20

Inverting the collector current equation: $$i_C = I_Se^{\frac{v_{BE}}{V_T}}$$ yields: $$v_{BE} = V_T\ln{\frac{i_C}{I_S}}$$ For example, let $$V_T = 25mV$$ $$I_S = 1 fA$$ $$I_C = 1mA$$ With these values, find that $$V_{BE} = 0.691V$$ Now, double the collector current and find that $$V_{BE} = 0.708V$$ Increasing the collector current 100% only ...


17

Sample Bipolar Schematic and Behavioral Description Let's look at the LM324. It's a bipolar opamp and it is also a lot easier to follow that some. But it is still fairly representative of the basic ideas related to your question: simulate this circuit – Schematic created using CircuitLab You asked about the diff-pair being either NPN or PNP. In this ...


15

The problem the author is trying to avoid is overloading of the source signal. simulate this circuit – Schematic created using CircuitLab Figure 1. A voltage source with it's source impedance, Rs feeding an inverting amplifier with input impedance Ri. Remember that the inverting input of an inverting amplifier is at virtual ground. Let's say my Vs ...


12

Vbe in a silicon transistor, acts like a silicon diode would. The Forward Voltage Drop, after a certain amount of current is passed, rises sharply. Increasing the current makes a negligible Vf difference at that point. Note that the Vf is different for Germanium Diodes, and Transistors, naturally.


11

The problem with this circuit is that the amplifier's output impedance is asymmetrical. When PWM input is low the transistor is turned off and the capacitor charges through R1 and R3. But when the PWM input is high and the transistor is turned on the capacitor discharges through R3 only. This has two effects. Firstly the voltage at the probe point does not ...


11

As I mentioned in the comments, you'll want two-quadrant drive. Assuming that your "load" is \$R_3\$ and \$C_1\$, which are just a simple RC low-pass filter for the output, then something this simple might be okay for hobby purposes: simulate this circuit – Schematic created using CircuitLab There are a lot of reasons why the above isn't ...


9

The main thing that is almost always mis-stated or mis-understood in all these examples is the fact that the inputs being close or even identical to each other is NOT a effect of the amplifier but is an effect of negative feedback. The amplifier is assumed to be ideal to simplify the discussion. An ideal op-amp without negative feedback will NOT have the ...


8

R5 equals R4 || R6 (resistors in parallel) calculated as $$R_5 = \frac{R_4 \cdot R_6}{R_4 + R_6}$$ and its purpose is to balance the two inputs of the opamp, it cancels the differences caused by the opamp input bias current http://www.ecircuitcenter.com/Circuits/op_ibias/op_ibias.htm


8

Your assumption "output voltage value is same either if I connect input signal into + port or if I connect input signal into - port." is wrong. Your left circuit does not work as a linear amplifier. Iuses positive Feedback. Depending on the input voltage it saturates close to one of the supply voltages (not shown in your diagramms; probably +10V and -10V; ...


8

The OP-07 is a great little device but not suitable for your application. The signal coming from the 555 and LC filter is about 10 kHz and has an amplitude of about 14 volts peak-to-peak and the slow OP-07 just cannot keep up. It's output is slew rate limited to typically 0.3 volts per μs: - And that means a 14 volt change in output amplitude will take ...


7

You can verify it using the common techniques for ideal opamps: 1) Having negative feedback makes the voltages on both terminals equal 2) There is no current flowing into or out of the terminals. So, if \$I\$ is the current flowing through \$R_i\$ and \$R_f\$, $$V_n=V_p$$ $$I=(V_n-V_i)/R_i=(V_o-V_n)/R_f$$ Rearranging: $$V_o=\frac{R_f}{R_i}(V_n-V_i)+V_n=\...


7

In short, every op-amp has a differential amplifier at the input. Thus, every transistor needs a "base" current (input bias current) to flow to work as the amplifier. So for example in the inverting amplifier (when \$V_{IN} = 0V\$), this input bias current will cause a voltage drop across the resistor and this drop will be amplified by the ...


6

The gain of an op-amp is usually very high (in the order of tens or hundreds of thousands at DC at moderate to low frequencies so, if the op-amp output is not end-stopped against the power rails, it's a reasonable assumption to say that the input voltage difference is Vout/100,000 for example. If |Vout| is maybe 10 volts or less, the input voltage ...


6

There are a couple reasons (at least) for choosing a particular configuration: If you need a high input impedance, then you are forced into a non-inverting configuration. This is commonly required in a buffering situation. An inverting configuration has an input impedance equal to the input resistor which may load the source circuit. If you need a summing ...


6

I normally would just rush forward with the obvious voltage divider equation. But I'm going to assume, because you reference it, that you want each voltage source activated one at a time while all other voltage sources are "shorted." It's more busy work. Oh, well. You only have two voltage sources, \$V_\text{IN}\$ and \$V_\text{OUT}\$. So this isn't too ...


6

The thing here is that you are looking (and very likely your simulator) at those two opamps circuits from a 'static' point view. If you actually go through the math, you indeed find that the gain for both circuit (even the positive feedback one) is the same. You start out with: $$V_o = A_{ol}(V^+-V^-) \tag1$$ For the one with positive feedback, you get ...


6

If VI is an ideal voltage source then it will have zero output resistance in which case R1 doesn't have to have a high value. In the real world VI will have an output resistance (RI) which forms a potential divider with R1. Thus if RI has a high value and R1 has a small value most of the signal will be lost across RI. As a rule of thumb, R1 should be at ...


6

The op-amp cannot output a voltage lower than its supply voltage (which is 0V for the V- rail). So the circuit you show will output slightly more than 0V since the op-amp output is sinking current (and it's a RRIO op-amp that works well from 3.3V). In the case of the ancient 741, it cannot work from 3.3V (+/-5V is the minimum recommended), cannot output ...


6

You might find the following illustration helpful (but might not!). simulate this circuit – Schematic created using CircuitLab Figure 1. Reference circuit for Figure 2. Figure 2. Picturing the inverting op-amp as a see-saw with it's pivot point being fixed at "virtual ground". How it works: The gain of the see-saw is -2. That means that if ...


6

No. When \$V_E\$ increases, \$I_E\$ decreases, because \$V_{BE}\$ decreases. Any extra current that flows through \$R_E\$ is supplied by the signal source, not the transistor.


6

Let's take the second circuit first. What you are seeing is a pathological condition. Letting the +in float is asking for trouble, and you're getting it. Just as an exercise, try triggering the scope on the middle of the rising edge of your waveform. When you do this, you'll probably see the measured frequency change to 120 Hz. Whenever you see 120 Hz on a ...


5

This is basically a classic window comparator, with stuff around it to make is actually useful in the particular application. PIR sensors report changes in IR accross the sensor area. C2 removes the DC bias, and the circuit around IC1D amplifies the result and also does some frequency filtering. This is probably in part to reduce frequencies that aren't ...


5

Here's a typical op-amp inverting amplifier: The input impedance is simply \$R_{in}\$, so for your requirements, \$R_{in} = 10k\Omega\$. \$R_f\$ is then whatever it needs to be to realize the desired gain. You want a gain of -10, so: $$ -\frac{R_f}{10k\Omega} = -10 \\ R_f = 10 \cdot 10k\Omega = 100k\Omega $$ Why does the input impedance depend only on \$...


5

DC balance, aka input offset compensation. Some opamps take a significant input current on each input pin - let's say 1 microamp. Work out the input impedance on the -ve input : 2 * 100k in parallel, giving 50K input impedance. Now that microamp input drops 50mv across that resistor, giving an output voltage of 50mv * the gain. R5 compensates for this by ...


5

You cannot analyze the circuit to explain this effect assuming that the op-amp is ideal. The DC open-loop gain of an LM324 amplifier section is typically about 1E5. It has a gain-bandwidth product of 1MHz, meaning the the open-loop gain at 1kHz will be only about 1000. The analysis you need to do will depend on whether your input frequency is a few Hz or ...


5

Try using negative feedback and not positive feedback. The circuit you have shown is nonesense unless you are trying to make a comparator with rather a lot of hysteresis. When using a simulator it might theoretically "settle" on what seems to be an improbable scenario - If the output is -1V and the input is +1V then the voltage at the non-inverting input is ...


5

No operational amplifier can actually achieve an output voltage of the negative power rail, although some can get close. This is due to the finite resistance of the output stage; even at no load, some current must flow in the output stage (although it may be miniscule). If you want to get all the way to 0V, then I suggest reading this excellent application ...


5

Yes - you can combine two opamps with the aim to improve the overall peformance (not only for enlarging the open-loop gain). HOWEVER, as outlined by Andy_aka you must NOT simply combine two "naked" opamps. Instead, the second one must be equipped with an internal negative feedback (reducing gain). In each case, one of the opamp must be inverting and the ...


5

I know you have the answer already but for this type of configuration, the best is to use the Extra-Element Theorem or EET from Dr. Middlebrook (see https://en.wikipedia.org/wiki/Extra_element_theorem). This is truly the best tool which can get you straight to the answer without almost writing a line of algebra. First, you identify the element that bothers ...


5

Stability is a function of NOISE GAIN, not strictly the same thing as gain... Noise gain follows the formula for the gain of a non inverting stage $$NG = 1 + Rf/Rg$$ For an inverting unity gain stage this will be 2, making the part stable in this configuration.


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