Bootstrapping is a clever technique where you use feedback to increase the input impedance (mostly in an emitter follower). Why is it named "Bootstrapping"? Just curious.

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    \$\begingroup\$ The term comes from outside engineering, ie "to pull oneself up by the bootstraps", and within engineering is by no means uniquely associated with the example you cite. \$\endgroup\$ – Chris Stratton Nov 5 '16 at 17:14
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    \$\begingroup\$ phrases.org.uk/meanings/290800.html \$\endgroup\$ – Andy aka Nov 5 '16 at 17:48

The term is derived from "pulling yourself up by your boot straps". In engineering it usually means to start with little, use that to make a little more, use that to make even more, etc, until eventually to get to full operation.

From that, it is sometimes used to refer to any startup process. When a computer is first turned on, it goes thru a "boot" or "bootstrap" process to eventually load the operating system and be ready for use. Now, "booting" a computer is synonymous with starting it up.

The computer actually goes thru several startup phases before being ready for normal operation, so this is a valid use of the term. I remember computers from the 1970s where you had to first toggle in a small program of maybe a dozen instructions. That was just enough logic to read a more complicated loader from a paper tape or something. That load contained the more substantial routines to read from the disk, etc. Modern computers still go thru a number of startup phases. The initial one is read from a ROM, which is why you can simply turn it on and wait. Booting a computer has actually gotten more complicated and takes longer than it took the much slower and smaller computers of the 1970s to boot up.

Personally, I wouldn't call a emitter follower a bootstrap circuit. I think that's a misuse of the term.

A valid example of that term in electronics is referring to the circuit that provides the initial internal power of a line voltage to DC power supply. Ultimately the supply will be a switching regulator that transfers significant power from the AC input to the DC output. However, that switcher needs to be powered somehow. In other words, you need a power supply to power the power supply.

The bootstrap circuit in a power supply provides the initial slug of energy for the switcher to run until it is able to make low voltage DC efficiently and at significant power levels. This includes providing the switcher's own operating power. The bootstrap supply is usually quite inefficient, but simple enough to run without itself needing a power supply. Either that bootstrap supply uses very little power, or it is shut off once the main power supply is up and running.

  • \$\begingroup\$ Olin.... Your description if I understood correctly, I think fit to explain the bimetallic starters used in the common florescent light. Am I right? \$\endgroup\$ – soosai steven Nov 6 '16 at 1:50
  • \$\begingroup\$ Going through the (link) provided by @Andy_aka, it kinda sorta makes sense. The bootstrapped emitter follower uses its own output to increase its input impedance and thus reduce load on the previous stage. Thanks a lot for such an informative answer :) \$\endgroup\$ – Vinam Arora Nov 6 '16 at 7:18
  • \$\begingroup\$ Olin Lathrop, I am with you: An emitter follower works because of NEGATIVE VOLTAGE feedback and - in addition - the bootstrap element causes POSIVE CURRENT feedback. Both effects increase the input impedance. \$\endgroup\$ – LvW Nov 6 '16 at 9:05

The fictional Baron Munchausen https://en.wikipedia.org/wiki/Baron_Munchausen once fell into a swamp, and only saved his life by pulling himself out by the straps of his boots.


Only if you imagine you can lighten your load by picking yourself up by the boots. Unfortunately we lose ground connection when floating , so we cannot lighten our load. But transistors with common gnd. references can do this.

This is how emitter feedback ratio to the base raises the input impedance by this ratio using positive feedback with gain <1.

Also on the flip side. negative feedback lowers both input & output impedance..

  • You can bootstrap the Nch Gate source voltage voltage from the Pch PWM or
  • bootstrap for input higher impedance or
  • bootstrap yourself to elevate your mood and more....

Here is a voltage amplifier from a low Z source mic via Common base to an Emitter Follower output with Cap feedback from Emitter out to base where voltage gain is slightly <1. Thus raises the 1st stage collector by impedance or input impedance of the load and thus the impedance ratio of 1st stage yields a voltage gain of 100 or so with an 8 Ohm source.

This is a bootstrap to raise impedance by voltage positive feedback <1 and result in 1st Av >1000 using 22uF for positive feedback raising base load well above Rc load , so Av~ 10k/8R or a bit less.

enter image description here

Bootstrap means to boost something basic like Zin or Vb " without external assistance". In the case of MOSFET Vb bootstrap the feedback of PWM to a series cap and diode clamped to Vdd yields a high voltage generated than input Vdd, so we say without external assistance a passive boost above supply here (without external help) is called a Booststrap. Same name yet different function.

... not to be confused with flyback boost or other boost topics.

  • \$\begingroup\$ TonySteward, how can you say that "negative feedback lowers input&output impedance"? What about Re-feedback for transistor stages and non-inverting opamps with neg. feedback? \$\endgroup\$ – LvW Nov 6 '16 at 9:02
  • \$\begingroup\$ this only applies to the points where neg feedback is applied, not alternate inputs. e.g for common base, with 2nd stage inverting feedback to 1st stage emitter input, its emitter output z must be reduced with base cap to allow negative feedback into emitter or non-inv OA uses alternate input. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 6 '16 at 15:20
  • \$\begingroup\$ I think, to avoid misunderstandings, we should state instead: Negative voltage feedback always increases and negative current feedback always decreases the input impedance of an amplifier. \$\endgroup\$ – LvW Nov 6 '16 at 15:26
  • \$\begingroup\$ yes but as you see, negative current feedback is positive voltage feedback with gain <1... hence positive bootstrap \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 6 '16 at 15:27
  • \$\begingroup\$ I don`t understand the sentence "negative current feedback is positive voltage feedback". We have negative voltage feedback (due to RE) and positive current feedback (due to the bootstrap path). And both effects cause input impedance increase. \$\endgroup\$ – LvW Nov 6 '16 at 15:57

protected by clabacchio Feb 12 '18 at 13:22

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