Bootstrapping - Page 1
Forum section: electronics / beginners

author: Taho
author_rank: Contributor
timestamp: May 19, 2014, 01:39:18 pm
content: One thing that I couldn't unsderstand is bootstrapping. Here is an example, can you explain what that cap is doing there ?Why and where do we use bootstrapping ?Clear and simple explanation with different examples please.

author: han
author_rank: Frequent Contributor
timestamp: May 19, 2014, 01:46:32 pm
content: the capacitor is bypass ac current from the output to input, so less current needed to drive the input equal higher input impedance...similar thing is power steering in car.

author: Taho
author_rank: Contributor
timestamp: May 19, 2014, 03:08:24 pm
content: Then you are openning another gate for the ac current however transistors low base current make their high input impedance is not it contraditory? I can not understand.

author: ejeffrey
author_rank: Super Contributor
timestamp: May 19, 2014, 04:14:44 pm
content: The base of the transistor itself requires very little current. However, it has to be biased into operation. For AC amplifiers, the standard way is as shown: use a single transistor capacitively coupled, and then use resistors to set the DC bias. That is what the resistors R1, R2, and R3 do. Those bias resistors end up robbing quite a bit of the current, and lowering the input impedance. The bootstrap capacitor allows the amplifier output to drive the bias resistors at AC without messing up the static biasing condition.

author: Legion
author_rank: Frequent Contributor
timestamp: May 19, 2014, 05:13:13 pm
content: Quote from: Taho on May 19, 2014, 01:39:18 pmI feel your pain Taho. I'm reading the same part of Art of Electronics. The book is a poor source to learn from when you're starting out. The student manual explains things much better, but unfortunately in this instance, it doesn't talk about bootstrapping. I'm going to breadboard the bootstrap circuit when I get to labs 4 & 5, sometimes direct observation makes things much clearer. If this thread doesn't answer your question by then and I figure something out, I'll be sure to post my findings.

author: LvW
author_rank: Frequent Contributor
timestamp: May 19, 2014, 06:20:46 pm
content: Quote from: Taho on May 19, 2014, 01:39:18 pmHi Taho,I´ll try to explain something to you:1.) The task of the "bootstrap effect" is to enlarge the signal input resistance of the common-collector configuration. This is necessary because this circuit mostly is used as a buffer (unity gain, large input resistance, low output resistance)2.) The principle is as follows: As you know the common collector stage has a gain of (slightly less than) unity.Let´s first assume that is unity.In this case both ends of R3 see the same signal voltage (due to the large C2 value). As a result: No signal current through R3.This would mean: Signal input resistance is infinite.However, in reality, the signal voltage at the emitter node is slightly less than at the base (gain slightly less than unity). Hence, we have an increase of the input resistance, but not up to infinite but to a value which is much larger than without this "bootstrap effect".3.) By the way: The name of this principle is derived from the fact that the signal voltage at the lower end of R3 is "bootstrapped" (connected) to the upper end of R3.Hope this helps a bit.LvWRemark: The opposite effect - however, with areductionof the signal input resistance - is the MILLER-effect (Resistor between base and collector node).

author: retrolefty
author_rank: Super Contributor
timestamp: May 19, 2014, 07:59:58 pm
content: Is it not just simpler to say it's providing a little positive feedback to raise the input impedance?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 20, 2014, 06:53:33 am
content: Quote from: retrolefty on May 19, 2014, 07:59:58 pmIndeed, and since the gain of a follower (of any type, be it BJT, FET or op-amp) is always slightly less than one, the positive feedback can never be enough to generate negative input resistance, let alone oscillation.Indeed, the advantage is inversely proportional to the difference from ideal (i.e., Av = 1 - epsilon, for epsilon ~= 0), so that a gain of 0.5 is able to double its input resistance, a gain of 0.99 improves it by 100x, and a gain of 1.0 improves it infinitely. Assuming, of course, the coupling capacitor is large enough by the same factor. The input resistance starts rising at F1 = 1 / (2*pi*Rin*C) and levels off around F2 = 1 / (2*pi*Rin*C/epsilon).Note also, an impedance which rises with frequency looks like a certain component...which one?Indeed, the input impedance has aninductivecharacteristic between F1 and F2, and a circuit designed specifically for this (called a gyrator) can be used to synthesize inductors from capacitors, among other things.Positive feedback is neat stuff; if you add just a little bit of gain to the circuit, say, suppose you place a noninverting gain-of-2 amp after the follower and bootstrap its output to the first input: now the output goes the same direction as the input,and then some, so that the output wants to drag the input along for a ride! If the source is able to keep control over it and it remains stable, this is simply negative resistance in action! If you add a potentiometer before that gain stage, so the output is variable from 0 to 2 gain, you can clearly observe the input resistance rising, from its initial value, up towards infinity (at exactly Av = 1), and wrapping around to negative numbers, until settling at -R_initial at Av = 2 (or, even smaller resistances if you keep adding gain, of course).I like this example, on a fundamental level, because it demonstrates the continuity of the real numbers, from zero, to positive finite, to infinity, to negative infinity, to negative finite, and back to zero. There are indeed many physical and mathematical situations where this becomes apparent; this example plays merely one humble testimony to that deeper truth.Negative resistance is hardly a useless, unstable gimmick; it can be used for a number of (stable) purposes, like compensating for voltage drop along transmission lines. A direct example regarding the bootstrapped emitter follower would be, overcompensating to account for other loading and loss and leakage at the input, so that when the whole circuit (source, follower, and other loads attached) is considered, it indeed achieves a near-infinite impedance.Tim

author: amyk
author_rank: Super Contributor
timestamp: May 20, 2014, 10:55:30 am
content: When Q1 is off the output is at ground and C2 charges up to some voltage through the voltage divider. When Q1 turns on the negative end of C2 goes above ground and adds to the charge on C2, boosting the voltage at the base.I don't think the explanation gets simpler than that.

author: LvW
author_rank: Frequent Contributor
timestamp: May 20, 2014, 12:51:40 pm
content: Quote from: Taho on May 19, 2014, 01:39:18 pmFor a good understanding of the principle - and thelimitations- of "boot-strapping" it is perhaps helpful to add the following:Because the task of bootstrapping is to avoid signal input resistances which are too low for some applications, one could think that it could be sufficient to increase the R3 resistor to - let´s say - 100k or even more. This is NOT recommended because of two reasons:* At first, the DC current into the base (Ib) will create a crresponding large DC drop across R3 (which, perhaps, seems to be acceptable)* Secondly, such a drastic increase of R3 means that we more and more depart from the principle of DC voltage "stiffness" at the base node. However, this principle is necessary to cope for Ic variations due to temperature changes. The principle of Ic stabilization (stabilization of the DC operating point) due to Re feedback works only if we have such a dc voltage "stiffness" at the base (I hope that this is the correct expression).* As a consequence, something like a trade-off is necessary between (a) signal input resistance and (b) Ic stabilization.(This trade-off results in R3 values in the lower kohm range up to 10...20 kohms)By the way: This is typical in analog electronics as we always have to find a trade-off between conflicting requirements.(I like to add the following comment: I am not sure if it really helps to understand the bootstrap-principle for linear circuit operation by considering just ON/OFF situations)

author: Taho
author_rank: Contributor
timestamp: May 21, 2014, 09:12:23 pm
content: Sory for the late reply, I was thinking about your words now I decided that I can not understand. I am thinking the circuit without C2 but I can not see any difference. We want higher input impedance, what may be the effect if I play with the value of R4 with or without C2 I can not see any difference.Emiter follower is a voltage divider for low loads am I wright ? Here R3 has same voltage at both ends OK it behaves like infine impedance current source but what is the point ?Can't you tell the the circuit with and with out C2 ?by the way thanks for the advise I will use manuel tooQuote from: Legion on May 19, 2014, 05:13:13 pmand an extra thanks for your clear explanation Lvw I was so close thinking about yours but soryQuote from: LvW on May 19, 2014, 06:20:46 pmThanks...

author: Alexei.Polkhanov
author_rank: Frequent Contributor
timestamp: May 21, 2014, 09:58:45 pm
content: You guys forgot to mention Baron Münchhausen somewhere in your explanations.

author: LvW
author_rank: Frequent Contributor
timestamp: May 22, 2014, 07:45:37 am
content: Quote from: Taho on May 21, 2014, 09:12:23 pmTaho - the answer is relatively simple:The capacitor C2 is necessary for DC decoupling the emitter node from the DC source.For normal common-collector operation only the DC current out of the emitter node goes through Re.

author: Taho
author_rank: Contributor
timestamp: May 23, 2014, 12:59:29 am
content: Quote from: LvW on May 22, 2014, 07:45:37 amOK, Thanks LvW.