This idealised version would be easy for us to simulate without doing any complicated calculation; simply clip at the saturation thresholds. In reality, an op-amp has a slight curve as it nears saturation, and it becomes non-linear, similar to the below which is a little exaggerated :.
A slightly more realistic graph of Op-Amp saturation. It has a fairly lacklustre performance characteristic as written; there is almost no linear region. However, I noticed that if I generalised the formula ie to the formula below , then as n increased, the characteristic became more like the idealised op-amp saturation characteristic.
If we apply the same input to this new circuit, we now get one transition of the output, because of the changing threshold voltage. When the output is high, the threshold is high -- when the output is low, the threshold is low.
Note that the output now only goes up to 2. Make sure you understand how this circuit works. It is a bit tricky both because the circuit is non-linear, and because the value of the output is not a single-valued function of the input. This is manifested by the fact that the output can either be high 2.
Hook up a comparator as shown in the figure below, and drive it with a 2 V peak to peak triangle wave centered at 2.
If your input goes outside these bounds, the circuit may not behave as expected. You will need to use the DC offset on the signal generator, and make sure that the oscilloscope is DC coupled. If you are not sure what this means, please ask me. Predict, then measure, the output waveform.
Though not shown on this diagram, pin 1 should be connected to ground. Repeat for the Schmitt trigger circuit shown below. Get a printout of your results. Notice that the output of the Schmitt trigger remains constant if the input stays below or above some critical threshold. You can demonstrate this by decreasing the amplitude of the input. Why does this occur? Rewire the circuit to be an oscillator as shown below from the LM datasheet , figure 44, page Note that this circuit has no input.
Use a LM even though the circuit calls for a LM Convince yourself that this circuit will oscillate assume the output is constant - either high or low - and show that the input must change. Predict and measure the frequency of operation. In particular, predict and then show the output voltage as well as the voltage at the positive and negative node of the comparator. Before leaving the lab, make sure you put away all equipment and electrical components.
Although this takes a few minutes, it is much easier to do than trying to sort through a huge pile of components at the end of the semester. There is a meter to read capacitor values if you have find that difficult see page on reading resistor and capacitor values.
Present a carefully considered response to each of the following numbered parts. Make sure the individual responses are clearly labeled. There need not be a lot of prose i. Comparator Schmitt Trigger. The voltage that you apply to the diode will result in a current according to this curve. Similarly a current that you apply would result in a voltage.
The diode is still limited by a forward voltage and a breakdown reverse voltage. In your circuit a reverse voltage of 5V is applied in normal operation and this could increase up to 10V if Rf and the opamp saturation voltage allow it.
However it should not be more than 5V because in the dark the diode behaves as another diode an the opamp feedback will function and the '-' input will settle to 0V. When the Rf resistor will be limiting the current because of saturation, the diode will be in forward bias due to the internal current. So the anode is higher than the kathode and it would be at about 6 V depends on the forward diode voltage. All the current that the diode can not deliver is internally consumed which results in a heat up of the diode.
It will not heat more than any other similar object that you would subject to the same light intensity. It will heat slightly less as part of the energy is consumed by Rf. The current flowing through the diode to the circuit is a reverse current from the diode's point of view.
From the datasheet, we can see that when the current through the photodiode is 0, but when there is enough light, then the diode voltage is at the forward saturation voltage. The current becomes more negative if the forward voltage of the diode gets lower. The photodiode will absorb this excess current internally and the diode voltage will be in forward bias.
Thanks to the negative feedback the '-' input voltage level will also decrease until the opamp saturates. The opamp can not lower the output voltage far enough to get 0V on the '-' input. The '-' input will be well above 0V, and even at about 5. This is coherent with the equivalent photodiode model which predicts that the excess current is essentially consumed by an internal equivalent "normal" diode.
I hope that the mystery is resolved. It is not "easy" to understand at first sight, you surely have to work on this explication a bit.
It won't be mAmps that will be flowing through the diode. When the opamp is saturated, you can just consider it as its grounded output and leave out the inverting input. There will be 5 mAmps flowing across the diode, the resistor and the low-side output transistor of the opamp.
Though the diode would let more current to go through, it gets limited by the voltage of the diode supply and the value of the resistor. There are two current paths for a hypothetical mA photocurrent; one is through the feedback resistor Rf, and the other is through the op amp input pin. That's important, because there are protection diodes, from each input pin to the power rails, and those will conduct at circa 1V; if they conduct more than 1 mA, you have exceeded the maximum allowed input current, and can expect chip failure.
The op amp will not only saturate, it will burn up. A current limiting resistor in series with the op amp input would be a good start, and a feedback resistor that WILL handle all the expected current, even in high-input cases, would be recommended. Keep inside the data sheet 'absolute maximum' range for applied power at the power pins, AND current at the input pins. Sign up to join this community.
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