"we end up spending a ton of money on batteries, because half of them go dead from shorts within 10 minutes of being handed out."
Is it save to short the 5 Volts to ground from a USB port?
Will it damage the port if shorted for a minute or more?
I think the computer will be damaged if the current is higher than 500mA.
Don't short the USB connector. (But the kids will short it, won't they?)
Uncle $crooge
If I'm not mistaken over current protection is part of the USB spec.
"Because I be what I be. I would tell you what you want to know if I
could, mum, but I be a cat, and no cat anywhere ever gave anyone a
straight answer, har har."
Well we hope you are right. If not, batteries are the cheaper option. However, it would seen that not so many people are involved. A decent protected bench supply might be best.
The class could build it, perhaps. (not my job, just saying)
well after i change the 22k to 47k this circuit sounds better...
at least i can "on" my output of my PIC and the transimitter do send the signal...
but now im facing the receiver error part as it does not receive any signal to "on" my PIC...
My resistor values were calculated for a 9V supply. If your supply is only 5V then they need to be changed.Originally Posted by disney_snoopy
"Sounds better??" The circuit turns on an LED when there is sound at the mic. It is not designed to produce an audio output since its output is DC not AC.
There is no receiver. The circuit is just an amplifier to light an LED.but now im facing the receiver error part as it does not receive any signal to "on" my PIC...
The output is a max voltage of about +7.3VDC when the supply is 9V.
The max output is about +3.3V when the supply is 5V.
Uncle $crooge
hmmm... i use the 47k for my 5V... can i know how does u design or calculate out the resistor value?
Actually my output is not an audio but is to trigger a high voltage for my PIC as input of my PIC. So this circuit does trigger more than 2.4V from the LED and flow to my PIC input. Hence, my PIC output does trigger as i put an LED on the output PIC port.
The only problem is my receiver part cant get the signal... i was not sure whether it is my transmitter problem or the receiver PIC programming problem...
With a supply of only 5V then you do not need the two resistors at pin 2 of the LM386. Remove the two resistors and the amplifier will be more sensitive.
Uncle $crooge
hi sir,
the two resistor that u call me to remove, i remove and i found that it was too sensitive and cause my RX side output always on.
Deleted...twice.
Last edited by flat5; 5th April 2009 at 05:45 PM.
This is what i said before, "The part at pin 3 is not a resistor, it is a variable resistor which is called a volume control. It adjusts and reduces the input level so that the LED is not turned on all the time".
Uncle $crooge
oh... i get what u mean alr...
However, the 1k and 10k that is above the microphone is for sensitivity also?
or any reason for putting this resistor?
The 10k resistor powers the FET transistor inside the electret mic.
The 1k resistor and 100uF capacitor decouple the supply to the mic so that there is no low frequency positive feedback.
If the 10k resistor value is less than about 4.7k then the sensitivity of the mic will be reduced. If it is more than about 22k then loud sounds might be distorted by the mic.
With the 10uF capacitor at pin 1 and pin 8 of the LM386 IC then its gain is 200.
Uncle $crooge
sorry. i still doesnt get what does u mean as in this circuit there is no 4.7kOhm.
It only have 47k but i have replaced it by 22kOhm. Can explain further???
Anyhow how to calculate the gain in theoretical?
As i find book it does have a formula
Gain = 30k/ (150 + ((1350 x R)/(1350 + R)))
However, can tell me where is the 30k, 150, 1350 come from? Im very blur of this...
You don't understand technical talk.
Use 47k.It only have 47k but i have replaced it by 22kOhm. Can explain further???
The datasheet for the LM386 shows its schematic and the 15k, 1350 and 150 ohm resistors that determine its gain. I was wrong before, its DC gain is 20, not 10. The datasheet for the LM380 explains why its gain is twice as high as calculated.Anyhow how to calculate the gain in theoretical?
As i find book it does have a formula
Gain = 30k/ (150 + ((1350 x R)/(1350 + R)))
However, can tell me where is the 30k, 150, 1350 come from? Im very blur of this...
Since it does not drive an LED in your application I reduced the values of its capacitors and used a 100k load resistor.
Uncle $crooge
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