New answers tagged

0

From an AC perspective both +5V and GND are both "AC Ground" because they have no AC voltage, only a DC voltage. So in effect they are in parallel from an AC perspective.


1

Ccm is just a circuit model capacitance, the sum of all capacitances affecting the input node. It includes e.g. op-amp capacitance and PCB wiring capacitance that are present in the circuit so it is not a capacitor the user provides. Cd is model for photodiode junction capacitance. It also is not a capacitor that the user provides.


2

Which diode capacitance value do I use for the input source capacitance (Cs)? I am having trouble understanding the difference between the 400pF vs 1.2nF. The 400 pF capacitance is when the diode is biased at 5 V. 1200 pF is for zero bias on the diode. What voltage do you intend to bias the diode at? Maybe the difference in capacitance, and thus speed of ...


4

I even tried auto set but the oscilloscope cant stabilize the waveforms. Not until both 1 kHz signal and 2 kHz signals are frequency locked as a perfect 1:2 ratio will this happen. Might I suggest you use your o-scope triggering in single shot mode rather than using it to display multiple images over and over again: -


4

The two separate function generators are not synchronized in any way. The generated 1 kHz and 2 kHz are not in any phase or frequency relation to each other, so the resulting waveform reflects just that, a sum of 1 kHz and 2 kHz sine waves that are not in perfect phase sync.


1

You ask "What's the performance difference between this two differential amplifier?" but I think what you really meant to ask was "What's the performance difference between these two current sources?". The first circuit performs better because it amplifies the sensed voltage directly. Though, the OP07 may not be the best choice. They make ...


3

A SPICE simulation would tell you this information. Figuring this out by hand is not pleasant because the impedance is complex and varies with frequency. In LTspice and PSpice, you can run an .AC analysis with the transducer's admittance numbers placed in a table. PSpice and LTspice will interpolate these number appropriately. From that, you can probe what ...


4

In your top circuit Zab plays no role at all, except to pass some pointless current. It's connected directly across two voltage sources, one (the input) that doesn't care one iota what's connected to it, and the other (the opamp output) which will adopt whatever voltage is necessary to bring its inverting input potential to zero (to equal the voltage at its ...


0

First of all, remember that the 634P is in end of life. The A part is the current one (usually they are simply improved but check the differences). Also be careful of your supply values, at full power it can only reach 2.5V to the rails. At the recommended -15…+15V dual supply this is enough for your 20VPP output signal. ±12V is not, you would be banging ...


1

ACS714 measure DC and AC current. See characteristic Vout vs Current. With DC current, if "logic" is not good, swap IP pins... But only in "some cases" ... See pictures ! Simulated with microcap v12 , http://www.spectrum-soft.com/download/download.shtm Offsets may change results. Text "Sweeping ..." Step = 1. Sorry.


0

It seems that your model is probably "bad" or "bad use" (I see DTC is grounded ?, see my schematic, others things missed ?). Here is a simulation with another simulator (operational test circuit from TI, datasheet of TL494). In this model, frequency changes with R and/or C. In TRANS analysis, click on component and use arrows keys up or ...


2

I understand I'm missing something in my basic understanding of how an op-amp works, but despite looking on several sources I wasn't able to find what I'm missing. I believe you are just confusing yourself. The simplification that you set the difference voltage, \$v_d\$ to zero, then solve for the output voltage, \$v_u\$, is only for stable, closed-loop ...


3

Whether input offset voltage (and input offset current) is included or not (and what polarity, if any, they pick) is up to the creator of the model. It's not unusual to see some there- in this case the 'typical' value is +/-1mV from the datasheet. If you look at a NS model, you can see that's what they have used: EOS 7 1 POLY(1) 16 49 1E-3 1 *Input offset ...


4

You want ideal results yet you simulate with a non ideal 741 opamp? What you see is the DC offset of the opamp. This opamp has a small input current. This current flows through resistors R1 and R2 and that causes a voltage drop. You don't have the same voltage drop at the + input of the opamp. If you want ideal / theoretical results then only use ideal ...


2

You can use Inverting Charge Pump Generates.LTC3261 is a good choice.Its output current capacity is 100mA.Please see the circuit below.


1

I believe the issue is how you are connecting your power supply to your op-amp. Most of the DC bench power supplies Ive worked with do not supply V_supply on the + terminal relative to ground and -1 * V_supply on the negative terminal relative to ground. Instead the voltage displayed on the supply is the voltage across the + and - terminals. If the + and - ...


0

Here is the behavior of this circuit. S1 is a switch 1 Ohm / 1 Meg, for this purpose, closed at start (Vc=0). It has to be analyzed in two steps. 1- Until Zener is well reversed polarized (current in Dz), output is linear -> constant current source for capacitor. 2- When Zener "switch" off, regular charge to 15 V. v(Vo) is behind v(Vc) (~ same). ...


0

Your schematic design is just fine providing you don't want high current output. At +/-5 V for the AD4084 you need to provide +17 mA and -24 mA if you want the supply to remain at +/-5 V while the output is shorted. If your load current is much less than the Isc rating, then the supply voltage will collapse under short circuit condition. Since this is self ...


0

You are right to think that the closed-loop inductive output impedance is not responsible for the potential instability. The stability of the loop is determined by the loop gain, which does not contain the inductive element that the opamp looks like after closing the loop. Instead one must look at the open-loop output impedance. As a final note, the open-...


1

I reckon you're missing the fact that whatever's connected to the output of an opamp, actually does form part of the feedback loop. It's not obvious from the classic opamp follower diagram, because that diagram does not show such an influence "inserted" into the loop itself: simulate this circuit – Schematic created using CircuitLab However, ...


3

Positive period amplified (by 2 in this case), negative period without amplification, all transfered to negative. Vf of diode is neglected. Blue is input and green the output.


0

The schematic is a well known application of op amps (bipolar supplies). Output voltage negative but can be changed by a classic inverting amplifier. A active virtual ground point can be used for unipolar supply (independant !) for the measurement system. Proposed also with very low offset op amp GS8333 (?) (not tested, offset simplified) or OP2189. Shown ...


1

Depending on the current draw on the -12V rail and how strict requirements you may have on the accuracy you may try to search for a switched capacitor converter. This is a class of ICs by various manufacturers that take positive voltage at the input and with only few external components (three capacitors and no magnetics) generate negative voltage at the ...


-2

You can design one using an IC like the one mentioned here: https://www.analog.com/en/technical-articles/inverting-dc-dc-controller-converts-a-positive-input-to-a-negative-output-with-a-single-inductor.html


1

A quick search on Digikey shows over 500 DC-DC converters that will take 12V and supply + and - 12V output. Without knowing more about your circuit and its requirements, that's about as specific as I can get. They make chassis/DIN mount converters that will do the same thing, in case your boards are already made.


1

With this small shunt resistor you get tiny voltages, which needs special care to measure. As already mentioned, you need amplifiers with very low offset voltage and drift, e.g. chopper amplifiers like LTC1049 or LTC1050 (I once used these kind for a similar application to measure 400A precisely for a MRT magnet). Measuring such low voltages is possible, but ...


13

You have an amplifier with a gain of about +2000 and an offset voltage of as much as +/-3mV at room temperature. It might typically be +/-2mV (onsemi datasheet). Since the output swing with no load and a 3.3V supply is from some mV up to maybe 2V, the zero-current output could be about anything within that range. Your entire full-scale input signal is only 1....


5

First of all, if you are measuring a sense resistor that is connected to wires, then there will probably be ground offset. You will need a differential measurement to handle that. The op-amp circuit you have shown is configured for a single-ended measurement. It needs to be set up for a differential measurement to account for the ground offset caused by ...


4

Realistically, you need something better than what you're working with. tl, dr: you need a better op-amp and a better design. We'll get to that in a moment. On the other hand you don’t need an instrumentation amplifier, which is a specific setup composed of 3 op-amps (2 followers and one differential) that is used for amplifying high impedance signals. ...


12

You don't have to use a differential amplifier providing you are careful about the voltage drops that occur along the grounds. If you return the bottom end of the 500 ohm resistor to the same point as the ground end of the sense resistor it should avoid most of those errors. However, one of the first things to do with this sort of problem is to ensure that ...


1

I know that the transistors act somehow as a limit of the amplifier, but I am not sure how this works. Can someone explain this? I agree, it worked somehow, if it did any. My best guess is that the schematics drawing is incorrect. If that was going to work, wiring has to be like this, ignore the component values: simulate this circuit – Schematic ...


3

Looks like what you're trying to do is make a +1/-1 gain switchable buffer. Your circuit needs a bit of work though. Try this (simulate it here): This gives a gain of +1 or -1, depending on the switch, and has a constant input impedance of 47 kohm to GND. If you need a different impedance change all the resistors to the same value (e.g., 49.9k). Switch 'up' ...


2

Without \$C_2\$ the input resistance seen by the signal source is equal to \$R_1 + R_2 = 200kΩ\$ But for the signal frequency, the capacitor will act just like a short circuit (\$X_{C2} \approx 0\$). So the situation will look like this: simulate this circuit – Schematic created using CircuitLab And finally, we have: simulate this circuit As you can ...


1

A bit of a hand-wavy answer but: The input capacitor should be chosen to appear as a short to the signal of interest and an open circuit to DC. After the capacitor, you have the parallel combination of the opamp input impedance and the bias resistance. Since the opamp input impedance is assumed to be very high in comparison to the bias resistance, the bias ...


-1

Capacitor C2 is necessary to separate the DC bias that is coming from the ground and the copy of the AC input signal that is from the output of the amplifier. If C2 was not present (ie replaced by a short-circuit) the ground reference would be overridden by the DC at the output of the amplifier and the boas would not be stable. In cases where the amplifier ...


0

Use a op-amp in the non-inverting configuration as shown in the picture. The gain will be $$A = 1 + \frac{R_2}{R_1}$$ so just choose the resistor values to meet the necessary output voltage that you need based on the input voltage. For example, if you had a 100 mV signal, you would want a gain of 50 to get to 5V. This is pretty low speed, so most op amps ...


3

They isolate RC filters. Each RC filter is a 1st order system with its transfer function. When you put opamp in between you do isolate, so that any forward section doesn't affect the backward section. The equivalent transfer function in Laplace domain is the product of all separate transfer functions. It's the way for study, you could have many other types ...


2

There is a path through R28 and R32 to give crosscoupling between Diff1 and Diff2 that is giving the small signal in OA4. The two opamps share the same offset adjustment. Any signal coming in the D input goes through R31, the R32 to the offset adjustment. Because that acts like a 2.5k resistor to ground there is about a 4:1 attenuation to this point. (expect ...


4

Your clipping depends on U4 being able to sink enough current to overcome what U2 can source. So, one output is fighting another. You can add some resistance before the clamp to limit the current to a few mA, maybe 5-10mA maximum. That would also prevent U4 from going into current-limiting on the negative side. Forcing the output into current limiting causes ...


4

TI says that the TL074H and TL084H are identical. Older datasheets of the previous-generation TL074 and TL084 devices (without "H") have shown the same noise specification of 18 nV/√Hz for over thirty years, so it's likely that TI ceased to do separate testing for a high-noise version a long time ago. (When the manufacturing process is good enough, ...


1

With +/- 2.5V or single +5V supply, the OPA2134 input will barely be able to work with your DAC output. And if you used the datasheet example circuit, it is set to output a 2.1Vrms line level output, which is about 5.9Vpp, so if it were an ideal component, it would require at least 5.9V of supply voltage at minimum. It might be able to drive a line input ...


0

If, as you say in your comment The output Voltage from magnetic pickup rises as more number of teeth cross it when speed increases this also increases the frequency -however the increase in the voltage may not be linear but yes it does rise with higher speed and frequency is equal to the number of teeth that cross so that rise as well then use a high pass ...


0

The Phoenix Contact module is an industrial device, for 24VDC. So it accepts many industrial sensors, but not a magnetic pickup. The easiest way is to replace the magnetic pickup with an inductive sensor with HTL output and 24VDC supply. There are many of them. Otherwise there is LM2917 that does amplification of the pickup and frequency to voltage ...


10

You could try this - most op-amps are not well-suited to driving speaker loads directly, though some may be okay driving headphone loads. In order to minimize the shutdown current you also likely need to drive the PWM output to 0V or the internal circuitry of the op-amp may draw current from the output, perhaps partially powering up. As shown you’ll lose a ...


5

No, they are not useless because you don't buffer power supplies with a voltage follower. You buffer signals sources. If your input is a sensor one with a very high output impedance, it isn't going to be able to drive much current through its outputs without completely distorting its output voltage signal. It might not even be enough to drive an ADC. You ...


6

Does the schematic make sense and am I hooking things the proper way? Do it like this: - You wouldn't use an NPN emitter follower to do this; use a PNP common-emitter circuit. I wouldn't recommend trying to switch the 0 volts to the amplifier with an NPN either if your amplifier is driving a loudspeaker. Is a pin's output stable enough to drive a op-amp ...


6

Amendment: Andy aka has pointed out that the input bias current is so small that my inital thought about R4 is incorrect. I stated, wrongly, that: R4 is to balance the opamp's input bias currents, as explained in this article: Opamp input bias current compensation I agree with Andy aka, but I'm not yet sure how R4 could protect the opamp's input from ...


4

Judging from the “2.65 to 2.70V” note on the output node, this circuit appears to be a 2.7V power supply, trimmable to the desired voltage with R1. R3 and R5 are inline with the opamp rails and also across base emitter junctions of Q1 and Q2. For Q1 (or Q2) to do anything at all there must be at least 0.6V across R5 (or R3). At 560 ohms that would be 1mA ...


9

To understand this circuit, you have to imagine the push-pull transistor pair at the pins 7, 6, 4 inside the OP-AMP. It uses the supply pins of the OP-AMP as inverted dual-outputs to simplify the next driver stage. It's a hack. Replacing the OP-AMP for another type requires thorough testing because this kind of circuit is a hack.


3

R4 may balance input offset caused by the op amp's input bias current and the non-zero source resistance that is R1, R9, and R10. Regarding R3 and R5, they are used together with Q1/Q2 and the class B output stage of the op amp to create a Sziklai pair (aka compound pair) for greater output current capability. The Sziklai pair may function without them, but ...


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