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236 Threads found on Linear Region
1-Those are semiconductor capacitors but I don't know what type they are.. 2-The MOS are not diode connected, they have been used instead of resistors ( may be voltage controlled, depends on Vint ) or they are simply Pull-Up resistors.Vgs=Vds so the MOS operates in linear region being as resistor
You'd still need a driver to excite the RC tank and once you have a driver it's best to use it to drive the actual gate directly. Assuming the mosfet is switching power, you want a strong driver to limit time in the linear region. Though there are plenty of applications that use basic oscillator circuits in power applications, there is few bene
Is the small signal model still valid?
I am told that power amplifier should be biased at saturation region. For the bipolar process, as shown in the figure, the saturation region( linear region) is at the left side, which is different from the saturation region of CMOS. 133664 So I need to bias the power amplifier to the B node, right? (...)
These differences pertain to mode of operation mostly. "On" resistance is resistance, desired to be low, from drain to source in the hard-gate-driven linear region. Of course a weakly driven FET has an on resistance too. Just lousy. "Output" resistance -is- "on" resistance, when the FET is "on". But it's more of interest (i.e. different interest)
133280 1)#1, which is an alway turned on transistor and that acts as a resistor? it's Not degeneration resistor. what is that for. If it's acting like a resistor then it much operating at linear region. 2)#2 why there's a cross coupled nand gates? 3)#3 Is that a cross coupled element used for as latch, for stoarge?
Yes, you can. Yes, people do. But there are a couple of things to watch out for. One is, the ramp time through the linear window of the front end (that is, in the second chart from ata_sa16, the region where the ouputs show any non-flat-line quality, plus some) must be much less than the delay time through the amplifier or you will pick up an er
Cascading NF specification of a system is valid only and only if the system is linear.( or the since the system is in linear region)Otherwise, non-linearity will bring more or less additional noise and this relation will not be valid anymore as in your system that includes a high nonlinear element frequency (...)
For a triangle waveform to produce the same result as a DC hysteresis sweep, the delay of the path in question must be << the ramp time -through the region of interest- (e.g. the linear window of an inverter chain). A 1uS total rise time and a 10nS prop delay will give you 1%-ish error if the whole range is linear, 10%-ish if only (say) (...)
So, do you agree that the PMOS connected its gate to ground is a linear device? do you agree that the PMOS connected its gate to CLK is a saturation device? No. PMOS w/ G=gnd! is linear provided that the drain is at some point above gnd!. At D=gnd! it is possibly still in saturation as |Vgs-VT| < |Vds|. Without def
OTAs are simple, easy to compensate (shunt C on output) and easy to slap around with their current limited output (e.g. soft start, current limit functions). Their gain tanks with any resistive load, but internal to a CMOS IC there may be none (just C pretty much). LDOs operate the pass FET in linear region -when low supply headroom- but this is
A sequence $x$ is the output of a linear time-invariant system whose input is $s$. This system is described by the difference equation $(1.1)$ $$x=s-e^{-8\alpha}s$$ $$\alpha>0$$ a) Find the system function $$H(z)=X(z)/S(z)$$ and plot its poles and zeros in the z-plane. Indicate the region of convergenc
126451 The region where the Vout is some constant times the Vin is called the linear region and where the output saturates due to slew limits or the maximum deliverable output current limit is called the slew rate limited region.
... how can i expect rds for a specific vds , how LAMBDA influences for currents (becauseit varying very drastically with high curretns? Find here an Id vs. vds plot of a MOSFET: 125969 Here you can see rds vs. vds: in the linear region rds is essentially constant for
s-parameter is small small signal but provided you are operating in linear region, it should give you good estimate. since you have nothing other then s-parameter, its good to simulate it cos you have no other options available!
I still want to get a linear change in resistance even below the cutoff/threshold voltage (0.6-0.7 V), is there any mechanism or other circuitry which can be used to achieve this ? The resistive part of the Ids vs. Vds characteristic (called triode or linear region) has nothing to do with the MOS
Hi i know translinear principle is nothing but the relation of transconductance and the collector current and also MOSFET in the sub threshold region have a linear behaviour. So my doubt i am not able to find if a topology of transistors build is in the translinear closed loop. Will transistor only in closed loop form the (...)
Hi, there are several manufacturers for the 4N25. Some just specify a minimum CTR. So 74 is within spec. Toshiba datasheet says ctr= 100typical. *** Maybe this is thae cause for the confusion. Anyway, it is a optocoupler for digital informations. Here you typically use a lot of overdrive causing saturation and don´t go to the ctr limit. T
If operating as a switch, you go from saturation to cutoff (or other way round) as you mentioned. As Vbe increases, for a short amount of time, the transistor will be in active (linear) mode until you enter saturation. But you want to design such that it spends as little time as possible in the linear region.
The resistance variation for 1°C increase in temperature is about 390 mΩ. for I mA current and accuracy about 0.1 °C the voltage will be 39 ?V You need to calculate the bias current so that the curve fit within the less non-linear region of the sensor. Keep in mind that PT100 sensors are not totally linear
You should take an IP3 measurement at ~10dB backed off from the PA's / RF Chains P1dB point. Thus reduce the I/P pwr till the O/P pwr is ~ 10dB lower than the P1dB point. You need to be in the Amp's linear region. If you put too much power you will be compressing Amplifiers, and the IP3 will get worse. Cheers
In the linear region a rds can be descript by rds = 1/(mu Cox W/L (VGS-VTH)). In a mos-cap VDS is equal to zero but what's the value for rds when VGS 0 --> rds -> ∞
Maybe you are thinking about an ordinary transistor saturation and linear. A Mosfet is the opposite: 1) It is turned on hard like a switch when it is a linear resistor. 2) It is an amplifier with plenty of drain to source voltage when it is saturated.
I expect Ib to be Ic/hFE (while its much lesser than this value) because bjt is clearly in active region with Vec=3V When you do not follow design guidelines for normal switch non-linear operation, you can expect unusual results. The BJT has operated as expected because your base bias current far exceeds your load
Hi, How can i model an opamp with its linear region area and saturation effect in ICAP/4? The problem is the saturation effect
how to mesure un*cox for nmos3v in TSMC? You can calculate it from an Ids current measurement in strong inversion and linear region from this equation: I_{ds} = \frac {(\mu_n \cdot \, cox)}{2n} \cdot \, \frac {W}{L} \cdot \, (V_{gs}-V_{th})^2 For the substrate factor
Output rise/fall will depend on output buffer drive strength and loading. Input rise/fall will contribute to jitter / PN, the slope there and the input linear region width determine your voltage noise to time noise (jitter) transform. The two are not specifically related if you have any significant stage-count and are not up against the speed limi
I think if the output stage is ensured to remain in triode region throughout its operating range, then the distortion will be less. I'm sure you thought of saturation region ? ... if we drive the output stage into linear I think that the signal will be distorted at it'
The gain of transistors may vary in function of the biasing current. The electric characteristic of inner components certainly are optimized to meets the optimal performance at the recommended operation conditions far from linear region.
Hi Dario I am not sure that Cadence/spectre can do the linear interpolation. What you should do is a interpolation of your points in matlab, and then feed this to Vdc in cadence. hope it helps
Two stage Opamp with miller compensation was designed with gm/id =13 and circuit is functioning properly for open loop. However, when this opamp is simulated as a unity gain buffer a few transistors start operating in linear region. What might be the cause for this behavior? Please help You probably changed the D
Class A depends on avoiding the region of saturation with around a 2V margin where linearity from current gain drops rapidly. Negative feedback improves this linearity and also lowers output impedance, but also lowers input impedance by the feedback ratio. Class D depends on avoiding the linear region by (...)
In this graph see the change of current when the voltage is increasing the region 5v to 40v current is almost linear
Turn on time of triacs is usually specified in a few ?s range. I agree with Dan Mills that critical dI/dt is the important parameter to determine usable switching speed in triac applications. The reason for rating dI/dt isn't that the triac goes through a linear region. The positive feedback of the four-layer structure causes fast transition to
Hi FvM I would like to operate the diode in the linear region, so sinewave modulation. I have not yet decided on the laser diode as I wanted to see if the idea was plausible before going further. It seems from the above answers that the basis it correct so I will contruct the circuit and see what the result is. Ma
The resistance rds in saturation mode models a behavior of the channel, it is not a physical resistance. The channel's noise is already modeled by In, so adding noise for rds would be like double counting. In the other hand, in the linear mode, the channel is an actual resistance and the noise is modeled as such.
Thanks, dick_freebird & FvM but I want to know what is limited voltage range("But a current source that only works right over a limited voltage range") and MOSFET Power dissipation P_dis = Vds*Ids, the need Vds is smaller working in linear range than saturation. so the power will be reduced, isn't it? BTW, In my testing board, there will be a pe
I have never seen an ID-VG curve that looks like constant-gm. There's always a peak and it's right below VT. By positioning yourself near the peak you can get close to constant gm, if that's what you're after, but across a broad range and especially in linear region, I would not expect anything close to constant.
Anything with a depletion region on either side of it will be voltage dependent. MOS channels swing from accumulation through depletion to inversion. Cgs follows gate voltage directly; Cgd, only when the channel is well lit near the drain (linear region - when it goes constant current the capacitance is sort of stood off from the (...)
The general rule of thumb for heating MOSFETs is either from linear region use, not enough current for switching, or actual conduction (on resistance). 1.) Can you show the circuit? 2.) What is your load that you are driving? 3.) What is the switching frequency? The problem could lie within the speed you are switching at and the 500mA the chip i
How to represent mosfet in triode and sub threshold region. like self cascode transistor one transistor is in saturation and another once will be in the triode region . i want to calculate output resistance and effective trans-conductance of self cascode MOSFET. can any give the brief note about this topic. thank you
According to SOA figure This is for, to operate in linear region. It does not imply here.
Check this IC You can use the alert outputs for the relay/mosfet control You can adjust the upper & lower threshold using the three resistors. - - - Updated - - - If this temperature control is only your requirement, you dont need any micontroller. You can us
Subthreshold (or weak inversion) actually isn't a region of a MOSFET's output characteristic like off, linear = triode, saturation or breakdown region - which are identified by a Cadence ADE analysis, (together with No.3=subthreshold) - but much more an operation mode, classified by the deg
The linear behaviour is real, present both in device equations and empirical behaviour. For most MOSFETs with source shorted to substrate, it ends when the drain-substrate diode starts to conduct.
Voltage gain is gm*rout, rout is the parallel sum of the gain device, its load device and the attached impedance (driven load or successor stage), and rout is lousy in linear region. If your driven load is high impedance then the gain stage needs to be high impedance (saturation) presuming you want your amplifier to, you know, amplify.
The output conductance of the transistor moving to linear region rises, hence the overall output resistance will fall.
Hi, Crest Factor = Peak Current / RMS Current. In a linear load, the Crest Factor is 1.414. However, in a general purpose SMPS, due to its non linear nature, this factor will be much higher - in the region of up to 4. See if you can measure the current using a CRO/DSO by putting a small resistor but having higher wattage in line. Then (...)
Hello Sir/Madam, I have simulated a MOS structure. I wanted to know the hole concentration in each region. When I performed the cutline I was able to get a graph in the log scale. Can I change the obtained cutline graph from the log scale to the linear scale. Thanks in advance, Vaish
EVERYTHING is non-linear