Problem

I am having problems with the 2nd question and the 5th question

Q2) Analyze the errors due to using a TLC277CP opamp

AND

Q5) What is the maximum number of bits (for another DAC having the same current O/P
range) which can be accurately used if the circuit is conditioned using an OPA177GP?
(Neglect sources of error other than the opamp)

Attempt at a solution

The general formula for finding the DC imperfection for op amp circuits would be Verror(RTO) = Ib*Rf + Vos*Gain. (and ignore Ios as a shortcut because it is insignificant in comparison to Ib error) But in this case, how do you find the gain when there is only one resistor in the circuit?

For finding the ENOB using the error from the op amp, my lecturer showed us this equation Vfs/2^N > Vos + Ib*Rf

where Vfs is full scale voltage, N is number of bits and right hand side is the DC error due to op amp. After substituting the Vos and Ib values from the datasheet into the eqn, i cannot get the answer she came up with - 15.xx

Here are my workings:

1/2^N > (60*10^-6) + (2800*10^-8)
1/(60*10^-6) + (2800*10^-8) > 2^N
log (1/(60*10^-6) + (2800*10^-8)) > log2^N
log (1/(60*10^-6) + (2800*10^-8)) > Nlog2
log (1/(60*10^-6) + (2800*10^-8)) / log2 > N
13.47(bits) > N

So i tried to find using algebra what was the missing variable that i did not account and derived this.

Verror t < (60*10^-6) + (2800*10^-8)

So the Verror is too big. It needs to be smaller to get a higher value of N.

Using a different datasheet, i used the typical Vos and Ib errors instead of the max Vos and Ib errors and managed to get the values. But i arrived at this answer purely using maths. Could anyone explain why this so to me?

I am not sure what i am missing and would appreciate any hints towards getting there..

 

What is this?

 

thats the datasheet for an OP177GP,

Vos max is 60 μv and Ib max is 2800 picoamps.

I have found the values that satisfy my lecturer's worked example, but it uses the typical values rather than the maximum values. Does anybody know why?

 

thats the datasheet for an OP177GP,

Vos max is 60 μv and Ib max is 2800 picoamps.

I have found the values that satisfy my lecturer's worked example, but it uses the typical values rather than the maximum values. Does anybody know why?

Stupid lecturer?

 

A lecturer that has never designed a circuit to meet worst-case requirements.

 

Sorry folks, but i still have more questions..

Just curious, if you always use the maximum values for Ib, Vos and Ios to design circuits, when will you find it useful to refer to the typical values at all?

Second, we were given the formula to calculate ENOB but i am unsure as to why the quantization error has to be bigger than the op amp errors. Does this mean that the quantization error has to be big enough so as to cancel out the op amp error, but isn't it the case that we don't know the polarity of the error voltage? What if they both add up together to cause a greater error instead?

 

Sorry folks, but i still have more questions..

Just curious, if you always use the maximum values for Ib, Vos and Ios to design circuits, when will you find it useful to refer to the typical values at all?

Second, we were given the formula to calculate ENOB but i am unsure as to why the quantization error has to be bigger than the op amp errors. Does this mean that the quantization error has to be big enough so as to cancel out the op amp error, but isn't it the case that we don't know the polarity of the error voltage? What if they both add up together to cause a greater error instead?

Using typical specs (a bad idea):
If I use Ib=1.2nA, I get
Error=20uV+1.2nA*10k = 32uV
2^N=1/32uV
N=14.14.932

If I use Ios instead of Ib, and Ios(typ)=0.3nA,
Err=20uV=0.3nA*10k = 23uV
2^N=1/23uV
N=15.408
The problem with this is that you only use Ios if there is a 10k resistor in series with the +in pin, which there is not.

In any case, the predominant error is due to the 0.1% feedback resistor. A 0.1% error equates to
2^N=1/0.1% =1/1000
N=9.966

Note that I did not include the resistor tolerance in the errors due to input current, because it is insignificant, and, in this case, irrelevant.