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. Am 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? 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. I'm trying 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 stable without compensation, how does one then calculate the bandwidth of the I-V gain? Thanks.

# TIA stability

Started by ●April 18, 2013

Reply by ●April 18, 20132013-04-18

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. Am 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. I'm not happy with the way that TI is doing their analysis. I find stability 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-. Do 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 loop is obvious: it's the op-amp's phase shift plus the divider's phase shift. If you use their numbers of a 22pF minimum capacitance and a 350MHz GBW amp, then the biggest resistance that you can use is something like 20 ohms -- and at that low of a resistance, you have to start taking the op-amp's possible internal resistance into account. When you put a capacitor in parallel with the feedback resistor, then you put a zero into the voltage divider response, which allows you more gain.> 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. I'm trying 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 stable > 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. Find the gain from current to voltage on v-, find the gain 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. -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com

Reply by ●April 18, 20132013-04-18

On 04/18/2013 05:02 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. Am 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. I'm > not happy with the way that TI is doing their analysis. I find stability > 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-. Do > 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 loop > is obvious: it's the op-amp's phase shift plus the divider's phase > shift. If you use their numbers of a 22pF minimum capacitance and a > 350MHz GBW amp, then the biggest resistance that you can use is something > like 20 ohms -- and at that low of a resistance, you have to start taking > the op-amp's possible internal resistance into account. > > When you put a capacitor in parallel with the feedback resistor, then you > put a zero into the voltage divider response, which allows you more gain. > >> 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. I'm trying 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 stable >> 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. Find the gain from current to voltage on v-, find the gain > 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. >Yup, algebra rules. And I agree that TI says some really stupid things about TIAs in their apps literature, mostly because they want to sell you uncompensated op amps especially the OPA657. That part is actually a poor choice for a TIA because of its huge input capacitance. Its little brother the OPA656 has much lower Cin, which makes it a much better TIA in general, despite being slower. The argument they use is correct, it just doesn't prove what they want it to prove. They say quite correctly that the capacitive loading on the summing junction makes the op amp run at high noise gain, so that you don't need unity gain compensation, and that's true. But the decompensated op amp they try to sell you with that argument stinks, because of the aforementioned high Cin. Try an ADA4817 for a better fast TIA. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net

Reply by ●April 18, 20132013-04-18

On Thu, 18 Apr 2013 16:29:08 -0400, bitrex <bitrex@de.lete.earthlink.net> 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. Am 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? > >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. I'm trying 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 stable >without compensation, how does one then calculate the bandwidth of the >I-V gain? Thanks.Take a look at Texas Instruments' Application Report SBOA055A. (Download the file and view the pdf file directly. The equations get fouled up when I try to view the file in my Firefox browser.) Eq 2 demonstrates that the zero frequency must occur at a lower frequency than the pole frequency. To ensure stability, the range of frequencies between fz and fp must not cross through the op-amp's open-loop gain curve. See Figure 4. fz is a function of the sum of the diode capacitance, Cd (plus the op-amp input capacitance, plus the board traces) and the feedback capacitance, Cf. Since the diode's capacitance is often large, any decent op-amp's open-loop gain curve will be to the right of fz. So you're generally stuck with ensuring that fp is to the left of the open-loop gain curve.

Reply by ●April 19, 20132013-04-19

On Apr 18, 4:29=A0pm, bitrex <bit...@de.lete.earthlink.net> wrote:> I was looking over the documentation about TIA stability from TI here:htt=p://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? > > 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 trying t=o> understand how the assumptions made by the equations are breaking down. > > If my intuition in the first paragraph is correct and the TIA is stable > without compensation, how does one then calculate the bandwidth of the > I-V gain? Thanks.What Tim and Phil said, That app note is not very good. IIRC adding the compensation cap (Cf) doesn't change the bandwidth, it just stops the ringing and some of the noise. The bandwidth is still the geometric mean of Rf*Cin and the opamp GBW. (Well you have to turn Rf*Cin into a frequency.) George H.

Reply by ●April 19, 20132013-04-19

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. Am 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. I'm > not happy with the way that TI is doing their analysis. I find stability > 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-. Do > 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 loop > is obvious: it's the op-amp's phase shift plus the divider's phase > shift. If you use their numbers of a 22pF minimum capacitance and a > 350MHz GBW amp, then the biggest resistance that you can use is something > like 20 ohms -- and at that low of a resistance, you have to start taking > the op-amp's possible internal resistance into account. > > When you put a capacitor in parallel with the feedback resistor, then you > put a zero into the voltage divider response, which allows you more gain. > >> 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. I'm trying 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 stable >> 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. Find the gain from current to voltage on v-, find the gain > 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. The application I'm interested in is doing some low resolution monochrome composite video output with a current output DAC. I'm hoping to get away with using a regular op-amp as the TIA/driver without using some kind of special video device. For 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.

Reply by ●April 19, 20132013-04-19

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.