# Tag Info

9

Step by step: The current, from left to right, through $R$ is $$I_R = \frac{V_{SRC} - V_{O2}}{R}$$ The current, from left to right, through the left-most 10k resistor is $$I_{10k} = \frac{V_{SRC}}{10k\Omega}$$ KCL at the input node yields $$I_S = I_R + I_{10k}$$ Using the well-known inverting op-amp gain formula, the two op-amp cascade has a gain ...

8

Start by forgetting about Vsrc and R, and just analyze the the two-opamp circuit by itself. You should be able to see what each stage does from inspection, which then easily gives you what both stages together do. Not even a calculator is needed here. Now consider the input impedance of this circuit. What is it? It should also be obvious from ...

7

Opamps can't usually supply 100 mA. But, they can still be used to control a power voltage if you add some current gain to their output. If you really want a opamp to control a power voltage that can supply 100s of mA, here is a simple way: The opamp still does the controlling and still provides voltage gain from Vin to Vout. However, Q1 provides most ...

5

I'd use a MAX999 comparator. Feed the signal into one pin via a series capacitor and biasing resistor to mid-rail then on the other pin have a resistor feeding a capacitor so that the average level can be followed - it's called a data slicer: - With the waveform in the OP's picture looking like it could undulate its average level a fair bit, using a data ...

4

Make the two input resistors significantly bigger (say 10xR) - the gain will now be lower but you can added an extra amplifier onto Vout to restore the gain you need. Be aware of common mode problems with this type of circuit - you'll probably need to use 0.1% (or better) resistors to get an accurate representation of your 1-5V signal.

4

What's wrong with studying the data sheet (and believing it) - it tells you virtually everything (unless the data sheet is not available). Here's what the AD8605 has to say about itself: - The graph at the top tells you what the output impedance is for various gain settings. Read between the lines for other gain settings. The input impedance is very high ...

4

It will be easier to see how to do this after distilling your requirements down to a real spec. You want 16-22 mV to map to 0-5 V. That's a gain of 833 centered around 19 mV. Let's presume you have a well regulated 5 V supply available. First, leave a little margin. Let's say we'll use a opamp with up to 1 mV offset voltage, and you don't want either ...

4

It is possible, but the CMRR (common mode rejection ratio) will be worse than a monolithic InAmp. In a monolithic InAmp, the resistors R1-R3 are matched. Also, they are on the same die at the same temperature. By the way, the above holds independent of the OpAmp model. A Designer's Guide to Instrumentation Amplifiers has more on inner workings and ...

3

Peter Bennett's got a point about the minimum and maximum voltages that can be present at the inputs, but I'm not certain it's the problem here. First off, we can look at the behaviour of your op-amp. With no input, it's being driven to the rails, as it's called. Basically, an op-amp's maximum range is it's positive and negative supply voltage, less a ...

3

I have read about using a shunt resistor and an op comparator (LM311), but I don't understand how I would do that, or if its even the current way to measure. The shunt resistor part is correct, it will crate a voltage drop that is proportional to the current that goes through it (and consequently the motor). In order to influence the motor current as ...

3

Refer to this literature from Texas Instruments, Understanding Op Amp Parameters Analog Devices, MT-041: Op Amp Input and Output Common-Mode and Differential Voltage Range Input Common Mode Voltage Range The input common voltage is defined as the average voltage at the inverting and non inverting input pins. If the common mode voltage gets too high ...

3

Actually, if I understand correctly what you ask, Vo is NOT the difference between the two inputs. Usually, a comparator works this way: If V1 > V2, Vo is equal approximately to +Vcc-0,7V. If V1 < V2, Vo is equal approximately to -Vcc+0,7V. (usually stated in the datasheet). So, it is a comparison (greater than or lesser than) between the two inputs. ...

3

It's a Howland current pump - a type of constant current generator: - Op-amp 2 in your circuit is, the same as the smaller of the two op-amps shown above. Out1 is, ostensibly the current output but it also shares this with R6 and R7 - as to what out1 and out 2 are used for I don't know because there is no detail in your question. Here is a decent article ...

2

You don't need that Sallen-Key filter before your amplifier if you are already filtering with the input RC high-pass filter. You can simply AC couple the input, bias the amp and gain it up. simulate this circuit – Schematic created using CircuitLab Some key things to note from this circuit: The input RC filter in your original circuit had ...

2

The results shown in the Bode plot are reasonable, and likely accurate. Here is the Bode response I got with a quick level 1 opamp model of the OPA3355. It shows slightly higher Q than your result, but I put no effort into the output impedance of the opamp model. A more realistic model would lower the Q. R1 and C1 combine to form a zero at about ...

2

See my answer here for an example of why you might want to use negative and positive feedback at the same time. The 598.3K resistor in the positive feedback path maintains a constant current through the variable resistor and the negative feedback path determines the gain (output volts per ohm of resistance of the variable resistor). To see how this ...

2

Let's start our with a single op amp and work out way toward a class AB design. Consider the voltage follower. simulate this circuit – Schematic created using CircuitLab In this case $V_{in}=V_{out}$. Why? Negative feedback. Negative feedback forces the inverting pin's voltage to match the non-inverting pin. In other words, the op amp will ...

2

If you regard V$_{SRC}$ as an input, the output of the 2nd op-amp will be 8 x V$_{SRC}$ and ideally the resistor R and the 10k resistor should form a potential divider that would recreate V$_{SRC}$ as if V$_{SRC}$ wasn't there. This would mean that when V$_{SRC}$ becomes connected, it would see an exact replica of itself and no current would flow ...

2

With the spec you have said (+/-5% regulation and output no greater than 250mA) there are fairly simple design topologies that you can use. Consider the synchronous buck regulator - it drives a top transisitor and a bottom transistor and produces a square wave output whose mark-space ratio largely defines the output voltage: - Output voltage = Input ...

2

As the datasheet shows, although this is a low-voltage opamp, it is not a rail-to-rail opamp. When powered by +12V and 0V, the output voltage can only swing between +1.5 and +10.5 volts. Also, if/when the input signal goes below ground, obviously the output can't follow.

2

Opamps are linear devices, and as such, they're going to be as (in)efficient as any other linear regulator. The load current at the output pin is supplied from the opamp's power supply pins, and any difference in voltage between the two pins appears across the driver transistors in the opamp's output stage. This voltage, multiplied by the load current, ...

2

That's not really a precision full-wave rectifier- the gain for positive inputs is +1 and the gain for negative inputs is -1.5. Add 300K to ground on pin 10 to correct that. It looks like it's oscillating due to stray capacitance and/or poor bypassing. Try bypass capacitors from +V and -V supply voltages to ground near the chip (100nF will do) and maybe ...

2

The inherent efficiency will be the same as any other linear regulator, give or take, depending on the op-amp quiescent current (could be uA to mA for the op-amp just sitting there with no load). $P_D = (V_{in} - V_{out}) \cdot I_{load} + I_q \cdot V_{in}$ 100mA will require an expensive op-amp or a booster stage on the output. There's another ...

2

This is a pretty fast op-amp. It's almost certainly due to layout or decoupling. Do you have ceramic decoupling caps right at the supply inputs to a ground plane? Is this on a PCB with short traces and a tight layout or is it on a breadboard? If it's the latter you might want to try a lower bandwidth op-amp. Since your differential gain is 1 but ...

2

In all the application circuits for this device I never saw a value of feedback resistor or input resistor that was bigger than 1kohm. They even have a section on the data sheet dedicated to giving this information and again, they do not recommend a resistance greater than 1kohm - take a look it's on page 14 of the data sheet. There are plenty of other ...

2

That is fine so long as the signal/power is isolated from the common reference/ground. In fact, using an isolated DC-DC regulator, it is common to swap the output leads to get negative voltage. The problem is that generally, one side of everything is connected to a common reference/ground. For example, on an ATX power supply, the negative side of +12V, +5V, ...

2

If the output of a power supply is isolated, meaning it can float over a range of voltage relative to its input, then the polarity of its output is only relative. Whether it is a positive or negative supply is only determined by how you think of it and how you hook it up to your load relative to whatever it is you consider 0 V. If a power supply is not ...

1

You won't be able to supply 100mA from all standard op-amps - maybe 20mA for most and up to 50mA for more specialist devices. There are such things as power op-amps that can deliver amps but these are specialist devices. Regarding efficiency - the power delivered to the load might be 3.3V x 20mA = 66mW and the power into the opamp will be 7.5V x (20mA + ...

1

The two statements need a small but important caveat. Statement 1: Band pass filters will output the fundamental frequency of the square wave multiplied by the gain at the center frequency - provided the center frequency of the filter is the same as the fundamental and the bandwidth of the filter is narrow enough to filter out the 3rd harmonic (3 x ...

1

If you are sampling at 1.5MHz, the time taken for your analogue amplifier to slew its output has to be a bit quicker than the reciprocal of 1.5MHz i.e. 0.667us. If it has to deliver a change of 10V in this period then the slew rate, as a minimum must be: - $\dfrac{10V}{0.667\mu s}$ = 15V per micro second. You also need to look at the op-amps settling ...

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