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Hello,
I am an analog circuit design engineer in the US working for a reputable company. If I choose to move to india, are there many opportunities to choose from? I know TI, ADI, Maxim, etc are all there. But, does a lot of innovation happen?
What is the avg salary range for an analog/mixed...
GDS viewer
Hello,
Does anyone know of any (freely available) windows based gds viewer?
What is the gds equivalent for a schematic? Is it ediff? If so, are there any such viewers for those as well?
Regards,
Transbrother
If this is a traditional compartor, you can run a DC simulation whereby you tie the negative input to the output and put a DC voltage at the positive terminal. The DC sim will converge easily if there isnt a latch (or hysteresis). If you can get the sim to converge then you can get the offset...
It is not shorted. Current first flows thru the 140 ohms and when the current is ~5mA, it creates a Voltage drop of 0.7V at which time the BJT will turn on.
The circuit ensures that the buffered output doesnt go much below 0.7V of Vout.
Output noise of LDO
tdf,
Yes. you are right about putting the AC source in supply and viewing the output to study PSRR.
For the noise, you need an an AC input source. You can use the noise analysis function in the analog artist window if you are using cadence. Define input and output...
Output noise of LDO
I can answer the PSRR Vs Output Noise part...I am not familiar with noise in HSPICE.
output noise occurs because of the noise present in the active components. E.g. the current and voltage noise (1/f, thermal noise, etc). The PSRR is the rejection of high frequency...
There should be no problem creating an opamp with PMOS input pairs that operate with an input of 0V.
Just make sure that it is a folded cascode type circuit whereby the drain of the PMOS input pair sees a current source (with VDsat across it) or even a resistor. If you use an NMOS current...
Bharathsmile,
it is hard to say in your study whether the startup time should be lower with or without the start up circuit. From what I understand of your findings, analyzing case 1, you are able to get the bandgap started up without the startup ckt. That means that there is some startup...
Bharathsmile2007,
If one cannot bring up the bandgap voltage unless a startup ckt is present then we cannot compare the case of bandgap voltage coming up with and without the startup ckt. Having said that, the startup ckt together with the amplifier and other ckts here that are involved in the...
AFter startup and in normal operation, MS1 has to be in triode region in order to turn MS2 off. If MS1 is in sat, MS2 will not turn off properly. The disadvantage is that current thru MS1 will not match current thru M2 or M3. However, the MS1 current will be lower than Id of M2 or M3. This is...
THe reason that the transient is starting up is probably because you have some leakage current path into V- and V+ starting up the ckt. You could put an ideal current source of 100nA to gnd pulling down on V- and V+ and turn this current off later. That way you'll ensure that V- and V+ is at 0V...
bharathsmile,
answer to where vbg will be different with and without startup:
If the circuit has started up, and vbg comes to the right voltage, taking the startup out of the picture should not change vbg.
On initial conditions:
Ramp up Vdd from gnd. Try different rise times. And, transient...
bharatsmile,
It looks like MS3 is a low W/L device because its vgs needs to be high enough to turn MS2 off.
I dont find anything particularly wrong in the startup circuit except that your condition to turn MS2 off could result in wasting some current. What i do like bout this ckt is that...
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