New answers tagged amplifier
0
What is this circuit called? A buffer?
Unity gain voltage buffer.
Is this possible? What needs to change to make this possible?
Look for an opamp that is rail to rail (input and output) compatible, this will ensure that the output can swing rail to rail. Op amps that don't operate rail to rail will have trouble reaching Vcc.
5
The LM358 is not exactly low noise.
You are using a lot of gain. 48000/180= 266. The amplification applies to noise as well as the signal.
The bias resistor for the microphone is rather large. This will also affect how much noise you have.
You haven't mentioned the power supply, but you have 5V for the microphone bias. The LM358 can operate on 5V, but ...
0
The text appears to indicate that current output is desired, so that summing is simpler. I believe based on the text that the diodes are added to indicate a conversion to current output for the summation. Note that it specifically states that "the detectors should be current outputs (not simple diodes)" indicating a linear voltage-to-current conversion. ...
0
"Negative gain" is an ambiguous term and should not be used.
"Negative Gain" could mean an inverting gain if the gain is specified as a ratio of output to input magnitudes or it could mean attenuation if the gain is specified in negative dBs.
Better to say that an amplifier has a gain or attenuation and also that this gain or attenuation is either ...
2
In the RF domain you already mention that what is important: impedance matching.
Fact is that RF signals generally have such a high frequency that the wavelength of the signal can come close to the length of the tracks on a PCB and/or wires between PCBs and connectors. That then means that we cannot use "just a wire" the wire needs to be a transmission line....
0
Only switched amplifiers (classes D, E, F, etc.) work in "saturation region". This kind of amplifiers doesn't switch the power supply rails because they are intrinsically efficient.
The only amplification classes that switch power rails or use some kind of dynamic power supply are class G and H, which are basically class AB amplifiers (thus working in "...
3
I also measure an offset of about 2.90V on those pins, but in the ds I
can't find this value neither.
The DS states the pins are biased at 3 volts (page 2).
I'm looking for the input range for pins INPR and INPL but I cannot
find this information.
You take the gain value (programmed at power up) and reverse back from the preferred maximum peak-to-...
1
A simple solution which also has approximately matched loading at the cells is:
simulate this circuit – Schematic created using CircuitLab
The resistance seen by each cell is just over 1M and the output span of all the amplifiers is 0 to 4V where 4V is the nominal maximum for each cell at 35V.
Given that this is a low speed application, a very ...
5
Just as an example, if you are seeing 49.3 degC in an ambient of 25 degC, the op-amps are warming up 24.3 degC. The thermal resistance of the dual op-amp (NE5532) is circa 90 degC/watt so that would imply the dual op-amps are dissipating about 270 mW each.
The op-amp spec implies a total (both in a dual package) no load current of around 8 mA and, on a ...
1
You've made an error here: -
The inverting input should go to the junction of R2 and R3.
Also, driving an inductor directly is somewhat frowned upon given that the likely offset DC voltage from the OPAMP is going to be in the realm of 20 mV and this might be trying to force several amps of DC through the 7 mH inductor.
0
If the LEDs drop 2.0 volts at 30 mA and the emitter produces 2.6 volts when driven by 3.3 volts at the base then the voltage across the LED series resistor is 0.6 volts and, with a current of 30 mA flowing, the resistance is 20 ohms.
With 25 mA of LED current, the forward volt drop of the LED might be a fraction below 2.0 volts but it will be a small ...
2
If you take only the output from the final amplifier, then the transfer function is:
At some value of input, the output stage will saturate and not increase any further.
If we then add the other outputs, we will get (not to scale):
As each amplifier saturates, then the slope reduces until the next amplifier in the chain saturates until all 4 amplifiers ...
1
If Rin is low in value, the voltages at the input terminals of the isolation amplifier will be "dragged" closer to each other compared to the scenario when Rin is high in value. This means that Rin's value does affect the conversion of common mode to differential mode signals. If you take it to extremes, if Rin is zero there can be no differential voltage.
0
I tried two ways: with a pre-programmed SparkFun OpenScale and with a custom circuit with Arduino Nano. Both ways do without the Sense cables, which serve for more accurate readings. They also use 5V instead of 10V, which reduces precision but seems OK for my application (see Elliot Alderson's answer for an excellent guidance on calculating precision and the ...
0
After researching a bit more, and trying to improve my MOSFET switch circuit, I realized that the MOSFET I used before (IRF series) isn't able to switch properly with 5 volts coming from the transistor. So I found a spice model for IRL540, replaced it with the IRF one. And also instead of the pull down resistor, I added a PNP transistor to pull the MOSFET ...
2
An LED strobe driver at that power level needs some significant attention to managing current during the pulse period to avoid frying the LED, and a fast recovery time after the pulse. Neither attribute is in the realm of an audio amplifier, which by its nature will be bandwidth limited so will do poorly on both counts.
A switch type driver like you're ...
4
I think you will find that an audio amplifier is a poor choice as an LED driver.
It is AC coupled so the output swings both positively and negatively with respect to ground. Therefore your LED will have to be protected against the reverse half cycles if it is to have a long life.
An audio amplifier output is a low impedance voltage source, LEDs require a ...
0
Due to the low impedance of your sensor, you should be able to just use one return resistor connected between either of the inputs and the device ground. As far as the value of the resistor being 10 MOhm, I would say that is a bit high given the typical recommendations, but it should be fine for most applications. It will only give a voltage error around 10 ...
4
It looks like a first-order passive high-pass shelving filter.
Its role is to attenuate low-frequency content of the audio signal without cutting it off. Mainly used to reduce (de-emphasize) excessive bass as a form of coarse equalization via bass-trebble (tone) controls.
Some features:
Attenuation of low-frequency content is: \$\alpha = 20 \log \big(1+\...
1
It is a high pass filter and I believe it is part of tone control. You can adjust the high frequency response of the amplifier. It might help to see more of the circuit.
2
Simulate it with a swept sine.
Assuming the source is low impedance and the following circuitry is high impedance, it's a 1:3 voltage divider at DC, and passes high frequencies unattenuated. There's a zero at \$\frac{1}{2 \pi \mathrm{(3.3nF)(40k\Omega)}} = \mathrm{1200Hz}\$, and thus a pole at \$\mathrm{3600Hz}\$.
So, it's an equalizer (or emphasis) ...
1
I recommend reading TI AN-31 appnote. You are probably looking for one of these
2
The Safe Operating Area curve does not tolerate arcing.
Using that plot Figure 16, at 150 volts from the plot, less than ONE AMP is allowed.
You may need to insert series current limiting resistance: non-inductive.
Different power transistors have different ability to tolerate pulse-energy overloads. NASA and the automotive manufacturers uncovered this, ...
0
Here are two solutions; first is discrete bipolar transistor circuit; 2nd is opamp circuit.
You could use a PNP emitter follower into a NPN common-emitter. Calibration and offset drift may be a bother. And the output will not give a solid 0.0 volts. But the opamp circuits cannot give a solid 0.0 volts output either.
simulate this circuit – Schematic ...
1
Your circuit is Ok from the point of view of the input signal polarity since it adds an offset to the input in order to avoid its saturation when \$V_\mathrm{in}\$ goes below zero: note however, from the picture you show, it seems that you need a voltage gain of at least
$$
A_v=\frac{V_\mathrm{out}}{V_\mathrm{in}}\approx\frac{1.5\mathrm{V}}{50\cdot 10^{-3}\...
0
You have some kind of misconception about op-amps. You just have to find a correct op-amp configuration to do it. They can add output DC offset and made not to amplify input DC component.
1
Use an opamp circuit, non-inverting config, with a gain of 30. I. E. R2/R1 = 29
Connect R1 to gnd through a large cap.
Make a resistive divider across your Vsupply to get midpoint of 1.5v
Connect your signal through sufficiently large cap to non-inverting pin.
Connect resistive divider centre point also to inverting terminal through large resistor (say ...
3
Upfront warning: amplifying the output of your wifi device is almost certainly illegal. There's strict legal limits on how much power you might output or how much power density you might emit in a specific direction.
I feel like the simplicity of the idea suggests that I am missing something. I think that amplifier chip could only give me a single signal ...
3
Class B has a maximum theoretical efficiency of 78.5% at maximum undistorted sine wave output. However at lower power output the efficiency drops because there is more voltage across the transistors and less across the load. At very low power the quiescent power consumption also becomes significant, so a high power amp will have lower efficiency than a low ...
0
A simple non-inverting amplifier with a single-supply or RR I/O precision (low Vos TCVos and high gain) op-amp will meet that requirement, at least in theory. The low output offset is not a problem if it is not required to sink current, otherwise you would need to make a negative supply rail to get nominal 0V out.
However, 100uV error at 10V is 10ppm and ...
2
Standard op amp with a gain of 2, you will need a rail to rail op amp, you will also want some pulldown, e.g. 2K to ground to ensure it really can get that close to 0V without issue,
I would recommend the 12V supply, to give you some head room to supply current for the pulldown resistor,
simulate this circuit – Schematic created using CircuitLab
0
kindly note that tsop works on a protocol.It can respond only when the received signal conforms only to the protocol not otherwise. you cannot simply generate signal by 555 and call for response.
2
You can't.
A swing of 20mV seems like a pretty small signal -- but it's not small enough to have a reasonably linear current through a typical silicon diode. So the emitter current of one of those transistor stages will be severely distorted with that large of a signal.
You need feedback to linearize the circuit. Emitter degeneration would be easiest, ...
1
You cannot use AVDD to create \$\overline{RESET}\$, since (circular dependency as OP states):
When RESET is released to turn on TPA3220, FAULT signal will turn low and AVDD voltage regulator will be enabled
From the bottom of page 24 of the TPA3220 datasheet:
The TPA3220 does not require a power-up sequence because of the integrated undervoltage ...
1
From this document:
Special consideration must be taken for the RESET pin. The RESET pin
needs to have a 5-V input. In order to do this R40 must be installed
to create a voltage divider network that will drop the voltage down to
5 V
It is certain that the RESET should be controlled.
The R40 (greyed out in the image) should be sized so that, ...
1
The highest temperature that your device shows is about 37 °C (set your camera to use Celsius; Fahrenheit is pretty much unused in engineering). That's pretty much "cold" for electronics. You'll be fine.
3
I know that heat is caused by current
Not really, no.
Temperature rises because of electrical power getting converted to heat. The amount of power converted determines the temperature rise, not the current alone. For example, a current of 0.1 A through a 1 MΩ resistor will cause 10 kW of heating – whereas a current of 10 A through a 1 Ω resistor will only ...
1
Yes it's case by case.
As a rule of thumb, if something catches on fire, desolders itself, or even just smokes, it's too hot.
Chips don't want to be over 55C on their surface, at least at ambient temperatures.
Resistors, probably the same unless they're obviously power resistors and are chosen to get hot on purpose.
Some variation within the above ...
0
Follow the traces from each damaged pad and connect a wire between that pin and the other end of the trace.
If it's a TRS jack (and not TRRS), the extra pin may just be for mounting (mechanical) and not need to connect anywhere electrically. It could also be a switched contact. You can figure that out with a continuity meter.
Be aware that this PCB likely ...
1
If your load is grounded you could add a BJT to a single-supply op-amp as below.
simulate this circuit – Schematic created using CircuitLab
The Zener and C1 and R3 form a 7.5V supply for the IC to allow it to drive the base of the transistor to 5.6V or so.
As shown this typically won't go all the way to 4.9V quite, but close. If you really need ...
1
Many opamps have a max current in the 30mA to 50mA, especially ones that are rail to rail. 60mA is hard to find, and opamps with a max current of more that 200mA are very few.
The best way is to use a rail to rail opamp in a voltage follower configuration. Because the op amp can introduce errors, select one that has a low input voltage offset. Either set ...
0
I had serious noise issues at first. Here is what I did and got excellent noise reduction -
Shielded the input wires with copper tape and grounded both sides.
Added 1k parallel to input lines to ground.
Added extra 10uf ceramic capacitor across vcc and ground pins
Wrapped the chip with few turns of copper tape to get a copper foil heatsink on the chip ...
1
This is just an illustration for LvW's answer, with two input voltages:
Note that the (small) open loop gain is the only non-ideal characteristic simulated. With this condition, the op. amp. does amplify the differential input.
As suggested in the comment below, an example with a more realistic open loop gain (100k), showing why we normally disregard the ...
2
Let me answer with a simple numeric example.
Open-loop gain (inverting) of the opamp Aol=-10³=-1000
Feedback resistors R2=10k and R1=1k (Design goal: Closed-loop gain Acl=-10).
Real closed-loop gain (taking Aol into account):
Acl=-[10/(1+10)]/[1/(1+10)+0.001]=-9.89
Comment: This equation results from the classical feedback formula.
Hence, for an input ...
0
Your question is over 1 year old, so you most likely have a solution already.
If not, here is my advice:
2
It turned out that the loud 60Hz hum was a result of using a footswitch whose cable wasn't shielded (not the original FS). This allowed the long cable to pick up 60Hz hum inductively from nearby power cables and couple them directly into the rest of the circuit. The high impedance of the V4A plate and surrounding resistors made it particularly susceptible to ...
3
You need to distinguish between large signal and small signal.
Suppose your input is a DC value VIN. This causes a DC current ID which is calculated with your equation.
Now, suppose you add a small signal to this: Then you can use Taylor's expansion to get an approximation, which will yield:
iD = (what you have) + (derivative, evaluated at the point you ...
0
I tried a simple LM725-based non-inverting 100x amplifier with precision resistors and offset-null trimmer, powered with a clean 2x15V supply, and reading out the result with a basic 3-1/2 digit multimeter on the 100mV range. It was not shielded but very close to the measured device.
It proved to be stable enough (in my non-controlled environment) to ...
29
It is an amplifier because you put a wiggly signal in, and you get a bigger wiggly signal out. The fact that the output wiggles down when the input wiggles up is a trivial detail, that -- if it's a problem at all -- can be solved in any number of ways (not least of which is following one stage by another, because two negative gains, when multiplied, result ...
0
The path of Vo2 (2 OPA) is obviously longer than Vo1 (1 OPA). Why there is no synchronization problem between them? Are these OPAs fast enough so we can ignore the asynchronous?
This IC is for audio signals so the opamps do not need to be that fast to be "fast enough". Look at the datasheet, page 8, figure 26 which shows the open loop gain over frequency. ...
Top 50 recent answers are included
