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Lab05: Operational Amplifiers 1

 

Inverting Operational Amplifier

Gain of Op-Amp

Build the following circuit in figure 1 with the power supply and a multimeter.

lab05:fig-20_inverting_op-amp.svg
Fig. 1: Inverting Op-Amp

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=100~k\Omega}$

Calculate the necessary value for $R_{\rm 2}$, so that the output $U_{\rm OUT}$ is +1.5 V.
Use the supply voltage of the operational amplifier for $U_{\rm IN}$.


$U_{\rm IN}{\rm~=}$
















$R_{\rm 2}{\rm~=}$


Analysis of inverting input currents


lab05:fig-30_inverting_op-amp_inv_input.svg
Fig. 2: Inverting Op-Amp: Analysis of currents of the inverting input

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=100~k\Omega}$
Use the values from figure 1 for $U_{\rm IN},~U_{\rm OUT},~R_{\rm 2}$.

Complete the reference arrows in the scematic of the circuit.
Determine the the currents $I_{\rm 1}$ and $I_{\rm 2}$ indirectly by measuring the voltage across known resistors
and calculate the algebraic sum of the currents at node $N_{\rm {12}}$ using Kirchhoff’s Current Law (KCL).

$U_{\rm 1}{\rm~=}$


$U_{\rm 2}{\rm~=}$



$I_{\rm 1}{\rm~=}$



$I_{\rm 2}{\rm~=}$



$I_{\rm N12}{\rm~=}$

Analysis of inverting input voltages



lab05:fig-40_inverting_op-amp_inv_input_virt_gnd.svg

Fig. 3: Inverting Op-Amp: Analysis of virtual GND of the inverting input

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=100~k\Omega}$
Use the values from figure 1 for $U_{\rm IN},~U_{\rm OUT},~R_{\rm 2}$.


Complete the reference arrows in the scematic of the circuit.
Take the values for $U_{\rm 1},~U_{\rm 2},~U_{\rm OUT}$ from figure 2.
Calculate the voltage $U_{12}$ using Kirchhoff's Voltage Law (KVL) within one of the possible circuit loops. Mark the chosen loop in the circuit.
Verify your calculated result by measuring $U_{12}$.


$U_{\rm 1}{\rm~=}$


$U_{\rm 2}{\rm~=}$


$U_{\rm IN}{\rm~=}$


$U_{\rm OUT}{\rm~=}$


Calculated $U_{\rm 12}{\rm~=}$


Measured $U_{\rm 12}{\rm~=}$


Analyse the physical significance of the potential at $N_{12}$ relative to GND (defined as $U_{12}$) in the context of the operational amplifier's input configuration. What do you observe?


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What happens if you short-circuit $R_2$ (the feedback resistor)?
Experimentally verify this effect and explain the observed behavior regarding the output voltage.


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Photodiode as current source




A photodiode is a special type of diode which, in the absence of light, exhibits a current-voltage relationship very similar to that of a standard diode (see the dark current characteristic in the \(I-V\) diagram).

When illuminated, it generates additional electron-hole pairs within the crystal.
lab05:fig-150_inverting_op-amp_photo_diode_function_principle.svg

Photodiodes are often operated in reverse bias, where the charge carriers (electrons and holes) generated by the incident light cause an increased reverse current flow (third quadrant of the I-V diagram). The higher the light intensity, the greater the reverse current. Forward bias operation is also possible, where the photodiode behaves like a solar cell (first quadrant of the I-V diagram).

Applications include remote controls (IR range), galvanic isolation (optocouplers), light measurement, positioning, and light barriers.

lab05:fig-160_inverting_op-amp_photo_diode_i_v_diagramm.svg

Fig. 4: Inverting Op-Amp: Operating principle of a photodiode






lab05:fig-120_inverting_op-amp_photo_diode_housing.svg

Fig. 5: Inverting Op-Amp: Photodiode BPW 34 S


lab05:fig-110_inverting_op-amp_photo_diode_diagramms.svg

Fig. 6: Inverting Op-Amp: Diagramms of BPW 34 S












lab05:fig-100_inverting_op-amp_photo_diode.svg

Fig. 7: Inverting Op-Amp: Photo Diode as current source

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V}$


We are assuming a well-lit room with an illuminance of 300 lx, lit by a white LED. White light is a mixture of many wavelengths across the visible spectrum, roughly 380 to 780 nm. For a typical white LED, the spectrum usually comes from a blue LED chip with a peak around 450 nm, plus a broader phosphor emission that spreads across green, yellow, and red wavelengths. For an easier calculation, we take a mean value of 500 nm which is close to the peak value of the blue LED and 300 lx for the illumination. (500 nm is in reality a greenish light and not blue)
The graph in figure 6 shows that the photodiode sensitivity at 500 nm is only 30%. The maximim current (100%) at 300 lx is 30 ${\rm\mu}$A.
We can now estimate the current we would expect from the photodiode at 300 lx:

$I_{\rm 1} = 30 {\rm~\mu A} * 0.3 = 9 {\rm~\mu A}$
$I_{\rm 1} \approx 10 {\rm~\mu A}$

30% of 30 ${\rm\mu A}$ is roughly 10 ${\rm\mu A}$.
We will assume a current of 10 ${\rm\mu A}$ at 300 lx for our calculations.

Complete the arrows in the circuit diagram in figure 7.

Calculate $R_{\rm 2}$ so that $U_{\rm OUT}$ = 5 V at 300 lx. Take a resistor from the E6 series that is as close as possible to the calculated value.
Also enter the values for $I_{\rm 1}$, $I_{\rm 2}$, $U_{\rm 2}$ and $U_{\rm OUT}$.





$I_{\rm 1}{\rm~=}$


$I_{\rm 2}{\rm~=}$


$U_{\rm 2}{\rm~=}$


$U_{\rm OUT}{\rm~=}$




















$R_{\rm 2}{\rm~=}$



What value would you expect for $U_{\rm D}$ in figure 7 and why?


$U_{\rm D}{\rm~=}$


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What value would you expect for $U_{\rm D}$ at 300 lx when the photodiode is not connected to the Op-Amp or any other electronic component (open-circuit voltage) and why?


$U_{\rm D}{\rm~=}$


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Measure or calculate the values given in the table below.

lab05:table-1_inverting_op-amp_photo_diode.svg

Tab. 1: Photodiode measured and calculated values














Non-inverting Operational Amplifier


Op-Amp as current source




An Op-Amp can not only amplify voltages and currents, it can also act as a current source itself. Here is the schematic of a typical Op-Amp current source:


lab05:fig-250_non-inverting_op-amp_current_source.svg

Fig. 8: Non-inverting Op-Amp: current source

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=100~k\Omega},~R_{\rm 2}{\rm~=10~k\Omega},~R_{\rm 3}{\rm~=100~\Omega},~C_{\rm 1}{\rm~=100~nF},~C_{\rm 2}{\rm~=100~nF}$





Measure the values given in the table below.

lab05:table-20_non-inverting_op-amp_current_source.svg

Tab. 2: Op-Amp as current source: measured and calculated values




Why does the current remain constant at the output of the Op-Amp?
Give a brief explanation of the circuit's operating principle.


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Op-Amp as non-inverting low-pass filter




We can also use an operational amplifier as an active filter. There are two typical circuit configurations: the Sallen-Key circuit, which uses a non-inverting operational amplifier, and the multiple-feedback circuit, which uses an inverting operational amplifier.
We use the Sallen-Key circuit as it is easier to understand and the gain of the operational amplifier can be determined independently of the cut-off frequency.
The cut-off frequency can be calculated using the following formula:


$$f_{\rm c} = \frac{1}{2\pi\sqrt{R_{\rm 1}~\cdot~C_{\rm 1}~\cdot~R_{\rm 2}~\cdot~C_{\rm 2}}}$$

Here is an scematic of an typical low-pass filter using the Sallen-Key circuit:


lab05:fig-350_non-inverting_op-amp_low-pass_filter.svg

Fig. 9: Non-inverting Op-Amp: Sallen-Key Low-pass filter

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=10~k\Omega,~C_{\rm 1}{\rm~=10~nF}},~R_{\rm 2}{\rm~=15~k\Omega},~C_{\rm 2}{\rm~=4,7~nF}$



Complete the schematic and calcutlate the cut-off frequency of the given circuit.




















$f_{\rm c}{\rm~=}$



What gain do you expect from the filter circuit, and why?


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Measure the values given in the table below.
Use the function generator with a sine wave of 6 V amplitude.
Plot the filter response on the logarithmic graph paper provided.


lab05:table-30_non-inverting_op-amp_low-pass_filter.svg

Tab. 3: Op-Amp as current source: Frequency Response Data




Golden Rules

Explain the Golden Rules of the operational amplifier.


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Op-Amp zero adjustment without feedback

Build the following circuit in figure 10 with the power supply and a multimeter.

lab05_en:fig-10_op-amp_no_feedback.svg

Fig. 10: Zero adjustment

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=100~k\Omega},~R_{\rm 2}{\rm~=10~k\Omega}$

Adjust the output of the Op-Amp to zero and describe your results.


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How can the circuit be improved?
Draw an improved version and explain it.





























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