On Apr 19, 7:10=A0pm, bitrex <bit...@de.lete.earthlink.net> wrote:> On 4/18/2013 5:02 PM, Tim Wescott wrote: > > > > > > > On Thu, 18 Apr 2013 16:29:08 -0400, bitrex wrote: > > >> I was looking over the documentation about TIA stability from TI here: > >>http://www.ti.com/lit/an/snoa515/snoa515.pdf > > >> and have a few questions. =A0Am I correct in thinking that an > >> uncompensated TIA where the pole formed by the feedback resistor and > >> input capacitance lies outside the GBW (0 dB open loop gain point) of > >> the opamp will be stable without adding compensation? > > > If you mean "outside" as in "significantly greater than", then yes. =A0=I'm> > not happy with the way that TI is doing their analysis. =A0I find stabi=lity> > analysis of op amps to be much easier if you treat the voltage at the > > minus terminal as being from the voltage divider from v_out to v-. =A0D=o> > that with this circuit, and you find that the voltage at v- is from a > > first-order lowpass filter. > > > Then (and this is why I prefer my method) the phase shift around the lo=op> > is obvious: it's the op-amp's phase shift plus the divider's phase > > shift. =A0If you use their numbers of a 22pF minimum capacitance and a > > 350MHz GBW amp, then the biggest resistance that you can use is somethi=ng> > like 20 ohms -- and at that low of a resistance, you have to start taki=ng> > the op-amp's possible internal resistance into account. > > > When you put a capacitor in parallel with the feedback resistor, then y=ou> > put a zero into the voltage divider response, which allows you more gai=n.> > >> For certain values of the parameters of equations 3 & 4 it seems that > >> nonsensical results are given, such as the break frequency of the I-V > >> gain being greater than the GBW of the opamp, which doesn't seem to be > >> possible for a compensated TIA by looking at the diagram. =A0I'm tryin=g to> >> understand how the assumptions made by the equations are breaking down=.> > >> If my intuition in the first paragraph is correct and the TIA is stabl=e> >> without compensation, how does one then calculate the bandwidth of the > >> I-V gain? Thanks. > > > I do it by going back to 3rd-year circuits and cranking through the > > equations. =A0Find the gain from current to voltage on v-, find the gai=n> > from v_out to v-, let the op-amp gain be (2*pi*GBW)/s, put it all into =a> > feedback equation, and turn the crank. > > > Or cheat, and use SPICE. > > Thanks for the reply. =A0The application I'm interested in is doing some > low resolution monochrome composite video output with a current output > DAC. =A0I'm hoping to get away with using a regular op-amp as the > TIA/driver without using some kind of special video device. =A0For an > opamp like the CA3140 with a GBW ~5 MHz, a feedback resistor of 1k, and > an output capacitance of 15 p the equation gives an I-V bandwidth of > something like 6.7 MHz, which doesn't make sense.- Hide quoted text - > > - Show quoted text -It's Rf and the input C that you have to worry about. I'm not sure where the 6.7 MHz came from, but if the opamp is only 5 MHz, and you want more BW... then it's time for a new opamp... Phil listed a nice one. George H.