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Help on opamp design (Read 4893 times)
Vabzter
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Re: Help on opamp design
Reply #15 - Feb 19th, 2007, 2:45am
 
robfox68 wrote on Feb 18th, 2007, 7:12am:
Do a fundamental calculation before you start. The bandwidth of any amplifier cannot exceed gm/CL. If all of your 50 uA bias current was used in a BJT, the transconductance would be Ic/Vth = 2 mA/V at room temperature. With CL = 35 pF, this gives 60 Mr/s, less than 10 MHz. The transconductance of a FET is always lower than that of a BJT for the same bias current.

Your specifications are not realizable.

Hi Robfox68,
                   Thats a good check go a feasible design..Thanks a lot for this tip. I guess I have to decrease the bandwidth.
BR
Vabzter
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RobG
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Re: Help on opamp design
Reply #16 - Feb 19th, 2007, 11:47am
 
Vabzter,
I agree with the other posters -- it will be difficult to achieve that bandwidth with that load cap and total current.  

Here is how you can quickly see that by simply knowing that gm≈2*I/Vod, where Vod=Vgs-Vt if we are in strong inversion.  (In weak inversion gm≈ 2*I/(100mV); that is, you can't invcrease gm by making Vgs-Vt < 100mV).  

We know for a single stage amplifier (folded cascode, telescopic, etc), the bandwidth is gm/(2*pi*C) (where gm is 2*I/Vod of one side of the diff pair).  In this case, the required gm is (2*pi*10MHz)*35pF = 2.2mA/V.  Thus, we need a Vod=2*25uA/(2.2mA/V) = 23mV.  This is impossible - remember Vod limits to about 100mV when the devices get into subthreshold.

A more reasonable assumption of Vod=150mV limits your bandwidth to (2*25uA)/(2*pi*35pF*0.15) = 1.5MHz.  

Note I've assumed nothing about process.... just that gm ≈2*I/Vod, where Vod = Vgs-Vt in strong inversion, and 100mV in weak inversion.  

You can do a similar analysis for a two stage (e.g miller compensated) and show that your non-dominant pole will be around gm/Cload, so even if you can get the bandwidth, you won't have phase margin unless you put a zero in the system - which will create doublets (read Kameth, Grey, Meyer, "Relationship between frequency response and settling time of operation amplifiers," JSSC, Dec 1974).  While you can cheat a little here... you need to gain a factor of 10 in bandwidth.... I don't think you can do it and drive that load.

Rob

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Vabzter
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Re: Help on opamp design
Reply #17 - Feb 20th, 2007, 1:19am
 
RobG wrote on Feb 19th, 2007, 11:47am:
Vabzter,
I agree with the other posters -- it will be difficult to achieve that bandwidth with that load cap and total current.  

Here is how you can quickly see that by simply knowing that gm≈2*I/Vod, where Vod=Vgs-Vt if we are in strong inversion.  (In weak inversion gm≈ 2*I/(100mV); that is, you can't invcrease gm by making Vgs-Vt < 100mV).  

We know for a single stage amplifier (folded cascode, telescopic, etc), the bandwidth is gm/(2*pi*C) (where gm is 2*I/Vod of one side of the diff pair).  In this case, the required gm is (2*pi*10MHz)*35pF = 2.2mA/V.  Thus, we need a Vod=2*25uA/(2.2mA/V) = 23mV.  This is impossible - remember Vod limits to about 100mV when the devices get into subthreshold.

A more reasonable assumption of Vod=150mV limits your bandwidth to (2*25uA)/(2*pi*35pF*0.15) = 1.5MHz.  

Note I've assumed nothing about process.... just that gm ≈2*I/Vod, where Vod = Vgs-Vt in strong inversion, and 100mV in weak inversion.  

You can do a similar analysis for a two stage (e.g miller compensated) and show that your non-dominant pole will be around gm/Cload, so even if you can get the bandwidth, you won't have phase margin unless you put a zero in the system - which will create doublets (read Kameth, Grey, Meyer, "Relationship between frequency response and settling time of operation amplifiers," JSSC, Dec 1974).  While you can cheat a little here... you need to gain a factor of 10 in bandwidth.... I don't think you can do it and drive that load.

Rob


Hi Rob,
            Thanks for giving a simplified explaination..Well so I have formed new specs for the opamp.
Bw=1MHz,
Vout =1.7V,
Vdd=1.8V,
Idc(max)=50uA,
Cl=35pF,Rl=5Kohm,

So with these specs I tried to design a simple 2 stage opamp. I am getting an open loop gain of 40dB(very less). The phase margin is 80•.

My application is it should amplify the voltage at ouput of DAC whose full scale output is 1.2V.

So I am not getting whether I should use a 2 stage simple opamp of Cascode one..What is the criteria for selecting the topology?

Thanks a lot for your help,
BR
Vabzter
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MTXamp
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Re: Help on opamp design
Reply #18 - Feb 20th, 2007, 5:59am
 
Hi Vabzter

Imho, your spec still seems quite tough to me. Since your Vdd is 1.8V, if I assume that your load resistor of 5K is going to be tied to 1/2Vdd of 0.9V...in order for the voltage to rise to 1.7V under DC conditions, your output stage would have to source out at least (1.7-0.9)/5K=160uA. Is this ok for you? This has not even taken into account the current that is going to flow in your feedback path, assuming that you are going to use a non-inverting configuration. With the 1MHz added in, you will most probably run into slew rate problem.

Although your BW has now been reduced to 1MHz, your load res and cap will still pose a big problem. I feel that 50uA is not going to get you through the typical and corner conditions.

A simple miller's opamp will most probably not be able to get the job done. If you were to use the classAB output as discussed in Hujsing's book, you will need more than 50uA at the output stage in order to stabilize the circuit.

My 2 cents worth.  ;)

Best regards
Quek
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RobG
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Re: Help on opamp design
Reply #19 - Feb 20th, 2007, 11:11am
 
Vabzter wrote on Feb 20th, 2007, 1:19am:
Hi Rob,
            Thanks for giving a simplified explaination..Well so I have formed new specs for the opamp.
Bw=1MHz,
Vout =1.7V,
Vdd=1.8V,
Idc(max)=50uA,
Cl=35pF,Rl=5Kohm,

So with these specs I tried to design a simple 2 stage opamp. I am getting an open loop gain of 40dB(very less). The phase margin is 80•.

My application is it should amplify the voltage at ouput of DAC whose full scale output is 1.2V.

So I am not getting whether I should use a 2 stage simple opamp of Cascode one..What is the criteria for selecting the topology?


As MTXamp stated, this is still a tough design.  The resistive load changes everything: you need an output stage to supply the DC current, so you are stuck with at least two stages.  It is impossible to say which topology will work best for you -- op-amp design is a bit of an art, not a recipe.  But I'd start with a Miller two stage amplifier with an NMOS diff-pair and PMOS output stage.  If the 5k-ohm is fixed, it will act to bias the PMOS output stage nicely.  Once you understand the Miller design, you may want to try other designs, such as using a folded cascode input stage, and a PMOS output stage.  

rg

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RobG
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Re: Help on opamp design
Reply #20 - Feb 20th, 2007, 11:14am
 
In addition to the resistive load, the output voltage being only 100mV from the power supply will be very challenging - you will get very little gain from your output stage.
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eng
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Re: Help on opamp design
Reply #21 - Apr 3rd, 2007, 1:38pm
 
Hi all,
I'm working on a similar circuit(fully diff). The replies were very useful for me. Thank you so much.
I have a question about the testbench of opamp. I see in some of the test benches that the bottom plate of load capacitor is connected to common mode voltage. (i.e. if vdd=1.8 and vss=0 and CM=.5 ; botton plate of load is connected to CM)

What is the reasoning behind it?


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Re: Help on opamp design
Reply #22 - Apr 6th, 2007, 1:52am
 
Hi, eng,

I think that configuration doesn't affect the function of the opamp. It get the same result as that the bottom plate is tied to gnd if you run ac simulation. Normally the fully differential amp has cmfb so that it doesn't matter where the bottom plate is tied to.


Yawei
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Re: Help on opamp design
Reply #23 - Apr 10th, 2007, 1:14am
 
If you connect the bottom plate of the load capacitor to the CM voltage there can be a bootstrapping effect. Is it connected to a voltage source,
or is it driven by a VCVS, so it follows the output of the common-mode feedback circuit ?
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eng
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Re: Help on opamp design
Reply #24 - Apr 12th, 2007, 10:29am
 
Here is the thing;
Since the purpose of simulation is to mimic the real environment, the output stage should be able to sink the same amount of the current from the load as well as it sources to it. When I looked at old opamp circuits the loads are grounded but these circuits have vdd and vss (=-vdd) and VoutCM is zero. The question is with a single supply should we connect them to Vdd/2 or CM(assuming they're different)?

imd1: Using VCVS seems a good idea but do you think even if we use fix CM bootstrapping actually happens and it affects the CMout?
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Re: Help on opamp design
Reply #25 - Apr 23rd, 2007, 10:02am
 
well done Grin
RobG wrote on Feb 8th, 2007, 10:29am:
Vabzter wrote on Feb 8th, 2007, 7:23am:
Hi,
     Thanks for the replies..But still I am not sure how to find the values of k(=un*Cox) for a perticular technology as we need these values for doing hand calculations..
BR
Vabzter



I don't worry about those things...  You can do a lot by simply knowing that gm=2*I/Vod, where Vod=(Vgs-Vt).   You don't need to know what K' or Vt is to tune Vgs-Vt.... if it is too big, make the devices wider or lengths shorter.  You can get Vgs-Vt from the DC operating point information.  Vdsat is close to the same quantity -- it is what I usually look at.  Hand calculations that use more than Gm have limited use in the design process.... everything interesting is determined by parasitics or nonideal effects (plus Vt and K' change depending on operating point, sizes, etc).

Opamp design is pretty simple (ok, that is a gross exageration)....  You give the diff pair high gm, and the load (current mirror) low gm.  If you don't, you get a bad op-amp.  End of story Wink.

For a given current, all you can do is change Vod.  150mV is a reasonable target for Vod for the diff pair.  Lower than that, the diff pair gets into weak inversion and wider widths offer only small improvements (i.e. diminishing returns).  2-400mV is a good target for the mirrors.  Since Vod =~ Vdsat, the upper limit is usually set by the amount of headroom you have.  

From that basic starting point you can play with the sizes to suit your particular situation (keeping current fixed and lengths as small as you can).  If your diff pair is causing frequency response problems you need to make it smaller, which will likely decrease gm (because Vod is increased).  Similarly, making the mirrors smaller (increasing Vod) will increase their frequency response (lower Cgs), but it may make other problems - like headroom or an inability to drive cap loads.  If you cannot fix things by changing Vod, your only other choice is to increase current (or change topologies).  

That said, the first thing I do when I get a new process is to plot the Vt vs. Width for various lengths of devices.  You get the Vt information from the DC operating point (just diode connect the devices and put 1uA thru them and do a parametric anlysis sweeping width and length).  You will find that Vt is a strong function of length and width for smaller geometries.  It is also instructive to plot gm/I vs Vgs-Vt for various sized devices, as well as Vt vs bulk-source voltage.  Armed with a few of these plots you can get most of the information you need.  Note that, in some sense, this is determining K' and other parameters, but since they are functinos of device sizes it put them in a more usable format.

rg

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