# Tag Info

14

You seemed to have actually found a reasonable circuit on the internet. I heard there was out there somewhere. The equations you cite are overly strict. Instead of just telling you the values, it's better to explain what each part does. R1 and R2 are a voltage divider to make 1/2 the supply voltage. This will be the DC bias the opamp will operate at. ...

8

If you have an ideal OpAmp the 10k resistor would carry no current and could be replace by a resistor with an arbitrary value, including a piece of wire. Therefore the answer is no, the 10k resistor plays no role wrt. the low pass filter.

7

The opamps work with single supply only. That means: To allow the output to swing above and below the quiescent value this quiescent dc value at the output must be somewhere in the middle between the limits (set by ground potential and Vcc). Hence, the input is biased with 50% of Vcc. This dc voltage is transferred to the output of the opamps because the ...

6

The input signal is present on the node at the right of R3, so R3 must be duplicated for each channel (otherwise you would be joining all the inputs together!). R1, R2 and C2 can be common to all channels. There are two possible issues with using a common voltage divider - bias current, and crosstalk. OP295 op-amps have a maximum bias current of 20nA ...

6

You've got it 800 degrees out of phase even at 100Hz where your amplitude is at 0 db. This is going to cause relative distortion between different frequencies of the music you play because your higher frequencies will be going through a different filter will likely less phase shift. The distortion may be less noticable because it's at the low spectrum. It ...

6

As you said, $$V_a = V_b = V_{in}$$ As $V_{in}=V_a$, there is no current through $R_1$. Since there is no current flowing into/from the negative terminal as well, we conclude that there is no current through $R_2$ as well. So if there is no current through $R_2$, the voltage on both sides of it is equal, and it is $V_a$ which is $V_{in}$. So ...

6

brhans got it already quite right: The TL082 is unsuitable for your problem, neither it supports input voltages close to the negative supply pin, nor it is able to output voltages near the negative supply. The TL082 is meant to be supplied with a negative voltage way below any signal voltage that occurs in your circuit. Typically, op-amps like that are ...

6

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: ...

5

The ultrasonic transducer is a capacitive element. This means it passes no DC current and has no facility to set the DC operating point of the circuit. R1 and R2 set the DC operating point. If you had a split supply op-amp configuration (+ and - 4.5 volt supplies) because the transducer is capacitive you would need a resistor from input to 0V to set the DC ...

4

You have a small mistake in the KCL equation at $V_b$: the denominator on the LHS is $R_1$, not $R$. This means $$V_b = \frac{R_2}{R_1 + R_2}V_1 + \frac{R_1}{R_1 + R_2}V_2$$ To prove this is a non-inverting summing amplifier, you need the relationship between the inputs and output. You know that $V_a = V_b$ so substitute the RHS of the above ...

4

As others have said, your load resistor should go to ground, not to the inverting input. Looks like you have fixed the power supply issues as of this post. The resistor ratio is correct, however your resistor values are WAY too low. The 200 ohm Rf (in your current schematic) goes to a virtual ground, so it's effectively in parallel with the load resistance ...

4

Redrawing the schematic for convenience: Since Z1 offers no DC path to ground, the voltage at the junction of R1 and R2 will be: $$E2 = \frac {E1 \times R2}{R1+R2} = \frac{9V\times 1M \Omega}{1M\Omega+ 1M \Omega}= 4.5V.$$ Then, since an opamp with negative feedback around it will cause its - input to follow its + input, and since R3R4C1 offers no ...

4

There is no need to make R4 that particular value since this is a CMOS op-amp (no matching of input bias currents). Capacitor values are determined by the desired lower corner frequency. C = $\frac{1}{2\pi f_c R}$ So if fc = 20Hz, and R1, R2 = 39K. Let's arbitrarily make R3 100K. Then C = 100nF is about right. C2 depends on what's on the power supply ...

4

The resistor limits the amount of current that the op-amp has to / can sink or source, ensuring that the output transistor/s are never driven into saturation even when the op-amp is charging a completely empty capacitor (which briefly resembles a dead short to ground, which will almost certainly cause saturation). A 'pole' is the frequency at which a filter ...

4

There are a billion such tools (well, that's hyperbolic, but there are many). They may not use the exact topology you have there, but they'll all get the job done. TI and ANALOG DEVICES are just two chip makers that provide good design tools online, and Microchip has a downloadable program Note that when you cascade two two-pole filters, your -3dB point ...

4

If you use this circuit on a regular basis I guess it would be worth to calculate its transfer function once and for all, so you can just easily evaluate it given the concrete component values. Doing the math you get (assuming an ideal OP) $$H(\omega)=\frac{j\omega R_1C_1}{1+j\omega R_1C_1}\left(1+\frac{j\omega R_4C_2}{1+j\omega R_2C_2}\right)\tag{1}$$ A ...

4

You can make a high-order low-pass filter with no op-amps at all. In principle you could obtain any roll-off rate you like, given enough pi or T sections. You could then add an op-amp output buffer to make your filter a "one op-amp" filter if for some reason you'd like to do that. But high-order filters tend to be very sensitive to small changes in ...

3

The problem with your circuit is that the load resistor is mismatched to the resistance range of the sensor you care about. The sensor in series with a resistor gives you a voltage that varies as a function of the full sensor range, from 0 to ∞. However, the resulution is highest in the middle of the range, which is when the sensor and the load ...

3

Yes. However, modern op-amps have VERY high open-loop gain, which makes the error (usually) acceptably small. The caveat to this is that as your operating frequency goes up, the open-loop gain goes down. A better way to express this is: an op-amp will behave ideally in closed-loop configuration so long as the open loop gain is >> than the closed-loop gain ...

3

I can resolder the pot and the error disappears only temporarily. I'm a sound engineer as well as a technician and I see that problem quite regularly. The carbon part of the control has a fine crack in it at the end where the terminal connects. When you apply heat to the control pins the carbon temporarily bridges the gap from the heat. There really is no ...

3

Here is what the inside of a different type of pot looks like: Your pot will have similar construction inside to this one, though it will be hidden by the molding. There is a metal part that connects to the circuit board. In the photo it's held by a rivet or similar crimped construction to the substrate, which has a silver conductive printed portion ...

3

The output calculates to -12V. Thevinize the - input which is at 0 volts, assuming the Av of the opamp is large. That turns the junction point "B" into 6V as "C" is almost 0.00V. Wow! Cut and paste doesn't work here. No MathCAD for you! 6K*4K/(6K+4K)=2.4K resistance to ground at "B" Thevinizing: 2.4K*11V/(2K+2.4K)= 6V with 6K series resistance (Point ...

3

I am really carious to know when can we add up and when we cannot? When two two-terminal elements are in series, the voltage across the combination is equal to the sum of the voltages of the branches. When two two-terminal elements are in parallel, the total current through the combination is equal to the sum of the currents of the two branches. ...

3

Simple triangle-wave generator using two op-amps. One op-amp is configured as a Schmitt trigger, other op-amp is an integrator. You mention 5V, so I'll include the bias network needed for single-supply operation. Note that you have to use op-amps with rail-to-rail inputs AND outputs. simulate this circuit – Schematic created using CircuitLab ...

3

Your average op amp might look something like this: simulate this circuit – Schematic created using CircuitLab That's a very basic two stage op amp: the first stage is built with M1-2-3-4 and is a differential stage, while the second stage is built with M5 and is a sort of common source stage. M6-7-8 just bias the whole thing. Looking at M3-M4 ...

3

The op-amp can't do anything useful in this configuration. You need to give it more supply voltage than the desired output voltage, even though it's "rail to rail". If you ask it to output 8V or even 11V with a 12V supply, it may be able to do that, provided the load is not too much for the op-amp to supply. In this case, that might be 10mA or so. I ...

3

Precision rectifiers at high frequency are surprisingly hard, as you are finding out... The tricky bit is, what does the amplifier do when the diode is non-conducting? In the second circuit, we can infer what's happening from the zoomed-in waveform. When the diode is off (V1 < Vout) the opamp Vin+ input is below Vin- and there is no NFB, thus the opamp ...

3

While I do understand the reason for limiting the bandwidth, I've seen a lot of schematics where there is no input filter. The inherent noise from an op-amp, as a first approximation (e$_{NOISE}$), cannot be reduced by using an input filter. It can be reduced by an output filter or filtering in the feedback loop to reduce the noise bandwidth of the ...

3

You are close to having a working solution. I attempted to edit your schematic but I don't see a edit link below the schematic in the preview window. So I'll draw a new one here. simulate this circuit – Schematic created using CircuitLab R5 & C2 filter noise from your 5V rail. You may need to increase the value of C2 if you are getting ...

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