it seldom is you can always assume .6 volts, tweak later. The voltage across RB is .6 volts + the voltage across the emitter resistor. Ohms law will get you the current in Rb and from there Ra will need to drop Vcc - Erb at Irb amps. Bear in mind that ending up with standard values is a plus. Rb and Ra can be tweaked to get the best compramise. As those values are varied to standard values, the collector Q point voltage will shift, as will the input impedance. There are other compramises that affect frequency response, noise, power dissapation, maximum ratings and so on. This is meant to show only how to juggle values around without resorting to tedious calculations, to achieve the DC operating point, set input and output z's, voltage gain and signal swing peak levels.
From: Jim Thompson on 12 Feb 2010 15:44 On Fri, 12 Feb 2010 13:43:37 -0700, "bg" <bg(a)nospam.com> wrote: >big snip - > > >This is an example of a common emitter voltage amplifier. It might be one of >the easiest stages to design. Normaly, I would start off by knowing what I >need for the stage, for example I need a voltage gain of 20, an input >impedance of 10K, Z out of ?? etc ---- > >An important thing to notice is that I do not labor over any tedious >calculations. All of the part values are based on simple resistor ratios. >Knowing how the ratios work, and how they affect each other, allows me to >adust just about any parameter with a quick mental estimate. > >Example - - - - >The circuit is based on the 2n3904 again, and there are probably hundreds if >not thousands of other transistors that could fit in this circuit and work >just as well. >The voltage gain of a common emitter stage is reduced to RL/ RE. The >emitter resistor, RE , has to provide enough feedback for this to be true. I >can't Make RE = 1 ohm and Rl = 1megohm and have a gain of 1 million. There >are limits. > >If I intended to run a CE amp with a 1 ma collector Q point current, and >with a 15 volt supply, I might choose 7.5K for RL which would put Vc at >Vcc/2. I know there will be an emitter resistor for feedback, and because >there is a voltage drop across RE, I will want to raise the collector Q >point voltage a bit higher to achieve the largest signal swing possible. For >a starting point, I'll make RL = 6.8k and RE 680 ohms. When the transistor >is cut off Vc = 15 volts. When the transistor is saturated RL and RE form a >voltage divider across the 15 volt supply, therefore Vc = 1.4 volts. The >midpoint between 15v and 1.4v comes out to 8.2 volts. >Notice I neglected Vce when the transistor is saturated. The overall picture >is more important at this point than details. I now can expect a voltage >gain of 10, a collector Q point of 8.2v at 1ma, and an output resistance of >about 6.8k. >Up to this point, here is the sum total of calculations - >The 1ma collector current was a choice because I know than a 2n3904 is >intended to operate at a collector current in that range. >I want a voltage gain of ten, no particluar reason, again it was a choice. >Re = RL/10 I do that in my head. >I did use a calculator to arrive at the desired Q point voltage. > >The next step - >The resistor RB which connects between the base and ground needs to be small >compared to the input impedance of the transistor. I know from experience , >and also because I read the book "transistor circuit design" , that RB >should typically be about 10 to 20 times larger than RE. So I'll go with >6.8K. I can do this because I know that the input of the transistor will >appear as an open circuit compared to RB. It isn't any different than a >simple voltage divider. For example if I had a divider using two 1K >resistors, I know that any load of 10K or higher isn't going to >significantly change the divider ratio. I'll still get a division of about >1/2 right? As the load dips below 10K, the output of the divider will be >less. Gut feel, and how much accuracy I'm willing to loose, determines if I >need to crunch some numbers. >The only calculation I made for RB is - - RB equals 10 to 20 RE > >The next step, is to select the forward bias resistor RA that connects >between the base and Vcc. >For this resistor I made an educated guess and hit the exact value on the >very first shot. I knew that the ratio of Rb to Ra would be close to 10 to >1. >68k is a standard value so I chose that and expected to tweak it later. But >in all fairness, this is how I arrived at that guess. >The voltage across RB should be Vbe + Vre. We know Vre is 1ma through 680 >ohms, and notice that I neglected the base current flowing through RE >becuase IE probably equals Ic + 1%Ic, the one percent doesn't matter. Vbe is >a tough one to call. Typically it ranges from .5 to .7 volts. Seeing as I >will probably miss the target, and I will want to use a standard value, I'll >do the math, see where I come out and tweak values if neccessary. The math >tells me that the current through RA is the same current as through RB + >Ib. The current through RB is Vbe + Vre / Rb = about 1.6 volts/6.8k = 235 >micro amp. Notice that I neglected Ib, and I can because I expect it to >insignificant. >Therefore RA drops 15 volts - 1.6 volts at 235 micro amp which works out to >57k. > >The next step >If you put all of this into circuitmaker , the collector Q point comes in at >6.2 volts if Ra = 57k. An obvious tweak is to raise Ra to the next standard >value, but for the model I used, Ra works out to be 68k for a collector Q >point of 8.37v which was my first guess. > >The beauty of this is that I can scale Ra and Rb up or down so that standard >values can be used. As long as I keep Rb around 10RE. > >Circuitmaker shows that the input impedance is 5.8k up until the input >capacitance becomes an issue. My 6.8k Rb is being shunted by Ra and the >transistor input impedance so I would expect the input Z to be less than >6.8K. The parallel equivalent of Ra and Rb is a hair under 6.8k, therefore >the majority of the shunting effect is due to the transistor. I could quite >easily lower Ra and Rb to make the transistor input Z even less noticeable. >Circuit maker also confirms that the output resistance is 6.8k and the >voltage gain is 19.571db (10x would be 20db). The peak to peak output signal >clips around 13 volts. > >You might have noticed that I never mentioned Beta. Over the range of >collector current that this stage operates , I can expect beta to be alot >higher than my target voltage gain. I have an open loop current gain >probably 10 times higher than my voltage gain with feedback. If I consider >the equation A' = A/ 1+AB, I could solve for B or just plug numbers in to >see that the stage gain is controlled by the feedback, and as long as Beta >remains fairly high, it is not a factor. From experience, I know that my >stage voltage gain will be much lower (due to feedback), than the transistor >current gain. Therefore I can ignore the transistor current gain. > >As a voltage amplifier for low frequencies, I would expect that my >preceeding stage will need an output resistance about 10 times lower than >this stage's input resistance. The preceeding collector load resistor would >need to be 680 ohms or less. The following stage would need an input >impedance of 10 times this stage's output impedance or 68k. Those values are >what I would be aiming for if I was putting this stage between two voltage >amplifiers. > >Summary - >#1 >Select a transistor that is meant to operate at the collector current for >which you intend to operate. >The collector load resistor IS the output impedance. That value combined >with the collector Q point current determines the Q point voltage. Any value >can be varied to zero in on the best compramise of Zout, Ic, RL, Vc etc. > >#2 >Select your voltage gain with, Gain = RL/RE >Again vary any value to achieve whatever compramise best suits your >situation. > >#3 >Consider what the peak to peak output signal needs to be. Make sure that >your Vcc and Q point can accomodate that signal swing. >Your peak to peak output signal is found by considering the collector >voltage when the transistor is open, and then when it is shorted. To avoid >clipping at the signal peaks, Vcc should be high enough so that the signal >does not need to swing to those extremes. In other words , leave some >headroom. More headroom will avoid distorting at the signal peaks. It is not >uncommon for Vcc to be 50% or higher than the peak to peak signal swing. The >only limit is that the transistor's maximum power dissapation or voltage >rating might be exceeded. > >#4 >Revisit steps one through three. These steps determine the voltage gain and >the maximum output signal that can be ahieved > >#5 >Select a value for Rb. Rb should be low compared to Re. Rb will be close to >the input impedance of your amp so consider that too. As you raise the value >of Rb, the transistor shunting effect becomes more significant and the input >Z of the amp becomes less than Rb. Again it is a compramise. There is no >limit for how low you go with Rb, but there would be benifits to using >something like a common base amplifier instead of a common emitter amplifier >if Rb needs to be very low. > >#6 >Select a value for Ra. This is best done by considering the bias voltage >needed across Rb. If an exact figure for VBE is not known, and it seldom is >you can always assume .6 volts, tweak later. The voltage across RB is .6 >volts + the voltage across the emitter resistor. Ohms law will get you the >current in Rb and from there Ra will need to drop Vcc - Erb at Irb amps. >Bear in mind that ending up with standard values is a plus. Rb and Ra can be >tweaked to get the best compramise. As those values are varied to standard >values, the collector Q point voltage will shift, as will the input >impedance. > >There are other compramises that affect frequency response, noise, power >dissapation, maximum ratings and so on. This is meant to show only how to >juggle values around without resorting to tedious calculations, to achieve >the DC operating point, set input and output z's, voltage gain and signal >swing peak levels. > > Son of a gun. Yesterday you couldn't even spell "engineer", now you are one (;-0 ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
From: Bob Monsen on 12 Feb 2010 16:21 "Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)My-Web-Site.com> wrote in message news:0m1bn5ln6ag64hdqnmr4te1qjeiec22i02(a)4ax.com... > On Fri, 12 Feb 2010 08:23:36 -0800, John Larkin > <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On Fri, 12 Feb 2010 10:23:24 -0500, Phil Hobbs >><pcdhSpamMeSenseless(a)electrooptical.net> wrote: >> >>>On 2/12/2010 9:52 AM, Jim Thompson wrote: >>>> On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs >>>> <pcdhSpamMeSenseless(a)electrooptical.net> wrote: > [snip] >>>>> >>>>> IIRC the LM395 is basically an LM309 with the voltage reference >>>>> removed. >>>>> Emitter-follower regulators are nearly bulletproof unless you >>>>> discharge a cap into the output. >>>>> >>>>> Cheers >>>>> >>>>> Phil Hobbs >>>> >>>> Close, but no cigar, LM395 = LM317 with some metal rearrangements. >>>> >>>I was thinking 317 but then I decided that the 395 was older than that. >>> I guess not. >>> >>>> I did this analysis for ICE back in 1980: >>>> >>>> http://analog-innovations.com/SED/ICE-LM195-LM117.pdf >>>> >>>> Additionally: Amusing myself with the thoughts of complementary- >>>> follower-style power amplifiers made from LM317/LM337 pairs, it fails >>>> because both, internally, are NPN's pass devices, so the LM337 has GBW >>>> and stability issues plus it needs substantial idle load to stay >>>> _vaguely_ stable. >>> >>>Non-LDO three-terminal regulators are so trouble-free that it's easy to >>>confuse them with Newton's laws. ;) >>> >>>Cheers >>> >>>Phil Hobbs >> >>LM1117 is an "MDO" regulator. It has an NPN pass transistor but a bit >>lower dropout voltage than an LM317. Its ideal as a 3.3-to-1.25 volt >>FPGA core voltage source... no resistors! My purchasing notes say "Do >>not buy Fairchild per JL" but I can't recall why. >> >>As with all vregs, one has to be careful about the output capacitors. >> >>John > > I'm puzzled why the big semiconductor houses don't turn out discrete > LDO's on a CMOS process. I do it all the time on complex CMOS > ASIC's... like a PLL chip, fundamental power fed from +3.3V, but > internal regulators producing +2.5V and +1.8V... at hundreds of mA !!! > ST Arm Cortex-M3 has an internal 1.8V regulator for the core, and can take any input voltage from 2.0 to 3.8V. Regards, Bob Monsen
From: bg on 12 Feb 2010 16:25
Jim Thompson wrote in message ... >On Fri, 12 Feb 2010 13:43:37 -0700, "bg" <bg(a)nospam.com> wrote: > >>big snip - >> > >Son of a gun. Yesterday you couldn't even spell "engineer", now you >are one (;-0 I finally did something right? Wow!!! |