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7

The concept of the "op-amp virtual earth" is very important to understanding why the op-amp is used as a mixing/adding circuit: - I've stolen the OP's picture and marked in red where the virtual earth is. Anything directly connected to the -input is called "virtual earth". OK that doesn't explain why so, here goes... If the voltage at the +input is 0V ...


5

There are off-the-shelf solutions to this problem, you don't need to mess with transformers and exotic circuits. You can take advantage of the fact that it's much easier to isolate a digital signal than an analog signal and use serial communications between your microcontroller and the battery monitor. You should search for a "multicell battery stack ...


3

I'm not sure if you'll be left with enough accuracy using this method, but I'd head in this direction first for its simplicity. At each junction point between the batteries, setup a voltage divider to ground where you get 1/16th of the total voltage from the junction. At a maximum of 50.4V this would result in a maximum input voltage to your system of 3.15V. ...


3

It is a 10uF non-polarized electrolytic capacitor. Here's a similar ECG front-end circuit that uses the AD620 as an electrode amplifier (from this site, en Francais): In that case, the designer used a 2,2uF film capacitor with a 2,2M resistor to form a 0.033 Hz high pass filter (\$ f_C = \frac {1}{2\pi RC})\$. As has been pointed out, the symbol is ...


3

Try this circuit with a dual op-amp. It uses the suggestion by @horta to move the voltage away from the rails. U1A produces a stable voltage at +3V. The input is capacitively coupled to that voltage. Since your pulses are -22.5mV, the threshold is set at 3V - 11.25 mV approximately (set by the ratio of R9 to R8 approximately) +/- 2mV of hysteresis (set by ...


3

The phase margin PM is a measure for the stability of a system with feedback. And, thus, it also applies to operational amplifiers. The PM is defined for the LOOP GAIN of the system - that means: open the loop at a suitable node and measure/simulate the gain and the phase around the complete loop. Then ,the PM is the DIFFERENCE between the measured phase and ...


3

I know that Opamp trys to hold the input voltages the same Not quite. The correct statement is "When negative feedback is present, the voltage across the (ideal) op-amp input terminals is zero" why we're not getting a virtual ground on the + node if we connect - node to ground? In the typical configuration, the op-amp output is connected in ...


3

The gain of the op amp is very high, typically 100k to 1M or more. The gain of the circuit in this case is very close to 1. For now, let's imagine a nice ideal opamp with 0 offset voltage, infinite gain, etc. Think of what would happen if the gain weren't 1. Let's say you drive Vin to 3.0 V. The opamp will then drive Vout to also be 3.0 V. Now imagine ...


2

As far as I'm aware the ADC on an arduino requires 0V to 5V as the range of input and because you have a signal that is -1V to +1V, you need to centre it about 2.5V. If the signal is purely AC then that is easily done with a potential divider and capacitor: - Note C1, R1 and R2. C1 blocks the average 0V level of the input signal and R1 and R2 set the ...


2

The most effective filter for receiving the magnetic part of the EM spectrum with a loop antenna is a tuning capacitor right across the input coil. It: - Minimizes unwanted large signals being amplified Improves the signal to noise ratio because you can "peak" the tuning right at the input Works at rejecting above and below your desired 5 MHz operating ...


2

Without an input, pin 3 will find its own level and the output could be anything between 5mV and about 7V from a 9V Vcc. OK that's based on the spec and you say +7.5V - I'm not surprised. When you connected the input to 1.5V, the gain of the op-amp circuit being about 8.7 will still mean the output voltage is end-stopped at 7.5V. With a 741 - you can't use ...


2

There is no guarantee what a opamp will do with one of its inputs floating. Your schematic shows a gain of 2, but your description implies a gain of 8.7. If the gain really is 8.7 (this kind of ambiguity is annoying in a question since we have to waste time dealing with it), then putting in 1.5 V should result in 1.5 V x 8.7 = 13 V. The opamp can't ...


2

As pointed out by Gsills, there is no negative power rail for it to go negative. The LM358 is a pretty old, and for general purposes where you would 'expect' rail-to-rail ( 0V -> VCC) outputs it doesn't cut it. Your comment of using an MCP6002 is a good idea, it will definitely get the job done. Just make sure it's slew rate and in general it's bandwidth ...


2

Omitting the supply rails and decoupling capacitors is a common shorthand in circuit diagrams. It's admissible, when supply rails are not important for the discussion. 0.001 μF in the original post is not a decoupling capacitor. It's a compensation capacitor between pins 1 and 8 of the CA3130 OpAmp. fig.8 on p.9 in the datasheet. Decoupling ...


2

I know analog guys don't like to hear this, but a small microcontroller could easily meet all of your requirements. It would be easy to modify its operational parameters even after the hardware is built, and it would have by far the lowest overall parts count. A built-in ADC could digitize the audio (as well as the voltages from your control knobs), ...


2

The input voltages need to be setup differently. You have one side increase while the other side stays the same. That goes against the entire point of a differential op-amp. You need the input to have V- go down while V+ go up simultaneously. That's why you're getting the weird knee in the output. To make the line ramp steeper, the fastest/easiest way to do ...


2

I'd look into putting a cheap microcontroller accross each battery. There are plenty of micros that can run from 3 to 4.2 V directly, and take very little current. The micro would measure the cell voltage, then send the information digitally over a opto-isolator. The output of all the optos from all the cells would be in parallel, each being able to pull ...


2

I don't really see the big deal about this. Most voltage regulators can be made unstable by one means or another. A perfect op-amp and a perfect darlington might be unconditionally stable (in all circumstances). On the other hand, an imperfect op-amp and imperfect darlington might be just fine for such a device as the 7805 voltage regulator. Take a look at ...


2

Analog Integrated Circuit Design David A. Johns, Ken Martin Copyright 1997 Chapter 5 To really understand op-amps though, you'll want to understand current-mirrors and single stage amplifiers. The book is fantastic in this regard as it takes you through the basics all the way up. This separate link has lots of examples of topologies of op-amps: ...


2

Probably: It's a fixed (approximately) frequency PWM or pulse width ratio modulator with high/low ratio of IC2_out controlled by Vin_AC voltage level. [See below for IC2 defn]. Certain conditions need to be met. See below. Because: lh (=left hand) opamp = IC1 rh opamp = IC2 Set Vin = 0 initially. Assume IC1 output high initially and V_C1 = 0V. IC2 ...


2

\$IC_1\$ is a configured as a comparator, \$IC_2\$ is configured as an inverting integrator. The output of \$IC_1\$, \$v_{OUT1}\$ is either \$V+\$ or \$V-\$. When \$v_{OUT1} = V+\$, the output of \$IC_2\$, \$v_{OUT2}\$ is a decreasing ramp. The voltage at the non-inverting input of \$IC_1\$ is given by $$v_{+,1} = v_{OUT1}\frac{R_4}{R_4 + R_5} + ...


2

First of all, stop thinking of transistors in terms of resistance, think in terms of transconductance, understand it, use it. First order effects: - current density, regime of operation, range of operation. Your W/L ratio and the bias current are the really the only things that you can change in a given topology. The characteristics of the devices' in ...


2

Actually your premise is not quite correct. The overall gain is not 1, it's \$ \frac{1}{1 + \frac{1}{G}}\$, where G is the gain of the op-amp. If G is very high, such as for a precision op-amp near DC, then the gain is very close to 1. But an LM324 (say) at 30kHz has a gain of only about 40, so the gain is closer to 0.975. \$V_{OUT} = G (V_{IN+} - ...


1

ICMRmax is set by keeping M5 in saturation. That implies that the maximum voltage at the drain of M5 is \$VDD - V_{dsSatM5}\$. M2 and M3 are PMOS, so their source will be their most positive terminal and their gate voltage must be less than their source voltage if they are to operate properly. That means: $$ ICMRmax = VDD - V_{dsSatM5} - V_{gsM2} $$ ...


1

The maximum voltage at the non-inverting input of U1 is about 0.15Vcc, so you need only bring the inverting input higher than that to ensure the output is low. Assuming a minimum Vcc of 5V, Vcc-4V is enough. So, you could add a series resistor as in your solution 3, and a 1N4148 from the comparator op-amp output to the inverting input and it should shut ...


1

If you go for shunting R5 down to ground at least the opamp remains in control and not saturated. Then add a load resistor to the output of the opamp to ground so that it does pull all the way down to within 50mV of ground, the leakage problem you mentioned should disappear. Let me know if you try this out and it's still not effective. The load resistor ...


1

The obvious method is to use a differential measurement, but getting acceptable accuracy with that method is quite difficult. A flying capacitor system would be less troublesome potentially, but in this case it would require quite high voltage switches. Individual isolation amplifiers are a bit expensive but will work with a minimum of issues. There's ...


1

why i can't add voltage without opamp? You can! Remove the op-amp and the feedback resistor. The voltage at the node connecting all four resistors is, by voltage division and superposition, $$V_{sum} = V_1 \frac{R_2||R_3||R_4}{R_1 + R_2||R_3||R_4} + V_2 \frac{R_1||R_3||R_4}{R_2 + R_1||R_3||R_4} + V_3 \frac{R_1||R_2||R_4}{R_3 + R_1||R_2||R_4} + V_4 ...


1

It would be wiser to amplify the signal far above the supply noise then use a Schmitt trigger or comparator with positive feedback as required. One solution uses a current sense chip on either 3.3 V rail or 6V rail, whichever is cleaner. http://www.maximintegrated.com/app-notes/index.mvp/id/1180 using a gain of 50 on a MAX4372.


1

I'll assume you want "average reading" equivalent to RMS for a sinusoidal input. If you want "true RMS" you'll have to add a more complex conversion circuit. This is the same principle used in non-"true-RMS" multimeters. The transformer gives you (60A/7000) * 47 = 402.8mV RMS AC full scale (full scale of the transformer) So you need to convert an AC ...



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