1
\$\begingroup\$

This is regarding an integrator op amp design, the rate at which the input voltage charges the integrator's feedback capacitor is 150V/s. I am trying to justify the minimum feedback capacitor value before the bias current (or whatever else) affects the minimum current too much! This is my understanding so far, the op amp that I am using has a bias current of 30pA.

$$\frac{dV_{\text{out}}}{dt} = \frac{i_B}{-C_f} = \frac{30\text{pA}}{1\text{nF}} = 30\text{mV/s}$$

The maximum output voltage of the integrator is 5V, so using 150V/s and 30mV/s. How can I determine how much the bias current is influencing the the output?

The way that I think of it is 150/30=5V and 30mV/30=1mV, so the bias current causes a drift of 1mV. Is this correct? I imagined it to cause worse errors than this.

The resistor at the input is 230kohm. and the input current is about 120nA but I am not sure if these are factored into the error.

I found online another equation for the error, which is including the resistor.

$$\text{Error} = \text{bias current} \times R$$

Is this factored in as well?

enter image description here

\$\endgroup\$
  • \$\begingroup\$ "resistor at the input is 230V"??? \$\endgroup\$ – Andy aka Apr 9 '18 at 16:54
1
\$\begingroup\$

By Kirchhoff Current Law applied to the x node in your circuit, we have $$ I_f= I_{in}+I_b. \tag{1} \label{1} $$ The bias current \$I_b\$ can be positive i.e. can flow from the inverting input of the OpAmp to the x node or it can be negative if it flows the other way around. Now, you want to calculate the absolute error caused on the rise rate of output voltage \$V_\mathrm{out}\$ by the presence of \$I_b\$: then, $$ \left|\Delta \frac{d V_\mathrm{out}}{d t}\right|_{I_b}=\left|\frac{d V_\mathrm{out}}{d t} - \frac{d V_\mathrm{out}|_{I_b=0}}{d t}\right|=\frac{|I_b|}{C_f} = 30\mathrm{mV/s} \tag{2} \label{2} $$ as you already calculated.

A few notes

  • You say that the slew rate of \$V_\mathrm{out}\$ is \$150\mathrm{V/s}\$ and also \$I_{in}\approx 130\mathrm{nA}\$: this cannot be even if you assume that the bias current \$I_b\$ has its maximum value. Perhaps the cause of this strong discrepancy is the tolerance of the value of \$C_f\$: a 15% tolerance can cause a slew rate error far superior to the one caused by the bias current of the OpAmp.
  • The error formula you report refers to the case of an inverting amplifier, i.e. when \$I_f\$ flows across a feedback resistor: in that case the error on the output voltage is proportional to the bias current through the value of the resistor connected between output and inverting input. In your circuit, a simple inverting integrator, there is not such resistor thus there is not such error.

An addendum

The so called "static errors" in an OpAmp are not limited to the bias currents required by the transistors in its input stages. The imperfect symmetry of these device models as an intrinsic more or less small input voltage which should be applied in order to have a zero output: this is the offset voltage \$V_\mathrm{IO}\$. Said that, the equivalent circuit of the OpAmp A above can be shown to be the following one

enter image description here

We have already accounted for \$I_b\$ (precisely \$I_{b-}\$) explicitly in the contribution to the feedback branch current \$I_f\$ (equation \eqref{1}): does the offset voltage contribute to it? Yes, but in a subtler way: the true value of \$I_{in}\$ is expressible as $$ I_{in}=\frac{V_{in}-V_{IO}}{R_{in}} $$ For a TL084 OpAmp, the typical value of \$V_\mathrm{IO}\$ is \$3\mathrm{mV}\$ up to a maximum of \$15\mathrm{mV}\$ (datasheet here). In our example, \$V_{in}\$ is \$\approx30\mathrm{mV}\$ thus, assuming a typical offset \$V_\mathrm{IO}\approx-3\mathrm{mV}\$ (its sing varies randomly, so it can add or subtract from the input voltage) we have that $$ I_{in}=\frac{30\mathrm{mV}+3\mathrm{mV}}{230\mathrm{k}\Omega}=\frac{33\mathrm{mV}}{230\mathrm{k}\Omega}\cong143\mathrm{nA} $$ The value obtained is very similar to the one the OP obtains by measuring the rise rate of the output voltage \$V_\mathrm{out}\$: this implies that the major contribution to the error on the output is due to \$V_\mathrm{IO}\$ and is precisely given by the following expression $$ \left|\Delta \frac{d V_\mathrm{out}}{d t}\right|_{V_\mathrm{IO}}=\frac{1}{C_f}\left|I_{in}-\frac{V_{in}}{R_{in}}\right|=\frac{|V_\mathrm{IO}|}{R_{in}C_f} = 13\mathrm{V/s} \tag{3} \label{3} $$ which is a far higher contribution than the one given by \$I_b\$ (equation \eqref{2}).

How to reduce the error caused by the offset voltage?

There are two possibilities

  1. Increase ten times \$V_{in}\$ and decrease ten times \$C_f\$: in this way the rise rate is kept constant while the output error caused by \$V_\mathrm{IO}\$ is decreased to one tenth. However, this means reducing the dynamic range of the input, and sometimes this is not possible nor auspicable.
  2. Change the OpAmp A with one with a lower \$V_\mathrm{IO}\$. Perhaps this is a more costly solution: however, OpAmps with a maximum \$V_\mathrm{IO}\$ \$<0.3\mathrm{mV}\$ (like this one) are easily found and not so expensive.
\$\endgroup\$
  • \$\begingroup\$ thank you for the thorough answer, regarding the first note. What do you mean that this cannot be? the input current is from a DAC that is outputting ≈ 30mV. The capacitor has a 5% tolerance if this helps :). the input current was calculated by I = Vout* C * f = 5 * 1n * 25.5 ≈130nA. The slew rate(?) was calculated using the equation I=c dv/dt ..... I/c = dv / dt = 5/36.36m (period of f = 27.5) \$\endgroup\$ – konobyBYnight Apr 9 '18 at 21:50
  • \$\begingroup\$ The formula for the input current \$I_{in}\$ you use is not correct: even assuming \$V_{in}=V_{in}(t)\$, i.e. time varying input voltage, the high gain of the OPAMP A forces the potential of node x to be the same of the non inverting input (the virtual ground concept), i.e. \$\equiv 0\$. Thus the input current is \$I_{in}=V_{in}/R_{in}=30\mathrm{mV}/233\mathrm{k\Omega}≡130\mathrm{nA}\$ and this is correct. However, \$dV\mathrm{out}/dt\approx−I_{in}/C_f=130\mathrm{V/s}\$ so there should be another cause of error. Maybe it's the OPAMP offset voltage \$V_{os}\$: what kind of OPAMP are you using? \$\endgroup\$ – Daniele Tampieri Apr 10 '18 at 6:46
  • \$\begingroup\$ I’m using a TL084(TI), the offset voltage is 5pA, but I imagined that to be almost negligible? So your saying that there be more error representing -Iin? \$\endgroup\$ – konobyBYnight Apr 10 '18 at 8:30
  • \$\begingroup\$ The parameter you have reported is the offset current, not the offset voltage. You can see the offset voltage \$V_\mathrm{IO}\$ value in the data sheet at page 6: for the TL084 it is typically \$5\mathrm{mV}\$ and this value adds or subtracts to the input voltage. However, it seems to me that you need some more information on the so called "static errors" of the OPAMPS: I'll update my answer accordingly in the next days, if you agree. \$\endgroup\$ – Daniele Tampieri Apr 10 '18 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.