New answers tagged audio
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Thank you guys, I just thought of a very simple solution: get 2 dual log pots, drill a hole at the armature on one of them and glue the end of the shaft of the other to the inner plastic part of the of the drilled pot. Just opened a pot up and figured out it ain't a very hard job to do.
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No this will distort the audio with quadratic variations for a fixed bias.
You want maybe a four OTA’s for a voltage controlled gain.
This is a JET designed for linear low noise differential amplifiers not log attenuators.
You will get quadratic distortion on the input signal.
< Three bucks for 1 quad log taper pot does not sound as bad as “hella ...
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First, the \$R_{DS(ON)}\$ vs. \$V_{GS}\$ of a FET depends on a number of quantities which are not normally defined in the datasheet, and which vary from part to part and over temperature. There is not a nice linear, unchanging relationship between \$V_{GS}\$ and \$R_{DS(ON)}\$.
Second, your FETs will never be matched close enough, probably not even if they ...
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A series of two GPIO ports can't supply 10V at significant currents. That's not what GPIO ports are capable of.
Also, it's not even clear whether you'd get the sum of voltages – if both GPIOs were fed from a power source with a common ground, you've just shorted one supply to ground with the other.
This all is a terrible idea and you might really need to ...
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The basic problems with your circuit are extreme sensitivity to DC bias point of the upper FET, interaction between DC bias point and gain of the input transistor and lower FET, and low AC gain.
I put your circuit into LTspice and made a few changes to improve DC stability and AC gain:-
To remove the interaction between M1 and Q1 biasing I AC coupled the ...
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Class A Complementary FETs will always be a struggle with 1 BJT driver
low open loop gain, thus >10% THD even with negative feedback since it is very quadratic.
only get 16Vpp from a 24V supply.
loss of low-frequency response with a single supply requiring large Cap may have poor characteristics.
poor supply ripple rejection (hum etc)
too many pots that are ...
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That isn't distortion, that is noise caused by a ground-loop. Like others mentioned, a 9V is not capable to deliver a huge amount of current. Use an USB powerbank or such to provide power. Besides 9V is a way to high to power the amp (and also I guess the BT-module), the amp accepts 6V at a maximum, better is to use 5V.
The problem of the noise is a cause ...
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You're thinking about this backwards. What you're describing is a full-matrix mixer, in which each of the outputs can have a mix of any of the inputs. In other words, treat each output independently as an N-channel mixer, where N is the number of inputs. Then build as many mixers as you have outputs.
You might need buffer amplifiers (unity gain, low output ...
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I am stuggling to find a component which can assign one signal input to 0..8 (or 0..16 if available, for stereo sound) outputs.
Because that doesn't typically exist – you'd use a system of analog switches to achieve that. So, basically, for every single connection you'd want, you'd have to have a switch that's either open or closed.
In essence, each of ...
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MOSFETs have very large input capacitance (several thousand pF).
To drive this effectively you will need to run the driver stage at higher current with a lower collector resistor.
Although feeding the collector resistor from the drain of the FET gives DC feedback to stabilize the operating point it will have the disadvantage of limiting the output swing as ...
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There is no "keeping the flow in one direction". You just hook everything up in parallel -- ground to ground, ring to ring, tip to tip. If you get some cable with the right number of conductors and female contacts to match, just hook up like to like and you should be OK.
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I don't know if your question received a proper answer and the problem is solved or it is still actual.
Anyway, your Fostex already appears to have an interesting bias loop control, due those parallel FET working as controlled resistors by a tension coming from an upper stage that is not full displayed. It is possible that this circuit to do a kind of HX ...
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Well, if "Audio quality is not a priority", why don't you keep the speaker and use a microphone for the recording? Choose a quiet room, night time and give it a try. Some post-recording noise removal in a software like Audacity would be a good thing.
It also checks all your requirements:
Simple
Galvanic isolation
No additional power supply
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One approach would be to solder a more manageable size wire to the shield. Then cover all the messy bare wire, including the solder join, with heat-shrink tubing.
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Below shows a high pass sallen key filter on the left, and a low pass sallen key filter on the right. The resistors and capacitors are swapped. (Both high and low pass filters form a band pass). If you want to convert one sallen key configuration to the other, then swap resistors and caps
Source:
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An actual design will usually be forced to balance various constraints. And it isn't always the case that you can follow only one recipe all the time. Often enough, you start out with some assumptions you feel are better and then find some reason why you can't go that way and you have to go re-assess your assumptions and make different trade-off choices. So ...
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I believe this is the circuit you are talking about
In order to get the maximum possible output voltage excursion at the collector without clipping, the transistor needs to be biased so that the quiescent collector voltage Vc is approximately half way between Vcc and Vre.
The design has to start somewhere and Iq of 1mA is chosen in this example as a ...
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The main reason that the 1.5V emitter voltage design point is chosen is to reduce the effect of temperature variations and device variations on the bias current.
The voltage across the base-emitter junction will change by about 2mV per degree C. Since the voltage at the base is assumed constant the voltage across the emitter resistor will increase by 2mV/...
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Your understanding of the Rbe is correct and the emitter R increases Rin to Re*hFE and thus reduces the exponential input voltage effects by making the input impedance more constant.
With V(Re) = 1.5V this reduces the nonlinear transformation of input voltage to emitter current and thus reduces sine distortion significantly. But is overkill for a 12V ...
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Try removing the resistors and connect the ungrounded side of the speaker to SPK +.
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I don't know if anything else is wrong, but always bypass your power pins. And I mean always, on everything. In this case it means a 100nF cap between pins 6 and 4 of each chip, placed as close to the chip and with the shortest leads that are practical.
You should also have some bulk bypass -- I'd go with a 100uF electrolytic cap from +V to ground, just ...
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Data sheet circuit for microphone: -
From this you can somewhat conclude that the microphone might need a minimum DC current of about 3 volts / 680 ohms = 4.4 mA. This means that using 2 x 10 kohm to feed DC to the microphone is likely to be insufficient.
I would configure the MAX9860 for single ended inputs to match the circuit shown in the microphone ...
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I gather you are talking about Figure 12.12 from the 5th edition. Here's the schematic redrawn, just slightly:
simulate this circuit – Schematic created using CircuitLab
You can see a very similar circuit to the portion of the above circuit within the right box in an answer I provided here. Note that I didn't include the equivalent of \$R_{17}\$ in ...
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I agree with all the answers so far but would like to add a different perspective.
The purpose of Q6 as a Q7 current limiter is to make the AC impedance of Q7 so high that it does not introduce any significant AC nonlinearity. This permits the next 4(?) stages of NPN/PNP Emitter Followers to buffer the AC load impedance scaled by each hFE.
This results ...
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Quite a lot going on in these 3 transistors.
View Q6/R113/R114 as a high-gain common-emitter amplifier; the addition of C5 serves to cut the ACgain in half, but greatly boosts the AC Power Supply Rejection. Common-emitter resistive-load stages have gains of Vdd/0.026 at the most (when Vce is very low, as in this case); assuming 50 volt rails, the gain is 50 ...
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Q6 and Q7 form a constant current source of about 10.3mA at Q7's Collector.
Q5 is simply a relative mirror of 7mA using the Base voltage of Q7 as its reference. Q5 has nothing to do with the Q6/Q7 CC source.
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A little more than a Q6 VBE across R18 turns Q6 on, which then shunts away the bias coming through R13, R14 to Q7. So Q6 and Q7 servo the voltage across R18 to be roughly constant. The voltage on Q7 base is then its VBE more than this, so after Q5's VBE, the voltage on R10 is the same as that on R18.
I'm happy with Q5 and Q7 VBE's tracking, but R18 voltage ...
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too many words for a comment:
You are most likely seeing a slight drop in the power supply to the microphone board at the time the write occurs. This board is designed for experimenters and they amplify the audio signal referenced to VCC/2 of the MAX 4466 breakout board. Even a tiny dip in the power supply subject to the gain of this board will be ...
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My guess is that the low frequency is inducing vibrations in the camera. Video stabilization software is unlikely to react fast enough to remove them. Do an experiment - put a video camera on a bass speaker and hit it with some low frequency notes eg 30-50Hz and see what happens.
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The mic amp gives out 2Vpp max. Audio outputs will vary between devices, but this does not matter as the devices also have user controllable volume control. The schematics given in the Arduino forum link are OK to start. The AVR can work with 3Vpp signal when powered from 3.3V. The 100k resistors will bias the ADC input to 1.65V or half-supply voltage. With ...
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Big difference. You cannot put a LPF after a HPF nor the reverse. These need to be parallel from source. Cascading filters also depends on the source impedance Z(f) and load R.
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This provides a lower impedance at Vcc/2 than is possible with a charge pump since it is DC coupled while the charge pump is created by AC cap coupled diode clamps and doublers.
There is no problem with connecting a headphone and I suspect it has a floating DC power supply so the DC midpoint is floating with respect to earth ground and would also have no ...
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