On 3/29/2013 9:04 PM, Phil Allison wrote:> "Phil Hobbs" >> >> As Phil A. says, the idea of bootstrapping is that if the input admittance >> of your circuit has zero swing across it, it draws zero current. >> > (snip) >> >> This is not a free lunch, because the SNR stays more or less the same, but >> it does give you a much nicer frequency response in general. >> > > ** One issue with bootstrapping input resistors is that it can dramatically > increase the noise compared to simply using a large value resistor. > > 30 odd years ago I attempted to build a JFET pre-amp for a condenser mic > capsule and not having any 1Gohm resistors handy tried bootstrapping a > 10Mohm one. The pre-amp tested fine, with an effective input resistance > close to 1Gohm - ie response was flat across the audio band when driven > via a 22pF cap simulating the capsule. > > When the capsule was tried, the background noise ( hiss) was about 20dB more > than with a commercial pre-amp and quite unacceptable for studio work. > > Thing is, with a 1Gohm ( gate bias) resistor, 22pF is enough to shunt nearly > all the audio frequency ( Johnson ) noise away - not so with 10 Mohms and a > bunch of positive feedback in place. >Yup. Bootstrapping doesn't help the SNR, it just flattens out the frequency response. It's the same with photodiodes--the noise floor due to the amplifier's voltage noise rises steadily with frequency. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 USA +1 845 480 2058 hobbs at electrooptical dot net http://electrooptical.net
How to bootstrap
Started by ●March 28, 2013
Reply by ●March 29, 20132013-03-29
Reply by ●March 29, 20132013-03-29
Jim Thompson wrote:> On Fri, 29 Mar 2013 19:53:44 -0400, "Tom Del Rosso" > <tomd_u1@verizon.net.invalid> wrote: > > > Jim Thompson wrote: > > > > > > Just analyze an ideal amplifier of gain +(1-delta) where delta is > > > small, but non-zero Apply a feedback resistor from output to > > > input, then calculate input impedance. > > > > What delta - just a change on the input? If gain is a function of > > that, even with the minus sign, I'd think it was an exponential > > amplifier rather than 'ideal'. > > GAIN = (1-delta) , delta small, but non-zero; so say that GAIN = 0.99, > for example. > > Draw yourself a picture/schematic of what I wrote in words.Ok I can solve for that, if I treat delta like a constant of, say, 0.01. But why do you call it delta? What change does it refer to? -- Reply in group, but if emailing add one more zero, and remove the last word.
Reply by ●March 30, 20132013-03-30
"Tom Del Rosso"> > But why do you call it delta? What change does it refer to? >** One is reminded of Humpty Dumpty's declaration to Alice: " When I use a word ... it means just what I want it to mean, neither more nor less" ... Phil
Reply by ●March 30, 20132013-03-30
On Fri, 29 Mar 2013 22:45:03 -0400, "Tom Del Rosso" <tomd_u1@verizon.net.invalid> wrote:>Jim Thompson wrote: >> On Fri, 29 Mar 2013 19:53:44 -0400, "Tom Del Rosso" >> <tomd_u1@verizon.net.invalid> wrote: >> >> > Jim Thompson wrote: >> > > >> > > Just analyze an ideal amplifier of gain +(1-delta) where delta is >> > > small, but non-zero Apply a feedback resistor from output to >> > > input, then calculate input impedance. >> > >> > What delta - just a change on the input? If gain is a function of >> > that, even with the minus sign, I'd think it was an exponential >> > amplifier rather than 'ideal'. >> >> GAIN = (1-delta) , delta small, but non-zero; so say that GAIN = 0.99, >> for example. >> >> Draw yourself a picture/schematic of what I wrote in words. > >Ok I can solve for that, if I treat delta like a constant of, say, 0.01. > >But why do you call it delta? What change does it refer to?Just call it "mary" instead ;-) ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | 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.
Reply by ●March 30, 20132013-03-30
"Phil Allison"> > ** One issue with bootstrapping input resistors is that it can > dramatically increase the noise compared to simply using a large value > resistor. > > 30 odd years ago I attempted to build a JFET pre-amp for a condenser mic > capsule and not having any 1Gohm resistors handy tried bootstrapping a > 10Mohm one. The pre-amp tested fine, with an effective input resistance > close to 1Gohm - ie response was flat across the audio band when driven > via a 22pF cap simulating the capsule. > > When the capsule was tried, the background noise ( hiss) was about 20dB > more than with a commercial pre-amp and quite unacceptable for studio > work. > > Thing is, with a 1Gohm ( gate bias) resistor, 22pF is enough to shunt > nearly all the audio frequency ( Johnson ) noise away - not so with 10 > Mohms and a bunch of positive feedback in place. >** Another sort of bootstrapping involves the output stage of an audio power amplifier. See low budget Germanium output stage typical of the late 1960s: http://1.bp.blogspot.com/_7q93wN0Pq1k/TRcwk1PjNZI/AAAAAAAAAA0/BvJcYYW0-FA/s1600/Picture+012.jpg The 180ohm collector load of the BFX88 is bootstrapped direct to the loudspeaker. Doing this improves both drive and drive linearity - as the speaker signal swings to 5 volts below the ground rail. Note how if the ( 3 ohm) speaker is disconnected, the OP stage is disabled. There is a double dose of bootstrapping going on at the input too. .... Phil
Reply by ●March 30, 20132013-03-30
On Fri, 29 Mar 2013 12:02:51 -0700, I wrote:>And the DC loading can be >increased because, although R1 and R2 still have their >Thevenin equivalent shunting effect, now you can add R3 >straight away to that, so that the DC loading is much lighter >than before despite a stiff bias pair.Very bad wording. What I should have written (and intended to write) was: "And the DC load impedance as seen by the input increases..." I think it's clear if you take the whole context in... but I wrote too abruptly and it could have changed the meaning. Sorry about that. Jon
Reply by ●March 30, 20132013-03-30
"Phil Allison"> See low budget Germanium output stage typical of the late 1960s: > > http://1.bp.blogspot.com/_7q93wN0Pq1k/TRcwk1PjNZI/AAAAAAAAAA0/BvJcYYW0-FA/s1600/Picture+012.jpg >** Seems there are a couple of drawing errors in the above schem: R6 should be 1.5kohms & R5 ought not be shorted out. ... Phil
Reply by ●March 31, 20132013-03-31
Reply by ●April 2, 20132013-04-02
On Sun, 31 Mar 2013 14:15:02 -0700, I wrote:>Left me worried, mikek. Was my explanation too poor to get >the idea across?I guess so. Is there anything you may want to ask about, then? Jon
Reply by ●April 3, 20132013-04-03
On 4/2/2013 2:04 PM, Jon Kirwan wrote:> On Sun, 31 Mar 2013 14:15:02 -0700, I wrote: > >> Left me worried, mikek. Was my explanation too poor to get >> the idea across? > > I guess so. Is there anything you may want to ask about, > then? > > Jon >Things for your info, I printed it out, it's on my bench waiting for me. I'm playing catch up at work, we're starting our busy season. One thing I'm pondering, the circuit I have is close to Version 2 about 1/2 way down this page. As you will note, the input wiring will have capacitance to the case, how can I best eliminate that? I think I need to build the second unit, so I can use KLeijer's method. Check Q with the first amplifier, connect the second amp with first still connected then retest* Q see if it lowers the Q. Mikek *using measurement data from first amp.
