On 3/28/2013 6:28 PM, amdx wrote:> I see the term bootstrapping, used when the designer wants a high > impedance and usually low capacitance. I know it involves feedback > but that's all I know. > > I want to learn enough to build a bootstrapped input with 10Meg/2pf > impedance. Those are ballpark numbers, the end use would be used > to measure voltage on a high Q coil and NOT load it. Frequency 100khz up > to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff) > > I would prefer a transistor circuit, that way I'll learn something, > but if there is an obvious IC circuit, I would like to know. > Mikek >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. A BF862 JFET follower has an output resistance near 30 ohms. (That is, assuming that the source current is somewhere near I_DSS.) If the current sink on its source is a 1k resistor, the voltage gain of the source follower stage is set by the voltage divider ratio: A_V = 1k/(1k + 30) ~ 0.97. Thus the voltage swing across the G-S capacitance is reduced by a factor of 30. The G-S current thus goes down by the same factor, so effectively the input impedance goes up by the same factor of 30. If you hang an emitter follower on the source of the JFET, and use that to move the JFET drain up and down to follow its gate, the contribution of C_DG goes down by about the same factor. 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. 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 28, 20132013-03-28
Reply by ●March 28, 20132013-03-28
On Thu, 28 Mar 2013 17:28:51 -0500, amdx <amdx@knologynotthis.net> wrote:>I see the term bootstrapping, used when the designer wants a high >impedance and usually low capacitance. I know it involves feedback >but that's all I know. > >I want to learn enough to build a bootstrapped input with 10Meg/2pf >impedance. Those are ballpark numbers, the end use would be used >to measure voltage on a high Q coil and NOT load it. Frequency 100khz up >to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff)Maybe a jfet follower then an opamp or emitter follower. The second follower would be AC coupled into the jfet drain, so it goes up and down to track the ac gate voltage. That cancels most of the d-g capacitance. AoE has a bunch about bootstrapping.> > I would prefer a transistor circuit, that way I'll learn something, >but if there is an obvious IC circuit, I would like to know. > MikekBootstrapping can be used for DC applications, like constant-current sources. https://dl.dropbox.com/u/53724080/Circuits/Current_Sources/Isrc_Boot.JPG https://dl.dropbox.com/u/53724080/Circuits/Current_Sources/TDC_Bootstrap_Ramp.jpg https://dl.dropbox.com/u/53724080/Circuits/Current_Sources/Isrc_5.JPG Here's a 1940-ish Blumlein circuit that uses a bootstrap to reduce the effective capacitance of an early TV picture tube: https://dl.dropbox.com/u/53724080/Circuits/Blumlein/Blumlein_1.JPG I'm not sure how exactly to define "bootstrap". -- John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser drivers and controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation
Reply by ●March 29, 20132013-03-29
On 3/28/2013 7:02 PM, George Herold wrote:> On Mar 28, 6:28 pm, amdx <a...@knologynotthis.net> wrote: >> I see the term bootstrapping, used when the designer wants a high >> impedance and usually low capacitance. I know it involves feedback >> but that's all I know. >> >> I want to learn enough to build a bootstrapped input with 10Meg/2pf >> impedance. Those are ballpark numbers, the end use would be used >> to measure voltage on a high Q coil and NOT load it. Frequency 100khz up >> to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff) >> >> I would prefer a transistor circuit, that way I'll learn something, >> but if there is an obvious IC circuit, I would like to know. >> Mikek > > The easiest way (for me) to make a bootstrap is an opamp unity gain > buffer the output drives a shield 'capacitance'. But as Phil said the > jfet follower is a classic. A bootstrap in a High Q LC circuit is > hard to understand. What are you doing? > > George H. >Not really part of the circuit, it would me an instrument. Several things. Measure Q (3db points), signal strength meter that doesn't load the coil. As a start. Here's the circuit I have built, just curious about a bootstrapped input. http://www.crystal-radio.eu/enfetamp.htm Mikek
Reply by ●March 29, 20132013-03-29
On 3/28/2013 6:35 PM, Phil Allison wrote:> "amdx" >> Phil Allison wrote: >>> "amdx" >>>> >>>> I see the term bootstrapping, used when the designer wants a high >>>> impedance >>>> and usually low capacitance. I know it involves feedback >>>> but that's all I know. >>>> >>>> I want to learn enough to build a bootstrapped input with 10Meg/2pf >>>> impedance. Those are ballpark numbers, the end use would be used >>>> to measure voltage on a high Q coil and NOT load it. Frequency 100khz up >>>> to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff) >>>> >>>> I would prefer a transistor circuit, that way I'll learn something, >>>> but if there is an obvious IC circuit, I would like to know. >>> >>> >>> ** I think you have been told this already - but a single JFET ( wired >>> as >>> a source follower ) is ideal. >>> >>> BTW: a 2pF cap has an impedance of 8000 ohms at 10MHz. >>> >>> >> Don't recall being told about the JFET source follower. >> With your reminder, I do recall the circuit I built before >> had 0.3pf. So forget the 2pf, need 0.3pf or less. >> > > ** Don't invent silly specs. > > It's the mark of an utter fuckwit to do that. > >> I have a High Q LC circuit, I put my bootstrapped measurement device >> in parallel, > > ** FET follower OK. > > No bootstrapping needed. > > > Piss off. >Here's your fantasy, http://i395.photobucket.com/albums/pp37/Qmavam/inside.jpg I didn't design the circuit. Did layout the pcb. I'm looking for a different input circuit. I'm not happy with the mechanical part that I put together. I used polystyrene with 3/4" hole. Not good mechanical strength and it doesn't like the heat of soldering. Mikek
Reply by ●March 29, 20132013-03-29
On Thu, 28 Mar 2013 17:28:51 -0500, amdx <amdx@knologynotthis.net> wrote:>I see the term bootstrapping, used when the designer wants a high >impedance and usually low capacitance. I know it involves feedback >but that's all I know. > >I want to learn enough to build a bootstrapped input with 10Meg/2pf >impedance. Those are ballpark numbers, the end use would be used >to measure voltage on a high Q coil and NOT load it. Frequency 100khz up >to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff) > > I would prefer a transistor circuit, that way I'll learn something, >but if there is an obvious IC circuit, I would like to know. > MikekOkay, mikek. Here's your bog standard bootstrapped degenerative amplifier using a single BJT. I'll try and look over it according to my poor hobbyist mind. I will assume you understand much of it, already. But ask about stuff, where I'm wrong about that. +V | +V | | \ | / Rc | \ \ / / R1 | \ | / +----Out | C1 | | || | In----------------||--, | | || | | | R3 | |/c Q1 +----/\/\-----+--| | |>e | | | C2 | | || +------, +---------||-------+ | | || | | | | \ | | / Ra \ | \ / R2 \ / \ / Re | / \ | | / | | | --- Ca | | --- gnd | | gnd | gnd (The simpler pieces missing from the above diagram are RS, the source resistance, and RL, the load resistance. Ignore them for now.) As you probably know already, the biasing pair of resistors R1 and R2 (in a non-bootstrapped case) are supposed to be stiff enough for the task of keeping Q1's bias point from moving much. It's common to read suggestions that the current through R1+R2 should be about 1/10th of Iq, which is the current through Rc in the quiescent state. Making them that stiff also means that they load the source (by their shunting effect.) I liked the description I saw from Phil Hobbs. It's exactly how I learned to see this, as well. He said that if you can make the swing across a resistor (he said admittance) to be identical (or as close as possible to that), that this means no current is drawn. Look at R3 for a moment. Assume that there is a DC bias across R3 providing the necessary Q1 bias current for its base. Now imagine keeping that DC bias, but asking yourself what would happen if the biasing pair node can be made to move up and down (AC wise) in exact lock-step with the base of Q1. If that could happen then the bias current would remain intact, but there would be no "change" in the current with AC changes. So no AC loading. 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. So more degrees of design freedom. What C2 does is to implement that requirement that the biasing pair node moves up and down in concert with Q1's base. At AC, the emitter is "following" the base (with very slightly less than 1 gain, Q1's alpha.) Assume the gain is 1 for now. If C2 is designed to be a 'short' at AC frequencies of interest and if the emitter of Q1 is a good, low impedance "source" of these AC changes (it is such a good source), then C2 will bypass emitter fluctuations directly to the biasing pair and it will do so at low impedance, easily driving the biasing node up and down in concert with the base signal. What's neat about this is that it uses an existing low impedance replica of the input signal that is isolated (mostly) from the input by the beta of Q1. And it uses it to force the biasing pair node up and down in phase with the AC signal. Since it is obvious from the circuit that Q1's base itself is moving also up and down by the same amount (almost) then it follows that the AC signal on both sides of R3 is the same (almost.) So at AC signals, R3 "looks" like an infinite impedance. And at DC, R3 greatly reduces the bypass loading. Okay. So some analysis. Miller's theorem says that the impedance between two nodes can be resolved out into two different components: z/(1-k) and z*k/(k-1). In this case, k is the voltage gain, k = Av = Vo/Vi, and R3 is z. Since we are tapping off of the emitter, not the collector, Vo/Vi is nearly 1. For example, if Av=.99 and R3=200k then the effective resistance is 20M Ohm. Full analysis is, of course, more algebra and work. But I thought I'd just get this out there for you to get the main point across. I'd use the term bootstrapping anytime this particular kind of approach is being taken. Jon
Reply by ●March 29, 20132013-03-29
On 3/29/2013 2:02 PM, Jon Kirwan wrote:> On Thu, 28 Mar 2013 17:28:51 -0500, amdx > <amdx@knologynotthis.net> wrote: > >> I see the term bootstrapping, used when the designer wants a high >> impedance and usually low capacitance. I know it involves feedback >> but that's all I know. >> >> I want to learn enough to build a bootstrapped input with 10Meg/2pf >> impedance. Those are ballpark numbers, the end use would be used >> to measure voltage on a high Q coil and NOT load it. Frequency 100khz up >> to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff) >> >> I would prefer a transistor circuit, that way I'll learn something, >> but if there is an obvious IC circuit, I would like to know. >> Mikek > > > Okay, mikek. Here's your bog standard bootstrapped > degenerative amplifier using a single BJT. I'll try and look > over it according to my poor hobbyist mind. I will assume you > understand much of it, already. But ask about stuff, where > I'm wrong about that. > > +V > | > +V | > | \ > | / Rc > | \ > \ / > / R1 | > \ | > / +----Out > | C1 | > | || | > In----------------||--, | > | || | | > | R3 | |/c Q1 > +----/\/\-----+--| > | |>e > | | > | C2 | > | || +------, > +---------||-------+ | > | || | | > | | \ > | | / Ra > \ | \ > / R2 \ / > \ / Re | > / \ | > | / | > | | --- Ca > | | --- > gnd | | > gnd | > gnd > > (The simpler pieces missing from the above diagram are RS, > the source resistance, and RL, the load resistance. Ignore > them for now.) > > As you probably know already, the biasing pair of resistors > R1 and R2 (in a non-bootstrapped case) are supposed to be > stiff enough for the task of keeping Q1's bias point from > moving much. It's common to read suggestions that the current > through R1+R2 should be about 1/10th of Iq, which is the > current through Rc in the quiescent state. Making them that > stiff also means that they load the source (by their shunting > effect.) > > I liked the description I saw from Phil Hobbs. It's exactly > how I learned to see this, as well. He said that if you can > make the swing across a resistor (he said admittance) to be > identical (or as close as possible to that), that this means > no current is drawn. > > Look at R3 for a moment. Assume that there is a DC bias > across R3 providing the necessary Q1 bias current for its > base. Now imagine keeping that DC bias, but asking yourself > what would happen if the biasing pair node can be made to > move up and down (AC wise) in exact lock-step with the base > of Q1. If that could happen then the bias current would > remain intact, but there would be no "change" in the current > with AC changes. So no AC loading. 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. > > So more degrees of design freedom. > > What C2 does is to implement that requirement that the > biasing pair node moves up and down in concert with Q1's > base. At AC, the emitter is "following" the base (with very > slightly less than 1 gain, Q1's alpha.) Assume the gain is 1 > for now. If C2 is designed to be a 'short' at AC frequencies > of interest and if the emitter of Q1 is a good, low impedance > "source" of these AC changes (it is such a good source), then > C2 will bypass emitter fluctuations directly to the biasing > pair and it will do so at low impedance, easily driving the > biasing node up and down in concert with the base signal. > What's neat about this is that it uses an existing low > impedance replica of the input signal that is isolated > (mostly) from the input by the beta of Q1. And it uses it to > force the biasing pair node up and down in phase with the AC > signal. Since it is obvious from the circuit that Q1's base > itself is moving also up and down by the same amount (almost) > then it follows that the AC signal on both sides of R3 is the > same (almost.) So at AC signals, R3 "looks" like an infinite > impedance. And at DC, R3 greatly reduces the bypass loading. > > Okay. So some analysis. Miller's theorem says that the > impedance between two nodes can be resolved out into two > different components: z/(1-k) and z*k/(k-1). In this case, k > is the voltage gain, k = Av = Vo/Vi, and R3 is z. Since we > are tapping off of the emitter, not the collector, Vo/Vi is > nearly 1. For example, if Av=.99 and R3=200k then the > effective resistance is 20M Ohm. > > Full analysis is, of course, more algebra and work. But I > thought I'd just get this out there for you to get the main > point across. I'd use the term bootstrapping anytime this > particular kind of approach is being taken. > > Jon >Thanks for that Jon, I'm at work now, but I'll print it and look it over when I'm home. Mikek
Reply by ●March 29, 20132013-03-29
On Fri, 29 Mar 2013 15:30:46 -0500, amdx <amdx@knology.net> wrote:>> <snip> > Thanks for that Jon, >I'm at work now, but I'll print it and look it over when I'm home. > Mikek >No problem. The basic idea boils down to this: If you can keep both sides of R3 moving around in lock step then in effect R3 has infinite impedance (from the AC point of view) and completely isolates the Q1 base from the divider network. So the biasing pair (via R3) gets to perform its DC function of biasing Q1 but, at AC, R3 works to decouple the Q1 base from the biasing pair. Of course, in reality it's not perfect and so the isolation is imperfect as well. Jon
Reply by ●March 29, 20132013-03-29
Jim Thompson wrote:> On Thu, 28 Mar 2013 17:28:51 -0500, amdx <amdx@knologynotthis.net> > wrote: > > > I see the term bootstrapping, used when the designer wants a high > > impedance and usually low capacitance. I know it involves feedback > > but that's all I know. > > > > I want to learn enough to build a bootstrapped input with 10Meg/2pf > > impedance. Those are ballpark numbers, the end use would be used > > to measure voltage on a high Q coil and NOT load it. Frequency > > 100khz up to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff) > > > > I would prefer a transistor circuit, that way I'll learn something, > > but if there is an obvious IC circuit, I would like to know. > > Mikek > > 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'. -- Reply in group, but if emailing add one more zero, and remove the last word.
Reply by ●March 29, 20132013-03-29
On Fri, 29 Mar 2013 19:53:44 -0400, "Tom Del Rosso" <tomd_u1@verizon.net.invalid> wrote:>Jim Thompson wrote: >> On Thu, 28 Mar 2013 17:28:51 -0500, amdx <amdx@knologynotthis.net> >> wrote: >> >> > I see the term bootstrapping, used when the designer wants a high >> > impedance and usually low capacitance. I know it involves feedback >> > but that's all I know. >> > >> > I want to learn enough to build a bootstrapped input with 10Meg/2pf >> > impedance. Those are ballpark numbers, the end use would be used >> > to measure voltage on a high Q coil and NOT load it. Frequency >> > 100khz up to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff) >> > >> > I would prefer a transistor circuit, that way I'll learn something, >> > but if there is an obvious IC circuit, I would like to know. >> > Mikek >> >> 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. ...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 29, 20132013-03-29
"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. ... Phil
