New answers tagged

1

The current i need while the device is fully engaged is around 118mA peak. Are you using BLE or WiFi? With WiFi it's almost certainly impossible that you're peaking only at 118mA with all the other stuff. Also you need to account for surges. ESP32 on its own for some reason sometimes can spike up all the way to around 250mA. Especially if it's calibrating ...


1

I won't promise this is the "best, cheapest, smallest" way to power your bone conduction hat, but I recommend you consider zinc-air cells, better known as a "hearing aid battery". Why: Cell voltage is 1.4 V, just connect 4 in series to get 5.6 V. High current capability Disposable, instead of rechargeable -- easier for beginners ...


1

In principle, a polarized electrolytic capacitor acting as a "decoupling capacitor", can be connected to a supply rail instead to the ground. This solves the polarity problem but the noise will be higher.


1

Why does the output of the op-amp (purple trace) [initially] stabilise at -0.6V? The feedback loop (R at the top) compensates for the diode drop. Why does the output finally go to zero? This will depend on the diode model used. A good model will conduct some current at very low voltages and eventually have a drop of zero. Are the caps in the wrong direction? ...


-1

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course). Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab When we use and apply KCL, we can write the following set of ...


0

If I understand correctly, you have multiple sources for your car sound amplifier, and you want certain communications radio outputs to automatically override the main music source (the "head unit") in feeding that amplifier. One way to start is to make all those sources reasonably compatible. You can find "speaker to line level output" ...


0

Such a high-value capacitor can find two usage areas in a MoBo's audio subcircuit: Bypass capacitor for the audio codec chip's ADC or DAC supply. Coupling capacitor for Line Out channels -- for better bass response. Regarding your problem, most likely the 100uF capacitor that you have replaced was a coupling capacitor. Some old motherboard designs have ...


1

Is it possible? Yes. That's pretty much what bench power supplies are for. Set the current limit to something low. Set the voltage to 9V before you make the connection. Make sure you get the + and - the right way round. You need to use the + and - outputs. + and GND alone won't do anything.


0

On a first look It is a TRRS 3.5mm splitter. Being TRRS an abbreviation for tip, ring, ring and sleeve. But since your are going to use this in a headphone I think you will be better served with the results of a 3.5mm TRS splitter or stereo splitter search. TRRS will be hard to find because the existing cables split the audio from the mic and it seems to me ...


4

The end that appears closer in the picture is a 3.5mm TRRS (tip, ring, ring, sleeve) connector, while the other two connectors are both 3.5mm (most likely, it's hard to tell if they may be 2.5mm) TS (tip, sleeve). This means that if the cable is wired normally, the TRRS connector goes like this: Tip = Left Audio (The hot lead on one of the TS connectors) ...


4

That's microphone and headphone cable combined, having a TRRS 3.5mm audio jack of unspecified pinout, with 2.5mm mono audio jacks at the other end. There's no better name for it, and your picture sadly doesn't fully specify how the microphone and two stereo channels should be connected to the TRRS, which depends on the device you plug this into. So: "...


0

Some Bypass capacitors of 104 pf connecting to the power pins (+ve and ground) of the MP3 module solved this problem. Noise is quite low now.


0

The old-style canonical way to do this was with VCA. But, with good performance, they are horribly expensive. If you need it directly controlled by an analog voltage that's IMHO the only way to do it. Just be careful since most VCA have logarithmic scale while you said you need a linear scale (it's a crossfader). A cheap way would be to use a JFET as an ...


1

in effect you need to multiply one signal by the control input (scaled to be a number between 0 and 1), and multiply the other input by 1- the control input, then add the two together. There's two basic approaches, either you buy an expensive multiplier chip, or you convert the control input into a fast PWM signal and use that to drive an analogue ...


1

Sounds like an application for analog multipliers: For example, the AD633 You could need two, plus a 3.0V reference voltage. If you need the exact transfer function you state, you'd also have to multiply the result by 3.33 (so an op-amp and a couple resistors). So you'd feed Vc into X2, 3.0V into X1, 0V into Y2, V1 into Y1, 0V into Z on the first. Output ...


0

Datasheet noise level (typical only) is given: Reference voltage is not stated. If we assume it's 3.6V then the resolution is 0.8mV and we'd expect to see about 4 different values with the input shorted. However with a non-zero DC voltage applied then other noise sources come into play. Depending on your conditions (and layout) 10 different values is not ...


3

Let's analyze this circuit by inspection. We have two capacitors, meaning we expect a second order transfer function. If we're lucky, no poles are complex conjugate, so we can express the transfer function as: $$ T(s) = \frac{N(s)}{D(s)}= \alpha\frac{(1+st_L)(1+st_H)}{(1+s\tau_L)(1+s\tau_H)}$$ We can first of all derive the poles by inspection. Using a ...


1

Generally, electret mics with a 3.5mm TRS (tip-ring-sleeve) male connectors have a connection like the following: Tip and Ring are shorted and connected to Mic+. Sleeve is connected to Mic-. So if you use the adapters shown, L and R will be Mic+ and GROUND will be Mic- (i.e. GND). An electret microphone capsule has a FET inside, so there should be a load ...


1

If you realy had to use the device you show as an amplifier (although it would not be suitable) you could tap the single + output to feed into a dummy load of say 8 Ohms, and to your analogue input of the MCU, as long as you also connect the - amp ouput to an 8 Ohm dummy load. The problem is that the device you show is not really suitable for microphones. ...


2

I have no idea what it's called I searched using "stacked 3 way jack socket" and found this straight away: - Picture from here


0

If you really wish to not use a transformer, one possible solution is to use a isolation power supply for your amplifier circuit. However, your design is not going to work because you didn't isolate the ground. The whole point of using an isolated power supply is to isolate both the ground and power. A true isolation power supply will have two input, power ...


0

Thumbs up for trying to build the circuit out of discrete components, and for trying to understand how these things work. If you'd like to learn how these things are normally done, in principle, a good google query might be class AB output stage. For a start, look for topologies that have an output "totem" of NPN+PNP transistors connected by their ...


2

You are basically shorting out TDA2030 inputs to ground in AC via C18/C19. You need to add another resistors between pin 1 of each channel (IN+) and half of power supply (R3/R4 and R11/R12). Audio signal goes into pin 1 like you do now from a capacitor. Look closely at TDA2030 datasheet for single supply schematics.


0

Your goal is to have an amplifier that can drive a typical speaker given a typical audio source. The most important early steps to take are: Learn everything you can about your audio source (or range of audio sources.) Learn everything you can about your output system (speaker/headphone/etc.) Work through the details you've learned about your audio source ...


1

Your output stage has idle state Ic =0,8mA The AC amplitude of the non-distorted output can be max. 0,8mA AC current 0,8mA peak brings to 8 Ohm about 2,6 microwatts power. That's not especially loud although it can be audible in quiet environment with a sensitive speaker. Calculations: P=0, 5 * R * Ipeak^2 = 0,5 * (8 Ohm) * (0.0008 amps)^2 = 2.56 millionth ...


0

I made the circuit exactly as you drew it in a simulator and as you say, it doesn't do much of anything. But when I moved the emitter capacitor CE2 to the actual emitter instead of the base, it worked fine to amplify a 1 mV (peak) signal, but even a 10 mV signal would cause Q1 to clip. This indicates that the amplifier is amplifying too much if your goal is ...


0

Use a Motorola speakerphone IC. They made a few versions of them. When you speak, the other end's mic is muted. When the other end speaks, your mic is muted. Background noise is cancelled from messing with the switching.


0

The answer is considerably simpler than other answers have indicated. In this circuit, the second 1458, which is used as a comparator, really needs to be powered by a negative voltage on pin 4. Once you do that, you should ground pin 3. The problem is that the 1458 (which is basically a dual-unit version of the now-ancient 741) was intended for split-supply ...


0

It looks to me like the mic preamp (3904 circuit) has a gain of approx. 20, or around 26 dB. Seems a bit low to me, but I didn't delve into the modulator. Also, an LM358 is (((not))) a dual 741. The circuit page is incorrect. The 358 has more noise and more output stage distortion, but also has what were in 1971 amazing input and output stage operating ...


0

Microphone Signal Level The schematics are there to read. The microphone input looks like this: \$R_{66}\$ and \$C_{64}\$ help isolate the circuit from the main \$+12\:\text{V}\$ power rail. Think of \$R_{66}\$ kind of like a set of bed-springs for the circuit. \$R_{60}\$ tells me this input supports electret microphones. They pull up to about \$1\:\text{mA}...


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