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Bandgap Emitter

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16 Threads found on edaboard.com: Bandgap Emitter
I would use an isolated resistor (or one capable of well over 50V) and a few diode stack referred to GND, with an emitter or source follower (high voltage device or stack) making "just enough" working voltage for the bandgap. This will give you first order DC and HF PSRR on the bandgap supply although its temperature effects will be (...)
The bandgap circuit seems a little strange to me. I am not shure is that because I miss the basic function or I am stick with standard safe design procedures for bandgaps. The strange points I found: 1. The inherent kT-loop of the bandgap have a current loop gain defined by the emitter area ratio of the npns. That is (...)
Hello, Can someone provide me information about high temperature leakage compensation in a simple bandgap reference ? I will appreciate any diagrams or technical notes for it. Regards.
The bandgap voltage spread is defined by the offset voltage spread of the control amplifier and the current mismatch of T4/T5. The used bipolars could have an impact depending on the amount of voltage drop of the internal emitter resistance. That resistor re*(1-1/12) have to be compared to the 200 ohm resistor and does match in the process. It i
A voltage to current converter with an internal resistor and a ptat current sink (from the bandgap) in parallel. Then use final part trimming to center the current at room temp.
bandgap voltage spread depend on resistor spread for the kT-generator and IS spread of the bipolar. Both depend on process control. Vbdg=b*VT+Vbe=b*VT+VT*ln(VT*ln(a)/(R*IS)) a: emitter area of kT-generator b: kT-multiplier R: resistor kT-generator IS: saturation current Vbdg=f(R*IS) So the relative spread is Vbdg,spread,rel=26e-3*
hi,I have one question:how to determine the bjt current in bandgap reference?I mean the current in PTAT Have somebody the experience how to choose the collector current of PNP? and how's the mismatch? Thanks
In CMOS a N+/NWELL/N+ structure ist used. The layout is a N+ diffusion, the emitter, surounded by either a Poly-Gate or LOCOS-Isolation and then a N+-Ring, the collector. The base is the NWELL. The substrate is an additional terminal. So it is a 4-terminal structure. Most connect them only as diode but there are other bandgap circuits which use the
actually all the bandgaps need base-current cancellation because of the finite gm or beta of more the beta the less cancellation you have to do. in normal cmos process the parasitic bjt has very poor beta .. but irony is that because of the grounded base we cant do base current cancellation very easily.
tsmc/ umc or other Fab will provide 5x5 or 10x10um (emitter area) BJT gds layout . but you should know in CMOS no twin/triple_well , only PNP bjt be use beta is small , only use for diode or bandgap cell some high Volt cmos process provide "really Npn or PNP" device . by the way , parastic BJT spice model is simple even corner
Hi, I understand 1:8 layout. 888 818 888 However, how do the best layout look like according to E, C, B (let's say this is PNP and my Collector is connected to Base). Should I make it like the array with all the emitter face to one side? Or should I flip them so that emitter always face to outside world? Anyway, it doesn't matter how my lay
Genneraly, the two bipolar PNP follwer is formed to deduce the effect of OPAMP offset to the output of bandgap. But the current gain β of local PNP(in cmos process) is relatively small, therefore, the first PNP base current is inserting the second PNP emitter, so this causes the offset of output of bandgap. Maybe there is some (...)
i want to study bandgap reference source.But if i use 0.5um CMOS processing how can i set BJT parameter in HSPICE or PSPICE?Is there any website just like MOSIS providing BICMOS fabrication parameter for HSPICE or PSPICE?
I can not find this kind of bandgap architecture in books or papers. Who can tell me how it works. Thanks.
The main circuit is the PTAT core. It is enclosed into a regulation loop which try to equalize the currents I1 and I2. The resulting current IPTAT=I1=I2 is the PTAT current. The equations for the PTAT circuit is I1*R+VT*ln(I1/(a*IS))=VT*ln(I2/IS), "a" emitter area ratio the solution for I1=I2 is IPTAT=I1=I2=VT*ln(a)/R Issue1: I
You could look at my tutorial on bandgaps at: cheers Ody :)