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6

I'm offering this answer as an alternative saturation detector. Pass a direct current thru the coil that can be varied from a milli amp to possibly several amps. This is easily achieved with an opamp and BJT as a constant current generator. Next, generate a low amplitude sinwave and capacitive couple it to the inductor. Measure or view the sine amplitude ...


5

The non-inverting op amp input should be connected to ground, not +15V. Otherwise you saturate the output because V+ is always greater than V-.


5

Why are you putting +15V on the op amp + input? That will make the output attempt to center around +15V, but it can't output that much. Maybe you meant to ground the op amp + input.


4

The differences are numerous, but I'll attempt to summarise. An opamp is a differential amplifier. That is, it amplifies the difference between its two inputs. An ideal opamp has the following characteristics of interest: Infinite voltage gain Infinite input impedance Zero output impedance Real opamps deviate from this, obviously, but generally share ...


3

Something like this should do you. Note that the input common mode voltage range of the LM358 amplifier only goes to 3V so you can't just buffer a divider with voltage follower (which would be the easiest way, and would typically work around room temperature, but this is engineering so we have to consider worst-case and temperature). The output range, with ...


3

Actually, you don't need an op amp in this case. An LM35 has a nominal output of 0 to 1 volts for a 0 to 100 C range. If the ADC is set up for a Vref of 1.1 volts and a gain of 1, you'll get a nominal scale factor of approximately 0.107 degrees C per lsb. Since the LM35 is only rated for +/- 0.5 degrees accuracy, this is better resolution than you can ...


3

Try something more like this- you may have to fiddle with the value of the capacitor to optimize it if you need really high frequency response. simulate this circuit – Schematic created using CircuitLab


3

The MAX410 will not drive headphones (23 ohms or 32 ohms) as you expect - it's just an op-amp (not some small power amp) and might be expected to deliver a 2.3Vp-p output signal into a 2kohm load. It's got a short circuit current of 35mA so, you might get 20mA peaks with an output voltage of maybe 2Vp-p. That's a power of about 25 milli watts. Also, the ...


3

Having a capacitor directly across the op amp inputs always looks iffy to me. It can make the op amp trend to oscillate because high frequency feedback gets attenuated. If you are measuring the 1.5v "latch-up" voltage with a DC volt meter, you might simply be seeing the average voltage of an oscillation. Try placing C1 before R2 and R3 and see if that ...


3

As mentioned in another answer, the die diagram is helpful if you are buying the chip as a bare die and doing chip-on-board or hybrid assembly. The schematic is also helpful for understanding how to drive the inputs and how to load the outputs. It helps you know tings like whether pull-ups or pull-downs are needed, whether ac-coupling is needed. It might ...


2

I have a formula of 1+(R2/R1), but that doesn't seem to be valid here. It's perfectly valid; \$\dfrac{V_O}{V_I} = 1+\dfrac{R_2}{R_1}\$ Think about how an op-amp works when negative feedback is applied; the output, via R2 and R1 sets the inverting input to equal the voltage on the non-inverting input. In doing so the output must be higher because of the ...


2

Yes, the output power is almost always greater than the input power in op amps. Ideally, the input power is zero since the input resistances are infinite. Here's a pinout for an 8-pin 741 op amp: See the pins labeled +V and -V? Those are for power supply voltages. That's where the output power comes from. Conservation of energy is not violated. Here's a ...


2

An op-amp driving the gate of a MOSFET with feedback from a resistor connected between the MOSFET Source and ground is the basis for a voltage-to-current converter. If the source resistor is 1k ohm, drain current 4-20 mA corresponds to 4-20 volts at the source (the gate takes no current). When your temperature sensor delivers a 75mV swing over the range of ...


2

Consider the simple precision half wave rectifier shown. Let the open loop gain of op-amp be \$A\$. From the circuit, the voltage at cathode can be calculated as $$V_{anode} = A(V_{in} - V_{cathode})\tag1$$ Now the diode will conduct when $$V_{anode} > V_{cathode} + 0.7V\tag2$$ $$A(V_{in} - V_{cathode}) > V_{cathode} + 0.7V $$ $$V_{in} > ...


2

This is your precision rectifier. simulate this circuit – Schematic created using CircuitLab But before that, lets go back and review some opamp (ideal) basics. The opamp has no input current The output will do whatever it can do make sure the inverting and non inverting terminals are the same. For your question, the first point does not ...


2

Easiest method: simulate this circuit – Schematic created using CircuitLab Component values are left for you to calculate. Note that the TL081 is NOT an appropriate op-amp, given your desire for a rail-to-rail output swing. R1 & R2 set the op-amp bias point. R1, R2, C2 forms a low-pass filter that reduces noise from the power supply ...


2

Some practical concerns... Q1 should have some base resistance to limit the current feeding it (when U3 is behaves as a comparator, in constant-voltage regulation). The "gain" of the constant-current loop formed by R1, U2, U3, and Q1 will vary depending on load current (because Q1's transconductance is proportional to current). In particular, evaluate the ...


2

Let's talk about the analog side of things for a minute, without worrying about the specific parts. If you have a voltage signal that you want to send down a long wire, whether or not it arrives unaltered depends on a number of factors. DC signals don't particularly suffer from any of the effects of transmission lines, so you are left with three primary ...


2

A problem of amplifier gain less than unity is loss of usable number of bits (UNOB), since the full input range of the ADC will not be used. In fact, if the ADC has input range of 0V to 3.3V then any gain less than 1.44 will result in loss of UNOB. Gain actually needs to be increased. Here's a circuit for reference: With the proper voltage reference ...


2

LM358 is not a rail-to-rail opamp (neither input nor output). The input CM range includes the negative rail and the output can swing close to the negative rail under the correct conditions, however this configuration passes current If to the output, and if that current exceeds (typically) about 50uA then the output can no longer swing all that close to the ...


2

Without knowing your illumination levels, it's impossible to tell if that will do what you need. That said, the circuit ought to work, with a couple of changes. First, eliminate the resistor from pin 3 to ground and connect directly to ground. Next, increase the 15 pF capacitor to 100 pF. Your actual photodiode has much higher capacitance than the ...


2

Something seems wrong to me - you mention that you are currently using a 330 uF capacitor. What is the impedance of your load? Are you expecting to feed something like a speaker or headphones with this or is it feeding an amplifier? For example, many audio amplifier inputs have an input impedance of 10 KOhms. Your 330 uF capacitor would give you a ...


2

Your biggest problem is that you don't know what a virtual ground is, and you need one if you're going to power your op amp from a single 5 volt supply. Try simulate this circuit – Schematic created using CircuitLab with the proviso that, depending on the input impedance of the next stage, you may have to increase the value of C4. As it ...


2

If you don't mind measuring the parameters using an oscilloscope, you could use the technique that I use. I have a large MOSFET with a low-value current-sense resistor in the source pin to ground. The gate has a 33 Ohm resistor in series with a 10k resistor to ground. The drain goes to a terminal for one side of the inductor under test and to a clamp ...


2

This circuit, a non-inverting summing amplifier, can be used for your purpose. The left op-amp is a 2.5v voltage reference (derived from your 5v supply), and the right op-amp is the summing amp. With the values shown, it takes the ±2.5v Audio input and adds it to the 2.5v offset, resulting in a 0-5v output. If R3 = R4, then Output = Gain * (Audio + Ref) ...


2

If your input signal is within the constraints of +/- 2.5V then it just needs to be lifted from an average voltage of 0V to an average voltage of +2.5 volts. In doing so, the output becomes constrained to a range of 0 to +5V. If the input is generally audio/music then you can use a capacitor and voltage divider like so: - If Vref is 5V, the average dc ...


2

Chip topography is useful only if you can get the chip as a bare die. You need to know the precise locations of the bond pads in order to wirebond it to the board. Schematics are not exceptionally common, generally you will get block diagrams and simplified schematics of particular components, especially I/O circuitry. Schematics can be useful to ...


1

Absolutely, it is done all the time in practice, especially when you want a device in triode. I use it regularly in bias circuits or comparator preamps with triode loads. From a practical standpoint, if you want to use a really long device, you shouldn't have the width at minimum but it is usually better to have it closer to 2x or 3x minimum.


1

The question "What's my problem?" cannot accurately be answered, since there are at least 5 problems. 1) The problem you're worried about, op amp conditions, has been addressed by Spehro. You are using an inappropriate op amp. Either change the op amp or provide a -12 volt negative supply. 2) An op amp problem not addressed by Spehro is that a 741 is not ...


1

No opinion involved- the 741 was not a 'single supply' op-amp, so the input common-mode range does not extend to the negative supply rail nor can the output swing all the way to that rail. Replace it with half an LM358 and much will improve, though that is not really a good way to do 4-20mA- it's preferable to have one side of the load grounded. Edit: You ...



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