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Adafruit MAX4466 mic board connection to audio in on mini DVR

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I also have the option of supplying everything with 7.4v using two 18650 li ion rechargeable batteries.... would 7.4v to the opamp work as long as I divide the voltage to 3.7v+ into pin3? That could be the solution to this altogether.
Lithium batteries are dangerous (explosion or fire) if discharged to less than 3V then normally charged. Your idea has nothing to detect a low voltage then disconnect the load. I do not know if your charger will detect a low voltage then attempt to charge safely.
Their voltage is 4.2V when fully charged so two cells produce 8.4V that might zap your DVR and their low voltage is 6V that is too low for many 5V regulators But a "low dropout" 5V regulator will work.
 
An electret mic is not an LED. An LED sets its own voltage but the resistor value, the supply voltage and the current draw of an electret mic sets its voltage that your calculator did not calculate.
You calculated 19k ohms but used 22k ohms so if the mic current is 0.50mA then the voltage across the 22k resistor is 0.5mA x 22k= 11V so the mic gets only 1.0V and barely works. If the 22k resistor is 5% high at 23.1k and the 12V is 5% low then the mic gets no voltage and does not do anything.

Most electret mics have a maximum voltage rating of 10V so why not use a 15k resistor so that the mic has 4.5V across it?

The mic will work but the noise might not be gone.

I was not aware the mic could take up to the neighborhood of 10v for some reason. Yes, I will definitely reduce the resistance to 15k. I will be sure to only use that calculator for my leds from now on.

On the last point, I will build another circuit with the regulated 5v feeding everything and see what results I get. I may eliminate the resistor before the mic altogether for that test circuit. So, 5v to the mic, the DVR and the circuit. I guess I can leave the voltage divider as is and expect 2.5v to pin3...
thanks again for your help and patience.
 
Lithium batteries are dangerous (explosion or fire) if discharged to less than 3V then normally charged. Your idea has nothing to detect a low voltage then disconnect the load. I do not know if your charger will detect a low voltage then attempt to charge safely.
Their voltage is 4.2V when fully charged so two cells produce 8.4V that might zap your DVR and their low voltage is 6V that is too low for many 5V regulators But a "low dropout" 5V regulator will work.
I have a BMS pcb connected for charging the 2 li ion batteries.
 
I was not aware the mic could take up to the neighborhood of 10v for some reason. Yes, I will definitely reduce the resistance to 15k. I will be sure to only use that calculator for my leds from now on.

On the last point, I will build another circuit with the regulated 5v feeding everything and see what results I get. I may eliminate the resistor before the mic altogether for that test circuit. So, 5v to the mic, the DVR and the circuit. I guess I can leave the voltage divider as is and expect 2.5v to pin3...
thanks again for your help and patience.
You can't remove the resistor from supply to the mic.
There will be no signal from the mic without the resistor.
 
I was not aware the mic could take up to the neighborhood of 10v for some reason. Yes, I will definitely reduce the resistance to 15k. I will be sure to only use that calculator for my leds from now on.
You need to calculate the current-limiting resistor for an LED because an LED has a range of forward voltage. Look at the datasheet for an LED. A white or blue LED might have a minimum forward voltage if 2.8V and a maximum forward voltage of 3.6V and a "calculator" program does not know the voltage range of your LED. The calculator guesses that you bought thousands of LEDs, measured their voltages and picked one that is 3.2V. You must calculate the resistor value so that if the LED has a low voltage then it does not burn out or if it has a high voltage then it is bright enough.

I may eliminate the resistor before the mic altogether for that test circuit. So, 5v to the mic, the DVR and the circuit.
Then the mic will not have an output because its output is shorted to the +5V supply. The resistor is the load of the Jfet inside the mic allowing its output voltage to swing up and down with the signal.
 
You need to calculate the current-limiting resistor for an LED because an LED has a range of forward voltage. Look at the datasheet for an LED. A white or blue LED might have a minimum forward voltage if 2.8V and a maximum forward voltage of 3.6V and a "calculator" program does not know the voltage range of your LED. The calculator guesses that you bought thousands of LEDs, measured their voltages and picked one that is 3.2V. You must calculate the resistor value so that if the LED has a low voltage then it does not burn out or if it has a high voltage then it is bright enough.

Yes, I use several different types AND colors of leds. From 3mm to high power 'star'' types, red, green and infrared. They differ in voltage up to 2.5v and go from 20ma to 750ma. I do enter that info from the datasheets into the calculator and so far so good. All leds have been performing well.

Then the mic will not have an output because its output is shorted to the +5V supply. The resistor is the load of the Jfet inside the mic allowing its output voltage to swing up and down with the signal.

I understand. Thank you very much for explaining. It is the same principle as when I have a pull down resistor on the base of my 2N3904's for switching the gate on the transistor.......
 
Yes, I use several different types AND colors of leds. They differ in voltage up to 2.5v and go from 20ma to 750ma. I do enter that info from the datasheets into the calculator and so far so good.
Yes of course different LEDs have different voltages. But I am saying that one LED part number will have some LEDs with a low voltage and some will have a high voltage. A "calculator"simply guesses at one voltage that produces a current too high or too low. You are lucky that your current was not high enough to cause damage.

The same principle as when I have a pull down resistor on the base of my 2N3904's for switching the gate on the transistor.......
No. The load resistor of a transistor to its power supply is at its output, a base resistor to ground is at its input which produces a completely different effect.
 
Update... I have a 9v battery going to the DVR (regulated to 5v). I have a separate 12v supply to my preamp & mic. I get perfect audio recorded to the DVR. The preamp circuit is working great! However, I then connected my cctv board camera and monitor to the mix to add video, powered by the same 12v battery and got a noticeable hiss caused by the small 5" led monitor and a loud buzz caused by the camera. I then powered the camera and monitor with another 12v supply and got great audio and video recorded to the DVR. I need to get some wildlife audio/video recorded for a website over the next few days, so next week I will be troubleshooting to see if and how I can power the mic/sound circuit, DVR, monitor and camera all from one 12v, 7AH battery source.
 
Have you tried to filter the mic supply as AG said in a message #67 ?
Make the supply resistor R12 of series connected 10k and 4k7 resistors and put 100uF from their connection point to ground.
 
Have you tried to filter the mic supply as AG said in a message #67 ?
Make the supply resistor R12 of series connected 10k and 4k7 resistors and put 100uF from their connection point to ground.

No, I haven't yet. I will definitely try that one evening while down at the ranch. Thanks for the reminder. I recall a video omline where a guy had a steady strong buzz in his audio set up on a breadboard and he simply added a capacitor into the circuit and the buzz completely disappeared. Will try it for sure !!
 
Have you tried to filter the mic supply as AG said in a message #67 ?
Make the supply resistor R12 of series connected 10k and 4k7 resistors and put 100uF from their connection point to ground.
Like this? I can try it while in south Texas this week....
upload_2017-8-6_9-34-31.png
 
I need the resistor as a pull down so the electret can perform it's function?
No. The resistor is the load of the junction field effect transistor (Jfet) inside the electret mic and it feeds positive voltage and current to the Jfet. The Jfet in an electret mic uses about 0.5mA and is an impedance converter that converts the extremely high impedance of the "condenser" part of the electret mic to a usable lower output impedance. The electret material permanently stores 48VDC so that the condenser part of the mic works as a capacitive voltage divider that changes its extremely high impedance output voltage as the audio vibrates the diaphragm closer and farther from a fixed part of the capacitive voltage divider.
 
No. The resistor is the load of the junction field effect transistor (Jfet) inside the electret mic and it feeds positive voltage and current to the Jfet. The Jfet in an electret mic uses about 0.5mA and is an impedance converter that converts the extremely high impedance of the "condenser" part of the electret mic to a usable lower output impedance. The electret material permanently stores 48VDC so that the condenser part of the mic works as a capacitive voltage divider that changes its extremely high impedance output voltage as the audio vibrates the diaphragm closer and farther from a fixed part of the capacitive voltage divider.
Thanks for taking the time to explain. It's a lot to absorb, but I think I get the idea.

Did you see my schematic above based on jjw's input? Do I have it right?

Thanks again.
 
I have added the camera and monitor to my schematic, along with the mic filter. The DVR gets 12v stepped down to 5v. The mic & sound circuit gets 12v with the mic itself getting approx 4.5v. The cctv camera and monitor get the same 12v.
So everything is powered by the same 12v - 7AH lead acid or AGM battery. How does this look?
upload_2017-8-6_18-8-6.png
 
I am confused when you say "Aud in -" on the mini-DVR instead of saying "audio in ground".
 
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