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146 Threads found on Miller Capacitance
basically, miller effect is the small g-d capacitance multiplied by the voltage gain at the drain. a high impedance at the drain gives a big miller multiplication. however, looking into a source is a very low impedance so the voltage gain, and consequently the miller effect, is smaller.
Do a web search on Google for "What is miller capacitance in MOSFETs" .. miller capacitance is usually given as Cgd capacitance versus Drain to Source Voltage .. see example below .. Regards, IanP
Hi, What is miller capacitance? Why it is used in Differential Amplifier?
what is the effect of miller capacitance on stability? if the miller effect dominates, does it mean that the positive zero is at very high frequency?
Dear all, How to calculate miller capacitance from the model parameters of a transistor that's biased up? Another related question is that when I want to use simple equation to find a particular parameter, say use Cgs as part of ft expression, how to find this capacitance? Since there are Cgs, Csg (for Cgd, there is also Cdg)? Thank you.
Can anybody help me on this feed forward compensation with no miller capacitance with some explainatory notes or materials. thanks in advance.
cascode miller cap.
In Gray it is said single stage opamps,such as the telescopic cascode or folded cascode,have only one gain stage;therefore miller compensation is not possible. Why? Can I attach a miller cap between negative input and single ended output?
If you use Hspice, then it is probably to use LX19(mn) to get information of miller effect.
In simple terms, the miller effect allows a small capacitance to be "amplified", which is useful for compensation. Using a non-miller capacitance for compensation means it will have a larger value, hence larger area.
Hi I want to calculate the output capacitance Co of a transconductor as depicted probleme i dont know how to do to calculate only Co without the contribution of the overlap capacitance here is my
what is the difference between feedthrough and feedback capacitor? what is the meaning of this statement "CGD represents miller capacitance feedback while CDG represents miller capacitance feedback" There are quite substantial differences! 1. A"feedthrough" capacitor is used to pass DC Current through a metal wall
I have read about miller capacitance topic in odyseus's website but i dont understand what -G (nagative gain) mean. 180 phase shift??? How does this millter capacitance effect the input capacitance of a FET/transistor? thank you
Thanks for your advices! 1. the pole and zero cancellation is not required that the zero= pole, it only need the pole is less than zero, then the magnitude is decreased by 40db/dec, but the phase is not it is not reqired a accurate resistor. 2. In razavi's book about the freq comp, which is almost talked about in miller comp between
To further amplify (sorry for the pun) on the post above, look at the schematic for the extra lumped capacitance added to the circuit. It will usually be across the input/output of one of the stages. The miller effect stated in the previous post is used to make the time constant larger than if the capacitance was across a lumped resistor.
miller effect: if a capacitor (C) is placed between input and output of an inverting amlifier,the equivalent input capacitance becomes (1+gain)*C When the transistor is in common emmiter mode,the base-collector capacitance is this miller capacitor.Since the gain is high,the input capacitance is equals Cb-e+ (...)
my opinion is when a few limiting amplifier cascade together, the gain will increase from first stage to the last stage. As gain increase, the miller capacitance also increase according to formula cgd(1+A). This will form a even lower frequency pole as incresing stages, f=/2pi. RC. The output of the last stage will be the dominant pole or the firs
it could be because: at the cascoded node, you reduce the milled effect cap for the input. but at the same time you isolate the feedback resistor with that node. it means the output resistance is increase to (gm//gds). i am assuming the feedback resistor is much less than gm//gds.. then the bandwidth at the cascoded node will be reduced..
You are not doing this the standard way that most circuits work best at. You need to find which stage has the lowest pole frequency and add to the miller capacitance by putting the capacitor from base to collector on that stage. This will lower the dominant pole. I would guess that this is the second gain stage where you have the capacitor from
first - 200uV is exceptional! you can only hope that your chip has less than a few mV of ripple. try to look at it in terms of percent, and keep the percent lower than 1%... or 0.1% for extremely high accuracy. say your full scale voltage in lock is 0.5v, you can endure 5mV steady state ripple before you reach the 1% region. and if your f
the equivalent circuit of a miller item lets say a gate-to-drain capacitance of a MOSFET is well know. The effect of Cgs can be replaced by a gate to ground capacitance and a drain to ground capacitance with the equation given. However, intuitively, I think this is only an approximation of the miller (...)
The desirable property of common base is that the base-collector capacitance goes from the output to ground and does not cause a miller effect or feedback that causes oscillations. The collector-emitter capacitance is much smaller. In the early days of large junction areas the common base circuit was commonly used at VHF. Modern (...)
Hi zhi_yi, I hope the following helps. Voltage Divider Bias - commonly used for biasing BJT single transistor amplifier, especially in a common-emitter configuration to provide for medium current gain and medium current gain. - bias the base voltage so that you have a DC level for the input signal (AC) to swing. - the resistors are selecte
I have designed a 6bit Pipelined ADC. However, I got some problem when i design the reference voltage circuit. Because I have no output pad for VREFP(1.75V) and VREFN (1.25V) , so I add a large cap (about 400pF) on chip. I wonder which kind of OPAMP I should use. For a one stage( folded) opamp, the large cap let the bandwidth very low ( about 1K
because the circuit is for digital signal, so it is not small signal operation.also remember for large signal operation the CL is nonlinear. but in digital circuit for a quick simulation, we usually define the equivalent CL. therefore CL is not the same as differential amplifer. and also the equivalent CL changes depending on the operation of th
In the differential pair in the picture, the second pole is equal to gm3/Cx where Cx is the total capacitance in node X to ground. It is approximated to be Cgs3+Cgs4. Why do we not consider Cgd4? For small lengths, wont the miller effect from the gate to drain of M4 cause this capacitance to be on the order of the gate-source (...)
G35, yes, voltage spike on the drain can effect the gate voltage. The reason for this is the gate-drain (miller) capacitance of the FET. Have you tried to limit the spikes on the drain by using a snubber circuit from drain to source? The snubber would either be a RC or a RC+diode circuit. Before putting anything else around the circuit I would
I'm using miller capacitance sample and hold cirucit. For an 8-bit resolution my sample circuit must be able to detect and hold at least 12mV. Right now I'm only able to make the resolution to 0.2V @ 200mV. Can I get ant suggestion of how can i achive my targeted resolution. Thanks in Advance. Best Regards Syukri
I think gate capacitance not depend on gate voltage, both of Gate to Drain, and Gate to Source are fixed. But because of miller effect cpacitor of GD can be appear very greater than initial value. Regards, Davood.
What do you need it for? Maybe you can use a gain stage and use its miller capacitance,
It is said that in miller compensation , the input node 's pole is moved to low frequency , and the output node's pole to highfrequency in many materials , but my opinion is reversal . Is there some one can expain ?
...if anything, at the moment of the collector voltage increase, via the miller capacitance, the base of the transistor would see a current injection, which should turn the transistor on more. But this current overshoot indicates an opposite transient. When we increase collector voltage: Collector-Base junction
I have tried to design some distributed amplifier using the book by Virdee. I do have design examples using ADS. Let me try to find it. I have done it for few years. In Virdee's book they use a chip FET and use wire bound as inductors. I dont know this approach will be good up to 10 GHz or not. I would think that it is better if you can design it a
Mos (also BJT) while turn on (VGS>Vt) Drain Gate capacitance increase. VGS voltage stay fix because Q/C stay fix. This section known miller plateau. Vg increase after the Cdg capacitor variation stoped. (This effect is same at the turn off process)
It can also increase the bandwidth, due to reducig the miller capacitance. of the commen emitter/source amplifier.
I see that the miller theorem you use when you have a circuit element between input and output w.r.t. a common ground and you can split the passive element accordingly as you have done here by using the given formula ... So, try doing this .. split the capacitor from in 1/2 in first diagram .. so that they are in parallel and effectiv
Like to other two contributors said, follow the design rules. Typical electrode configuration is C-E-B because you can put the base electrode over the outside isolation thus minimizing your collector-substrate capacitance. In addition, keeping the base electrode away from your collector will also eliminate fringing capacitance to the two electrod
what is miller, lead and Lag compensation ? Thanks
Well, that depends on the application .For digital circuits ,for example, it is usually required to reduce the drain capacitance because the drain node is usually connected to output while the source is usually connected to GND or Vdd .Thus, by decreasing the drain capacitance, less delay ad power consumption can be achieved . For analog applica
First, concerning the capacitive load, usually you will have a minimum capacitive load as a spec and some times u will need to increase this load to enhance stability in load compensated amplifiers while u will need to decrease this load to enhance stability in miller compensated amplifiers. Second, concerning the slew rate, usually u will have
Hi, I've found some papers. Mainly poor PSRR in ur ckt is result of miller capacitance influence. Change output ckt. Choose solution by yourself.
I have read and searched google a lot about miller effect but could not find out 1 thing? i.e. The capacitance doubles when there is opposite swing or opposite polarity? I mean capacitance doubles at the tranistion or level?
what is miller effect?
A is the gain of the system without the miller capacitor.... so all that you have to do to find the value of capacitor to use at the input at that frequency is to remove that miller cap and find the value of gain at that frequency and then use the multiplication factor and place the capacitor....
MY Question what conditions we must take the miller's theorem into what conditions the miller's theorem will work???In a passive or in a active circuit?? I think miller's theorem will work only in a invert ciruit,Am'i right?
Hey As far as I know, this is the most common method and the aim is to use the miller capacitance to move the poles to stabilize the op amp. It is however not necessarily about the first and the second stage.
the miller effect thrives the input capacitance it a needed one or unwarranted..if it is not needed than how best it can be eliminated,
Hello everybody, I want to design sample and hold circuit using double sampling technique at 160MHz clock (320MS/s) and 70dB IMD. Here I choose the two-stage opamp, the schematics are here. (Two stage module and the double sampling). But slew rate and charge injection decrease the IMD sharply. I try to adjust the miller capacitance, but the effe
in wiley sansen book, while he was talking about nested miller compensation , i noticed that he connect Cc between two nodes that the gain is positive , according to miller ; we multiply the cap. by (1-Av) and so , seems like if the gain is positive then -ve capacitance, can anyone tell me what i am missing ?? thanks
During oscillation startup, an inverter is basically operating as an amplifier with a dominant pole (due to miller capacitance). What would you expect for the total gain and phase shift when cascading these elements?