15
This is not an amplifier. Q1 will clamp the signal at power-on to keep DC from propagating. If the input has +DC on it, the DC will propagate until C10 charges. Q1 will clamp the DC until C11 charges. After C11 charges, there is no base current path and Q1 is effectively out of the circuit. D1 keeps Q1's base from going negative on power-off.
9
All of this powered by a, say, 150W power supply with +/- 15v rails.
With 15V rails, a BJT output stage should output a peak voltage of about 13V. In a 8 ohms resistive load this is 21W peak, or 10W "RMS". Note "Watts RMS" really means "active power in a resistive load"...
So your amp needs to run on higher voltage rails, which means it can't use opamps ...
7
OK, I think there's many questions "hidden" in this one.
First of all, yes, lower volume means that the amplitude of your speakers vibrations is lower.
Why? Because the device with the knob is designed to change the way it operates the speaker when you turn the knob.
That can be as directly as "the potentiometer is used as a series resistor, so when that ...
7
How can I make sure my audio ground is totally protected against this
scenario?
The short answer is "with difficulty".
For instance, a static discharge (ESD) onto the enclosure will create a current pulse through to ground but your chassis is connected to real ground via an earth wire that has inductance. Inductance can be generally approximated as 1 uH/...
6
Not odd at all. Car stereos don't necessarily have ground and positive for the speaker, they can drive both wires in a H-bridge configuration. Both speaker wires idle at about 6V, and there would be 0V over the speaker, and this way it is possible to drive up to +12V to -12V over the speaker. They usually say this in manuals too that speaker negative is not ...
6
We can only guess, but it is most likely a buffer, both to speed it up and to reference it to the 3.3V supply going into L1.
If I were implementing something like that, I'd make sure that the 3.3V supply was exceptionally clean, possibly even regulated from some higher voltage for no other purpose than to provide power for that AND gate. The reason for ...
6
To get 150Wrms on an 8 ohm load you'll need 35xsqrt(2) volts on your DC rails... Plus and minus 50V. Plus a little bit for losses etc. 150 watts rms is a lot of power for home use. 150 watt amp will melt the voice coils of most home HiFi speakers. Use P=V^2/R, and remember Vpeak is sqrt(2)xVrms
Don't bother with toroids bridge rec etc... Making your own psu ...
5
That's an old site I had maybe 20 years ago. I had forgotten about it...
In retrospect I guess this memory distortion thing was just a red herring. This kind of stuff pop up periodically in the audio community, someone finds a magical reason that explains everything, it gets popular then is forgotten then it starts all over again with something else.
Sure, ...
5
Could an op-amp also be used for this, and how?
No need to use an op-amp if your signal is audio. Shifting a biased audio signal of 0 - 4 volts to an unbiased voltage requires only a RC high pass filter. It will remove the DC offset: -
You decide on the values of R and C so that the lowest frequency you wish to pass is: -
$$F_{LOW} = \dfrac{1}{2\pi RC}$$
5
A suggestion. Rather than approach this as one enormous project, approach it as a project with subprojects and learn along the way:
First buy off-the-shelf modules for a pre-amp, a power amp, pre-built power supplies for each (perhaps wall-warts), a case, volume controls, and make it work. Given availability of (normally Chinese) imports on Amazon etc., ...
4
The LM386 has two inputs; they are differential like so: -
Ignoring the ridiculous value of the input capacitor on pin 2 in your picture, you basically have both inputs fed from the same signal and that won't work very well at all. Use one of the standard circuits is my advice and watch out for power 0V currents flowing through input areas. A standard ...
4
The simplest way to fix your circuit is to build it as the datasheet recommends:
Notice the variable resistor connected to pin 3. You describe the problem as "noise when the music is playing," which makes me think what you actually have is distortion.
The most common sources of audio will have a line level output (PC) or a headphone output (smart phone, ...
3
Because the voltages are in rms. Remember that the rms voltage is the equivalent DC voltage source that delivers the same amount of power to a resistive load. The rms is given by
$$v_\mathrm{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2 \ \mathrm{d}t}.$$
In our case \$v\$ is a sum of pure sine waves. Hence it is easier to work in the frequency domain, if you ...
2
The AND gate's inputs are wired together, so it is acting as a simple pass-through buffer. This will probably be there because the source of the signal does not have a low enough output impedance to drive the RC filters that follow. Furthermore, it may also act as a level-translator for the input signal.
2
This is not really an answer, just help in identifying the problem.
the signal noise also vanishes if I disconnect the outputs of the USB sound card.
This means it could be a ground loop issue. So, please try this:
This means you desolder the end of each 470R resistor from the wire that goes to the signal wire of your TRS jacks, and solder them to the ...
2
Assume the output impedance of U2 is 0 ohms across the frequency band of interest. If so, then the impedance seen by the U3 circuit is 50 ohms at one end of the FREQ1 pot rotation, 0 ohms at the other end, and approx. 12.5K ohms in the center. Same for the U3 circuit. This introduces a non-linearity into the circuit's response to pot rotation.
Both pots ...
2
Frequency is proportional to 1/(R*C), so it won't be linear with R.
2
Yes, adding a base resistor (usually a 50 - 1000\$\Omega\$) is one solution to RF oscillation of an emitter follower caused by capacitive loading. It doesn't take much capacitance, tens or hundreds of pF can be enough, which is not much length of cable.
Often a ferrite bead is used, either in the base or emitter lead because it doesn't degrade the high ...
2
Your car radio's output amplifier is working in "bridge mode".
Figure 1. A typical bridge mode configuration.
Bridge mode amplifiers are popular in car audio systems. The output of each amplifier is biased to half-supply. (Since both sides of the speaker are at the same voltage in the quiescent state, no current will flow through the speaker.) One ...
1
Any common collector ( emitter follower) or FET source follower with a reactive load where the slightest resonant gain is >1 can result in emitter oscillation with any wire L and C, capacitance loads.
This can be avoided by attenuating the loop gain, improving phase margin or matching emitter towards load impedance. This may be done by;
adding ...
answered Jan 18 at 0:43
Tony Stewart Sunnyskyguy EE75
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1
I imagine that R6 prevents C3 from being unreasonable charged when the input is disconnected. It could charge up to around 15 volts given the presence of D1. It might charge up due to various reasons such as the presence of static electricity or mishandling.
So, if it charges up to say 15 volts (the op-amp input side being at 0 volts) AND someone connects ...
1
R6 is needed to define a proper DC voltage level (bias voltage) at the signal node. Since the other side of R6 is connected to ground, this voltage is 0 V (zero).
However if Vin has a DC voltage level of for example 1 V DC then R6 might not be able to pull the node back to 0 V DC. That's not an issue as C3 will prevent that DC voltage from reaching the rest ...
1
Given;
uncompensated discrete amplifier with closed loop gain of 20 dB , (R1+R2)/R1= 10x
high Q resonance at 5MHz,
~20dB/decade rolloff >10MHz
0dB gain at 50 MHz which becomes the GBW product.
We could compute the open loop gain
and determine the ω/RC breakpoint for some circuit Req and added Miller C to achieve 0 dB gain
and then expect 45 deg phase ...
answered Jan 8 at 19:16
Tony Stewart Sunnyskyguy EE75
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1
Is one of the output terminals of the power amplifier grounded internally? It often will be.
If so, then by connecting one of the outputs of the rectifier to ground, you are effectively shorting out the amplifier for one half of each AC cycle.
1
Your output is saturated open loop due to offset between sensitivity to R5, R9 values and hFE assumptions. I can tell by your choice of R9 = 595 Ohms and Vout = 14.3V
Increase R5 to say 4k7 until you get in the linear output range and test Aol with 1uVpp input. You don't have to null it.
Also your 4 Vbe output bias circuit needs to be more than 3:1 R ...
answered Jan 7 at 0:37
Tony Stewart Sunnyskyguy EE75
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1
Exposing myself to the danger of probably missing your intentions, my suggestion would be to use an OPAmp summation amplifier (or rather two OpAmps for the two of your channels) instead of the voltage dividers comprised of the two 470Ohms resistors.
Because the input impedance of an OpAmp is very high (at least megaohms), you can choose pretty much higher ...
1
It may be possible to hack into the audio amplifier by feeding your signal into the "top" tag of the volume control as this is a common point between the pre-amplifier and the "power" amplifier.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Tapping into the volume control.
Figure 2. A cassette adapter. Source: Wikipedia.
...
1
Such a simple passive summing circuit with too many inputs has another problem - its inputs are weighted and each of them depends on the number of the connected devices.
If all devices are connected, the input gain of each of them will be only 1/6. If some of them are connected (other inputs are unconnected), it will vary up to 1/2 (when only 2 devices are ...
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