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148 Threads found on edaboard.com: **Equal Gain**

Hello Friends,
I am designing a Charge sensitive amplifier and would like to study about the stability.
we know, A/1+AB where AB is the open loop **gain**. If AB is **equal** to -1 then the system will oscillate. How would I know that AB will be **equal** -1. Please tell me the process to calculate.
A= G sCdRf/ 1+sRf( Cd+Cf)
B= (...)

Analog Circuit Design :: 02-28-2017 13:55 :: tisheebird :: Replies: **0** :: Views: **1**

GBW of aux opamps has to be **equal** to dominant pole of main opamp. In other way You get zero (if GBW is lower) or additional pole if it is higher.
It results with auxilliary amps dominant pole location **equal** to main opamp dominant pole divide by aux opamp **gain**.

Analog Circuit Design :: 12-20-2016 17:56 :: Dominik Przyborowski :: Replies: **5** :: Views: **1091**

Hello, I had a question in op amp?
If we provide **equal** dc voltages to an op amp it will produce some voltage 'Veq'.
The op amp has a finite **gain**. My understanding is, if we use this circuit in a negative feedback loop of another circuit X, its input voltages come closer to each other.
However they do not necessary become exactly **equal**. I (...)

Analog Circuit Design :: 09-27-2016 06:15 :: Debdut :: Replies: **1** :: Views: **538**

Equation 23, page 14 , of AN57 by power integrations (below) shows a parameter called
?K TL31? which is the **gain** of the TL431.
From the attached TL431 datasheet , does this **gain** **equal** ??.which **equal**s 1/0.0014 = 714
TL431 datasheet:
AN57 by Power Integr

Power Electronics :: 02-28-2016 13:27 :: treez :: Replies: **8** :: Views: **712**

How OPAMP ensure both input terminal are **equal**?
Simply using negative feedback ??
Thanks

Analog Circuit Design :: 12-14-2015 20:18 :: Ltarek :: Replies: **2** :: Views: **461**

Yes, that means the op amp will be unstable if used as a +1 follower.
A minimum closed-loop **gain** of +2 for a non-inverting amp or a **gain** of -1 for an inverting amp is the same from a loop-**gain** point of view since in each case you have two **equal** value resistors going to the op amp (-) input.
The only difference between the (...)

Analog Circuit Design :: 12-09-2015 20:02 :: crutschow :: Replies: **3** :: Views: **505**

Normally all values are **equal** for a unity **gain** differential amp ( 10K) to minimize offset from input bias current.
inverting **gain**= -R7/R8 *7V = (-1) * 7V
Non Inv **gain** = 1+ |inverting **gain**| * V+in= (1+1) * 5V*R5/(R5+R6)= (1+1) 5v * 1/2= 5v
Using superposition rules the output-7+5 =2 the difference.
As a (...)

Analog Circuit Design :: 07-30-2015 20:45 :: SunnySkyguy :: Replies: **4** :: Views: **957**

Where do you see doubled current, in which situation? In steady state, the inner loop current setpoint must be obviously **equal** to output current, are you talking about transient states?

Power Electronics :: 06-25-2015 22:19 :: FvM :: Replies: **7** :: Views: **681**

In open loop configuration it's enough to use a single pulse source in series with one of the Vcm bias supplies (small signal stimulus).
In closed loop configuration (your application) you better use 2 **equal** anti-phase pulse sources in series with a common Vcm bias supply and (if so) the **gain** setting input series impedance d

Analog Integrated Circuit (IC) Design, Layout and Fabrication :: 04-09-2015 19:53 :: erikl :: Replies: **1** :: Views: **684**

hello everyone...
is it necessary to have the UGB of single ended differential amplifier used as CMFB of main opamp to be **equal** to the UGB of the main opamp ????
Is there any specific crieteria???

Analog Circuit Design :: 02-12-2015 15:03 :: pankaj jha :: Replies: **1** :: Views: **660**

It seems that you should use linear values for every specification. The reason 1 is subtracted from NF is because it is always greater than one in linear scale. Therefore, that performs is just how much better than 1 your design is. The same goes for the **gain**, if you double the **gain** while maintaining everything else **equal**, you should have (...)

RF, Microwave, Antennas and Optics :: 12-20-2014 07:43 :: csolis_GT :: Replies: **10** :: Views: **1142**

The loop **gain** at DC is 90dB. However, my PSRR at DC (taken at the top of the resistor) is 80dB.
Does this make sense? I would have thought the PSRR at DC should **equal** the loop **gain** at DC.
There's no direct connection between open loop DC **gain** and PSRR at DC (better: at low frequencies: DC is static, no cha

Analog Integrated Circuit (IC) Design, Layout and Fabrication :: 12-11-2014 20:47 :: erikl :: Replies: **2** :: Views: **758**

You can use a positive bandgap reference of any voltage and your negative supply scaled to be **equal** and opposite. Then your inverting input is now 0V ,( with +Vin=0V) your (-) must now use the same source impedance with a common feedback **gain**.
You may add series RC values for derivative and proportional **gain** control and common RC (...)

Analog Circuit Design :: 11-06-2014 18:28 :: SunnySkyguy :: Replies: **18** :: Views: **2301**

A Sallen-Key filter with **equal** resistor and capacitor values will have a Butterworth response when its **gain** is 1.586 and will oscillate when its **gain** is about 2.0 and higher. Try it with a **gain** of 1.8 times.

Analog Circuit Design :: 11-01-2014 14:57 :: Audioguru :: Replies: **5** :: Views: **1461**

The folded cascode has a higher **gain** than that of simple amplifier, but its **gain** is little lower than that of cascode. As far as I remember, the main idea of the folded cascode is that the input and the output common mode points are **equal**. Why do we need such a thing? in case you are designing multiple stages (cascading) in order to increase (...)

Analog Circuit Design :: 09-05-2014 17:44 :: Engineer4ever :: Replies: **1** :: Views: **484**

Hi,
I am finding Current **gain** of Common gate amp with gate biased and applying supply at source.
If I measure Iout/Iin getting 1 as current **gain** till mosfet is ON and when mos enters to cut off it is giving **gain** of sudden spike around 20.
and also by varying input sweep in steps 0.01/0.001/0.0001.
In this case source current is (...)

Elementary Electronic Questions :: 09-01-2014 09:05 :: raj_007 :: Replies: **1** :: Views: **487**

It's related to the fact that the **gain** of an inverting op amp is ideally **equal** to the Feedback-Impedance divided by the Input-Impedance.
As the frequency increases the impedance of C1 goes down. Since a reduction of the input impedance increases the **gain** of the op amp, the input RC causes the **gain** to increase with (...)

Analog Circuit Design :: 07-14-2014 05:56 :: crutschow :: Replies: **4** :: Views: **978**

If the current **gain** is not **equal**, a differential output voltage will exist for an **equal** input current.
This is even more important if only a single ended output is used and current mirrors are used with a different Vbe characteristic. Some more info here

Analog Circuit Design :: 06-18-2014 02:27 :: SunnySkyguy :: Replies: **2** :: Views: **629**

Correct - however, does this justify the dimensioning with **equal** components?
I don`t think so - because this has severe disadvantages.
Selection of **equal** component values requires a fixed **gain** rater close to the stability limit (which is "3") and the filter parameter sensitivity upon tolerances is rather large.
A better design is based o

Analog Circuit Design :: 03-18-2014 08:53 :: FvM :: Replies: **7** :: Views: **876**

Things are very simple.
Whatever **gain** the RX and TX antennas have and whatever loss is between them, the received signal power level (at RX input) cannot be higher than transmit power level (at TX output).
In the best case (and ideal) they can be **equal**.

RF, Microwave, Antennas and Optics :: 02-18-2014 08:33 :: vfone :: Replies: **11** :: Views: **657**

Comparing DCM and CCM flybacks, both in current mode, which has the higher **gain** of the feedback loop?...(assuming all other things **equal**.)

Power Electronics :: 01-04-2014 00:59 :: treez :: Replies: **1** :: Views: **742**

Hi, i'm trying to calculate small signal **gain** for common emitter amplifier with bjt diode connection load.
For diode connected bjt resistance is **equal** to parallel connection of Rbe and current source b*Ib. Rbe much smaller then Rcs and therefore R**equal** ~ Rbe.
For **gain** we have A = -gm1*Rbe2
Am i right?
Update: Sorry, (...)

Elementary Electronic Questions :: 11-20-2013 20:22 :: sarge :: Replies: **0** :: Views: **608**

You mean F=SNRi/SNRo but just at certain level of noise?
Yes, let's suppose we have a noisy amplifier having **gain** G. At the input we'll place a resistor having value **equal** to the input impedance of the amplifier. This resistor will generates a noise power given by N=k*T*B where k is the Boltzmann constant, T the tem

Elementary Electronic Questions :: 11-10-2013 22:39 :: albbg :: Replies: **3** :: Views: **1981**

... keep the same stage **gain**. I did try to use a non inverting op amp stage after it instead of two inverting stages but couldn't seem top get it to work correctly.
After the tap of your poti you could do it with one opAmp and 4 **equal** resistors (say: 100 kΩ) , and a positive reference voltage of 7.

Analog Circuit Design :: 10-02-2013 16:32 :: erikl :: Replies: **10** :: Views: **722**

Can inverters, like the ones found on 74LS04 ICs, operate like op-amps? I read somewhere that by connecting the output to the input, it can be made to function like an op-amp. If this is true, then the **gain** is **equal** to the feedback resistor value / input resistor value, assuming that that is all there is connected to the

Analog Circuit Design :: 08-28-2013 12:03 :: IanP :: Replies: **4** :: Views: **671**

Not quite right. Output impedance of common collector (emitter follower) is measured between the emitter and common, not emitter to collector. The output impedance of the common collector is approximately **equal** to the source resistance divided by the current **gain** (hfe).

Elementary Electronic Questions :: 08-13-2013 15:45 :: barry :: Replies: **1** :: Views: **660**

In simple terms if Noise is calculated using:
Noise Power Density = kTB + NF + **gain** (dBm/Hz)
Then for your case with the amplifier and transmission line:
Noise Power Density = kTB + NF(amp) + **gain**(amp) + NF(line) + **gain**(line)
For any passive device, like the transmission line, the Noise Figure is **equal** to the (...)

RF, Microwave, Antennas and Optics :: 07-06-2013 00:16 :: RealAEL :: Replies: **9** :: Views: **696**

Hello,
Can anyone recommend a high **gain**, low power BJT for use at 450MHz?
I have been using the BFT25 but I was wondering if there is something out there with more **gain**. 10LogS21/S12 is roughly **equal** to 14db.
Thank you

RF, Microwave, Antennas and Optics :: 06-14-2013 13:58 :: darcyrandall2004 :: Replies: **3** :: Views: **676**

Hello everybody ,why do we perform DC and AC analysis in mosfet amplifier ,is the voltage **gain** **equal** to the ac output voltage over the ac input voltage or the resultant of DC and AC voltages, and finally is Id the DC drain current or ac drain current such in this equation Id=.5 Kn*(Vov)^2 ?

Analog Circuit Design :: 04-19-2013 09:17 :: hoo :: Replies: **0** :: Views: **466**

You must show us your schematic. The easiest way is to use two **equal** value resistors directly across the 0-10V signal. At their center it will be 0-5V and no op-amp is needed.
Brian.

Power Electronics :: 04-05-2013 11:38 :: betwixt :: Replies: **5** :: Views: **1211**

breakfrequencie, 3dB Bandwith and unity **gain** bandwith are one of many things i am bit unsure of.
I know that the breakfrequency is the frequency for where the **gain** drops 3dB.
which means that unity **gain** should be the same, (which i know is wrong, since formula is different)
And at last 3dB bandwith should **equal** (...)

Elementary Electronic Questions :: 03-31-2013 15:33 :: kidi3 :: Replies: **0** :: Views: **676**

Could you please show me in any graph th -3 dB frequency of the buffer connected op-amp please
Junus - it is very simple:
For any first-order circuit the 3dB point is at a frequency where both imaginary and real part of the denominator are **equal**. This applies also to the simple RC lowpass.
The denominator of the

Analog Integrated Circuit (IC) Design, Layout and Fabrication :: 01-22-2013 13:06 :: LvW :: Replies: **7** :: Views: **2017**

vo = A ( v(+) - v(-) )
that is output voltage vo is **equal** to the difference in voltage between + and - inputs multiplied by the Open loop **gain** of the op. amp.
In an ideal op. amp the current at the inputs is zero (infinite input impedance) and A is very large (--> infinite). Output resistance is zero.

Elementary Electronic Questions :: 01-14-2013 19:43 :: albert22 :: Replies: **5** :: Views: **599**

it is really difficult to get an exact **equal** voltage at the input of a very high **gain** comparator, a comparator with a very haigh **gain** can switch the output with a very small difference. assume the **gain** is 10000(80dB) and the supply voltage is 3.3 and by asssuming that the offset voltage is zero then the minimum voltage (...)

Analog Circuit Design :: 11-08-2012 14:43 :: Junus2012 :: Replies: **4** :: Views: **490**

The Barkhausen criterion requires a loop **gain** of **equal** or larger than unity and a phase shift of 360 deg at a certain frequency.
In a typical ring oscillator the loop **gain** always is larger than unity. Each stage has an inherent delay, which can be increased using external capacitors.
There is one particular frequency Fo for which this (...)

Analog Circuit Design :: 07-20-2012 19:49 :: LvW :: Replies: **3** :: Views: **1084**

you can use a differential amplifier
77360
Apply 0.5v to VR, V1 to gnd and the input to V2.
If all resistors are **equal** then you will get 0-2v, if you use RF twice the size of R1 then you can get a **gain** of 2 and 0-4v output.
You can also use a VR of 1v to get some headroom in case the input is <-0.5v and then compens

Analog Circuit Design :: 07-17-2012 19:08 :: alexan_e :: Replies: **2** :: Views: **554**

paths add constructively and destructively with **equal** probabilities, thus the channel **gain** varies about the mean channel **gain** -> 0 dB (**gain** = 1).

Digital communication :: 06-12-2012 13:55 :: Ahmed Alaa :: Replies: **4** :: Views: **3102**

You cannot remove poles. But you can compensate those poles which is not a very efficient solution. A better solution is to make those **gain** boost amplifiers excess stable as said in Vadim Ivanov's book. If you are compensating you are reducing the speed that you have already paid for with power by using capacitors which is **equal** to area. Use only o

Analog Integrated Circuit (IC) Design, Layout and Fabrication :: 05-22-2012 20:06 :: kemiyun :: Replies: **2** :: Views: **581**

To summarize (and partly correct) some of the statements made before, I like to point out the following - applicable to each kind of harmonic oscillators (LC or RC):
* Each harmonic oscillator must be designed to enable the following condition (formulated in the 1930th by H. Barkhausen):
(1) Loop **gain** magnitude **equal** to unity for one single frequen

Elementary Electronic Questions :: 05-22-2012 12:33 :: LvW :: Replies: **15** :: Views: **954**

I want to connect between two stage PGA that the overall **gain** **equal** i have designed the first stage by telescopic cascode op amp and feedback resistor loop on it SO how i connect between them or any one can give me any materials that talk about this problem..!!

RF, Microwave, Antennas and Optics :: 05-07-2012 02:21 :: Solom_1010 :: Replies: **0** :: Views: **612**

I just came across this- We know that feedback **gain** is "β". I read that when β=1, we have worst condition. Why is that so?
Thanks in advance.:razz:

Analog Circuit Design :: 04-23-2012 16:01 :: iVenky :: Replies: **5** :: Views: **566**

I am simulating a waveguide in HFSS having 2 ports( rectangular dielectric waveguide) .I want to ask what is real meaning of S(1,2) because in one case i am getting S(1,2) plot showing 5db **gain** while the propagation constant (beta) is **equal** to 0. Field overlay also shows wave dying much before even reaching port 2 then how to interprete S(1,2) val

Electromagnetic Design and Simulation :: 04-21-2012 19:15 :: bhvk.ameta :: Replies: **0** :: Views: **673**

Yes you can do so if your Yagi-Uda balanced feeding is converted to unbalanced with correct impedance, as the whip you have seems to be unbalanced.
Plus, you'd need to ensure the PC also can transmit to the same range No.
If it result in antenna directional **gain** is this **gain** **equal** affecting both TX and RX, so it is no special need to

RF, Microwave, Antennas and Optics :: 03-23-2012 16:58 :: E Kafeman :: Replies: **5** :: Views: **1160**

If you have biased symmetrically and correctly, in-spite I should say, If it is biased symmetrically, you will get **equal** **gain**.
This is ac **gain** and there is no +/- sign for AC input signal.

Analog Circuit Design :: 03-01-2012 09:15 :: varunkant2k :: Replies: **3** :: Views: **1113**

You are correct but if the output is fed straight back to the inverting input (and assuming it is unity **gain** stable) you do not get any a**gain** amplification, you have created a signal buffer.
The virtual ground is a point where the incoming signal and the feedback are **equal** but opposite polarities so they effectively cancel out. You will (...)

Elementary Electronic Questions :: 02-03-2012 09:28 :: betwixt :: Replies: **1** :: Views: **1276**

Dear all,
I have designed fully differential symmetric SC-amplifier. When i load differential output with **equal** capacitive load (say, C1), then differential output is, say, y. But when i load differential output with un**equal** capacitive load (C1,C2), then differential output is y?Δ for the same input signal.
Does un**equal** (...)

Analog Circuit Design :: 01-12-2012 10:32 :: abacus004 :: Replies: **1** :: Views: **629**

If you are allowed to use as much as 3 opamps I would recommend the first of the circuits as shown in your document.
Both integrating stages may (and should) have the same time constant RoCo. This defines the frequency wo=1/RoCo.
The inverting opamp has a **gain** of -1 (two **equal** valued resistors in the kohm range).
Don't use the diodes as shown in th

Analog Circuit Design :: 11-26-2011 13:09 :: LvW :: Replies: **3** :: Views: **526**

Here you have a example of a headphone amplifier that that I designed few years ago.
The supply voltage was **equal** to 3V.

Analog Circuit Design :: 11-16-2011 17:10 :: jony130 :: Replies: **6** :: Views: **658**

Hi, Medium wave and short wave bands will have intermediate frequency stages scince due large bandwidth/frequecny range. the RF amplifier **gain** is not **equal** for all receiving radio chanels at different frequencies due to passive components. so what ever the desired frequency we are receiving we first convert it into IF frequency (intermediate freque

RF, Microwave, Antennas and Optics :: 11-08-2011 00:53 :: murali_dece :: Replies: **5** :: Views: **1232**

MCP601 is wired as a unity **gain** buffer, the output voltage will be **equal** to the input voltage of the positive input.
When PIN4 of the jtag plug is not connected then the positive input of the opamp will have 3.3v from the output of the TSP77030 so the output of the opamp will be 3v3.
When PIN4 is connected to the voltage of the target (it seems to

Analog Circuit Design :: 10-21-2011 06:39 :: alexan_e :: Replies: **4** :: Views: **766**

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