This post was last edited by cruelfox on 2018-9-8 09:31 I have been playing with circuits sporadically for more than ten years, mainly audio amplification. I have seen and experimented with some interesting circuit structures, and I have wanted to share them with you for a long time. This time I will analyze the bootstrap circuit in the amplifier circuit. As a DIYer who has never taken an electronics class and has learned by himself out of interest, if there is anything wrong with what I say below, please correct me! The term "bootstrap" can be seen in many fields (literally means to lift yourself up by the straps on your boots, which is of course impossible). In circuits, it is an ancient technology. And bootstrap is not only used in amplifier circuits, for example, it is also used in power supplies, but I don't know much about it, so I won't discuss it here.
* Bootstrap capacitor in audio power amplifier This bootstrap circuit is the earliest bootstrap I have seen. It is often seen in the power amplifier part of old discrete semiconductor radios (compared to the type with input and output transformers, it is actually not that old, but radios have long used integrated circuits), just like the part marked with a red box in the figure below.
Another example is in the classic JLH 1969 power amplifier circuit (R3, R8, C5 in the figure below constitute the bootstrap):
However, both of the above circuits include negative feedback. If we make it simpler (not practical) to analyze it, it will become the circuit I drew below:
This circuit is a common emitter amplifier (Q2) plus a complementary emitter follower (Q1, Q5). If we ignore the bootstrap capacitor C1, then R4 in series with R5 together form the collector load resistor (3.7k) of Q2. Of course, the input impedance of Q1/Q5 must be taken into account when calculating the voltage gain of Q2. The gain of the voltage amplifier of Q2 is roughly proportional to the collector load impedance (Miller effect and Early effect are not considered here). If the input impedance of the emitter follower is high enough, the larger the collector load resistance, the greater the gain. However, if we take the DC operating point into consideration, if we want to increase the collector load resistance while keeping the collector current unchanged, we have to increase the power supply voltage...so the choice of collector load resistance is limited. Okay, now split the collector load resistor into two sections and add a bootstrap capacitor to form the above shape, and the DC operating point remains unchanged. What is the collector load resistance of Q2 now? R4? It doesn't seem right. Let's look at the time domain simulation analysis:
The voltage waveform of the output node (R3, R1 common terminal) shown in the simulation: After the 100uF bootstrap capacitor is added (red line), the output signal amplitude is one order of magnitude larger than when no capacitor is added (blue line). Why is there such an effect? In the above circuit, one end of the load resistor R4 is connected to the collector of Q1 (according to the AC equivalent, ignoring the change in the voltage drop on the three diodes), which is also the input of the emitter follower, and the other end is connected to the output of the emitter follower through the bootstrap capacitor. Because the emitter follower is a common-phase amplifier, the voltage gain is slightly less than but close to 1, which is a positive feedback connection. Take out the part of the above circuit that acts as a bootstrap and draw it as the following figure:
The output of the first-stage transistor is regarded as a current source, and the emitter follower is equivalent to an amplifier with a gain G of approximately 1. It has a high input impedance Zin and a low output impedance Zout. Here, the impedance of the bootstrap capacitor is also merged into Zout. Although the bootstrap capacitor in the complete circuit above is connected to a resistor to the power supply (AC equivalent), it is equivalent to the amplifier using the Thevenin theorem (resulting in a decrease in gain). The output impedance of the first-stage transistor and Zin are in parallel and can be considered together. After the above simplification, the gain of the circuit is not difficult to calculate. According to the current balance, the voltage at the input of the amplifier can be solved to be
I*(R+Zout)*Zin/[Zin*(1-G)+R+Zout] = I*[(R+Zout)/(1-G)*Zin]/[Zin + (R+Zout)/(1-G)] That is, the load impedance seen from the output of the first-stage transistor is the value of (R+Zout)/(1-G) and Zin in parallel. In other words, the bootstrap technique makes the collector load resistance "increase" by about 1/(1-G) times. Let's verify it with spice simulation, as follows:
When G is 0.5, 0.9, 0.99, 0.999, the resistance is "amplified" by 2, 10, 100, 1000 times, resulting in a gain increase of 6dB, 20dB, 39dB, 54dB (limited by Zin, the gain cannot be increased infinitely). Finally, there is a small question: In the actual power amplifier circuit, negative feedback controls the overall gain, so does the bootstrap circuit still play a significant role? In fact, the negative feedback causes the closed-loop gain to decrease, which reduces the input amplitude of the common emitter circuit in front, and the gain of this stage alone is still increased by the bootstrap circuit. The increase in open-loop gain improves the frequency response and distortion rate after the closed loop, so such circuits were often seen in power amplifiers in the past.
* Using bootstrap to increase input impedance Ignore the common emitter amplifier part in the above circuit and look at the emitter follower alone? Oh, the circuit is missing something - the emitter follower needs to be biased. The bias resistor will also become a load for the input signal, reducing the input impedance of the entire follower. However, after the connection method of the bootstrap circuit is deformed to the following...
Using the "resistance multiplication" principle just derived, the input bias resistor in this circuit is changed from the original R3+R1//R2 to (1-G)*R3, and the impact on the input impedance is eliminated. However, to further increase the input impedance, the load of Q1 needs to be reduced. The following figure is an example of a high input impedance circuit discussed in Douglas Self's book
There are two stages of followers in this figure. The bootstrap capacitor takes the signal from the second stage. It can be considered that the total gain is closer to 1, and the input impedance of the first tube is also increased.
* Bootstrap capacitor in cross-emitter follower This four-tube two-stage emitter follower circuit is what I learned from Kuroda Toru's book (Japan), and I used it in a headphone amplifier module I made.
It is somewhat similar to the first classic bootstrap circuit, but the function is different. First, according to the previous conclusion, R4+R6 and R5+R7 are "doubled" respectively, and the equivalent impedance is increased. Note that Q1 is an emitter follower, and R4+R6 is now the emitter resistor of Q1. When it is equivalently enlarged, it is equivalent to connecting a constant current source to the emitter of Q1. At the same time, the bootstrap capacitor also changes Q1,The collector AC potential of Q4 makes VCE almost constant - the advantage of this is that it greatly eliminates the influence of the transistor collector junction capacitance, reduces the input capacitance in this circuit, and improves high-frequency performance. In summary, we have seen several uses of the bootstrap circuit so far: (a) Increase the gain of the common emitter amplifier circuit (can also be used in the common base amplifier circuit) (b) Increase the input impedance of the emitter follower (follower, common collector circuit) (c) Stabilize the emitter current of the emitter follower (d) Make the collector voltage of the emitter follower follow the emitter The common characteristics of the above circuits are: (1) The signal is drawn from the output of the emitter follower (voltage gain is about 1) using the bootstrap capacitor and fed to its front stage (2) Feedback from the low resistance node to the high resistance node, only one end of the bootstrap capacitor is a strong drive (voltage output) signal
* Eliminate the bootstrap capacitor The role of the bootstrap capacitor is to block DC, that is, it is only effective for AC signals, and the DC operating point will not be changed after bootstrapping. If the operating point is selected appropriately, bootstrapping can be achieved without using a bootstrap capacitor, as shown in the following circuit:
Note the connection of R6. Is it "doubled" by the bootstrap?
* Replace the emitter follower with an op amp The amplifier with a gain of approximately 1 in the bootstrap circuit can be realized by using a source follower, an op amp buffer circuit, etc., in addition to using an emitter follower. The potentiometer in the following circuit is used to adjust the feedback ratio, which is equivalent to changing the gain, so that the multiple of the resistance "doubled" can be adjusted to achieve an adjustable filter.
IC7a, together with R42 and VR5a, constitute the amplifier in the bootstrap. C33 is the bootstrap capacitor, which makes R41 equivalent to "doubled". Note that this "multiplier" is not linearly related to VR5a. I guess this is intentional, otherwise we would not need a bootstrap circuit and would just replace R41 with a potentiometer. Let's look at a more complex, less obvious circuit (the diagram in the pdf is just so unclear, there's nothing we can do about it):
Except for a resistor in series with the capacitor for frequency compensation, the remaining resistors are used to set the DC operating point and have nothing to do with gain. From the gate of the JFET to the output of the op amp, the voltage gain is 1. Note that due to the strong open-loop gain of the op amp, the G, D, and S poles of the JFET are all at the same potential for AC signals. Due to the bootstrap effect of the op amp, the current of the tube is almost unchanged, and the voltage at the DS terminal is also almost unchanged. In this circuit, bootstrapping is used to achieve a very high input impedance. In the above bootstrap circuit, the output of the key part - the amplifier (buffer that provides current) is also the output of the signal. It seems that adding a small amount of resistors and capacitors increases the bootstrap function. The following is a circuit that uses a buffer only for bootstrapping, but does not output a signal.
* FET Bootstrap Cascode This is a circuit I personally like very much. The figure below is a differential input stage of an amplifier I made, using bootstrap on two JFETs.
In this circuit, Q5 Q6 are differential amplifiers, and the quiescent current is determined by the common current source of the source. Q1 Q2 forms a mirror current source load, which is a very common circuit connection method. Q3 and Q4 are used for bootstrapping. Their bases follow the source voltage changes of Q5 Q6, so the emitters also follow the source voltage changes of Q5 Q6. The purpose of this bootstrap circuit is to keep the VDS of Q1 Q2 constant, thereby eliminating the influence of the parasitic capacitance Cgd (because the capacitance of JFET is relatively large, which is a disadvantage). The role of R1, D1, D2, Q7 here is to use the voltage regulation characteristics of the diode to provide a basically constant bias voltage. Here, if Q3 and Q4 are regarded as buffers, the signals they output are not used, but are only used to determine the drain voltage of Q5 and Q6. However, Q3 and Q4 do output signals, and they are output from the collector - the gain is much greater than 1 when applied. Taken apart, Q3-Q5 and Q4-Q6 are all cascode circuits, but the base voltage of the common base amplifier part changes with the differential pair tube - so it is a bootstrap cascode. The common base transistor biasing method of the cascode in the above circuit is a little complicated. If it is replaced with an appropriate JFET, the cascode can be made very simple, as shown below.
This connection method limits the choice of JFET, and it is not enough to just grab two tubes. Because it is necessary to ensure that under the set current, the VDS of the tube that inputs the signal (Q1 Q2 in the figure above) is equal to the VGS of the tube used for bootstrapping above it.
* Bootstrapping of the op amp Finally, let's look at a special use of bootstrapping: changing the power supply voltage of the operational amplifier so that the two power supply terminals follow the input terminals. In this way, it seems that the + input signal is constant to the bootstrapped op amp - that is, the input common-mode voltage is eliminated. This technique is used to reduce the distortion of the op amp input stage.
As shown in the figure above, the price paid for power supply bootstrapping is not small. In addition to adding an op amp as a follower, two transistors are used to expand the current because of the need to provide large current. To summarize the key points of "bootstrapping" in the above example circuit: there is a buffer that provides a low impedance output (an amplifier with a gain of approximately 1), which buffers the voltage of the reference node and applies it to another node in the circuit, so that its AC voltage changes in the same way as the reference node. The result is a constant voltage effect on the terminals of certain components or parts of the circuit, improving certain aspects of the circuit's performance.
This content was originally created by EEWORLD forum user cruelfox. If you want to reprint or use it for commercial purposes, you must obtain the author's consent and indicate the sourceThe S poles are all at the same potential. Due to the bootstrap effect of the op amp, the current of the tube is almost unchanged, and the voltage at the DS terminal is also almost unchanged. In this circuit, bootstrap is used to achieve a very high input impedance. In the above bootstrap circuit, the output of the key part - the amplifier (buffer that provides current) is also the output of the signal. It seems that adding a small amount of resistors and capacitors increases the bootstrap function. What will be introduced below is a circuit that uses a buffer only for bootstrap without outputting a signal.
* Bootstrap Cascode of Field Effect Transistor This is a circuit that I personally like very much. The figure below is a differential input stage of an amplifier I made, using bootstrap on two JFETs.
In this circuit, Q5 Q6 are differential amplifiers, and the static current is determined by the common current source of the source. Q1 Q2 forms a mirror current source load, which is a very common circuit connection method. Q3 and Q4 are used for bootstrapping. Their bases follow the source voltage of Q5 Q6, so the emitters also follow the source voltage of Q5 Q6. The purpose of this bootstrap circuit is to keep the VDS of Q1 Q2 constant, thereby eliminating the influence of the parasitic capacitance Cgd (because the capacitance of JFET is relatively large, which is a disadvantage). The role of R1, D1, D2, and Q7 here is to use the voltage regulation characteristics of the diode to provide a basically constant bias voltage. Here, if Q3 and Q4 are regarded as buffers, then the signals they output are not drawn out for use, but are only used to determine the drain voltage of Q5 Q6. However, Q3 Q4 does output a signal, which is output from the collector - the gain is much greater than 1 when applied. Taken apart, Q3-Q5, Q4-Q6 are all cascode circuits, except that the base voltage of the common base amplifier part changes with the differential pair tube - so it is a bootstrap cascode. The biasing method of the common base transistor of the cascode in the above circuit is a little complicated. If it is replaced with an appropriate JFET, the cascode can be very simple, as shown below.
This connection method limits the choice of JFET. You can't just grab two tubes at random. Because it is necessary to ensure that under the set current, the VDS of the tube that inputs the signal (Q1 Q2 in the above figure) is equal to the VGS of the bootstrap tube above it.
* Bootstrapping of the op amp Finally, let's look at a special use of bootstrapping: changing the power supply voltage of the operational amplifier so that the two power supply terminals follow the input terminal. In this way, from the perspective of the bootstrapped op amp, it seems that the + terminal input signal is constant - that is, the input common mode voltage is eliminated. This technique is used to reduce distortion at the op amp input stage.
As shown in the figure above, the price paid for power bootstrapping is not small. In addition to adding an op amp as a follower, two transistors are used to expand the current because of the need to provide large current. To summarize the key points of "bootstrapping" in the above example circuit: there is a buffer that provides a low-impedance output (an amplifier with a gain of approximately 1), which buffers the voltage of the reference node and applies it to another node in the circuit, so that its AC voltage changes in the same way as the reference node. The result is a constant voltage effect on the terminals of certain components or parts of the circuit, improving certain aspects of the circuit's performance.
This content was originally created by EEWORLD forum user cruelfox. If you need to reprint or use it for commercial purposes, you must obtain the author's consent and indicate the sourceThe S poles are all at the same potential. Due to the bootstrap effect of the op amp, the current of the tube is almost unchanged, and the voltage at the DS terminal is also almost unchanged. In this circuit, bootstrap is used to achieve a very high input impedance. In the above bootstrap circuit, the output of the key part - the amplifier (buffer that provides current) is also the output of the signal. It seems that adding a small amount of resistors and capacitors increases the bootstrap function. What will be introduced below is a circuit that uses a buffer only for bootstrap without outputting a signal.
* Bootstrap Cascode of Field Effect Transistor This is a circuit that I personally like very much. The figure below is a differential input stage of an amplifier I made, using bootstrap on two JFETs.
In this circuit, Q5 Q6 are differential amplifiers, and the static current is determined by the common current source of the source. Q1 Q2 forms a mirror current source load, which is a very common circuit connection method. Q3 and Q4 are used for bootstrapping. Their bases follow the source voltage of Q5 Q6, so the emitters also follow the source voltage of Q5 Q6. The purpose of this bootstrap circuit is to keep the VDS of Q1 Q2 constant, thereby eliminating the influence of the parasitic capacitance Cgd (because the capacitance of JFET is relatively large, which is a disadvantage). The role of R1, D1, D2, and Q7 here is to use the voltage regulation characteristics of the diode to provide a basically constant bias voltage. Here, if Q3 and Q4 are regarded as buffers, then the signals they output are not drawn out for use, but are only used to determine the drain voltage of Q5 Q6. However, Q3 Q4 does output a signal, which is output from the collector - the gain is much greater than 1 when applied. Taken apart, Q3-Q5, Q4-Q6 are all cascode circuits, except that the base voltage of the common base amplifier part changes with the differential pair tube - so it is a bootstrap cascode. The biasing method of the common base transistor of the cascode in the above circuit is a little complicated. If it is replaced with an appropriate JFET, the cascode can be very simple, as shown below.
This connection method limits the choice of JFET. You can't just grab two tubes at random. Because it is necessary to ensure that under the set current, the VDS of the tube that inputs the signal (Q1 Q2 in the above figure) is equal to the VGS of the bootstrap tube above it.
* Bootstrapping of the op amp Finally, let's look at a special use of bootstrapping: changing the power supply voltage of the operational amplifier so that the two power supply terminals follow the input terminal. In this way, from the perspective of the bootstrapped op amp, it seems that the + terminal input signal is constant - that is, the input common mode voltage is eliminated. This technique is used to reduce distortion at the op amp input stage.
As shown in the figure above, the price paid for power bootstrapping is not small. In addition to adding an op amp as a follower, two transistors are used to expand the current because of the need to provide large current. To summarize the key points of "bootstrapping" in the above example circuit: there is a buffer that provides a low-impedance output (an amplifier with a gain of approximately 1), which buffers the voltage of the reference node and applies it to another node in the circuit, so that its AC voltage changes in the same way as the reference node. The result is a constant voltage effect on the terminals of certain components or parts of the circuit, improving certain aspects of the circuit's performance.
This content was originally created by EEWORLD forum user cruelfox. If you need to reprint or use it for commercial purposes, you must obtain the author's consent and indicate the source
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