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It's the power source capable of delivering the 4 Amps that the fan requires?
In the datasheet says that "With control signal lead disconnected, the fan will spin at maximum speed". It also says "Start at low speed, after 10 sec running at full speed".
Best regards
Hi Nishanth, the capacitor acts as an open circuits because it's capacitive reactance (Something like It's resistence in AC) depends of the frequency of the signal in AC: \[X_{c} = \frac{1}{w{\cdot}C}=\frac{1}{2{\cdot}{\pi}{\cdot}f{\cdot}C}\]. As the frequency gets high in the previous equation...
Hi Nishanth !!
It's a classic (non inverting) integrator. For high frequencies the capacitor acts as an open circuit, while the resistor provides the needed feedback in DC to make it stable.
Best regards
What kind of problem do you have? (It is a good idea to add some kind of extra information to a problem. Because nobody will help you if you just paste the code and says "it doesn't work")
- It doesn't work at all?
- It works randomly?
- It sends garbage?
Best regards.
According the LM555 (google it) datasheet these are the two ecuations you must use:
\[f=\frac{1}{T}=\frac{1.44}{(R_A+2*R_B)*C}\] (1)
\[D=\frac{R_B}{R_A+2R_B}\] (2)
With (2) you can find the relation between RA and RB. You need a 1s pulse (On time) after 20s (Off time), this gives you a total...
I don't remember exactly where it it's located in the datasheet, but when you write to 16 bit registers (Like UBRR) you must write the high byte first.
Try inverting the order in which you write in UBRR, this way:
UBRRH = (BAUD_PRESCALE >> 8);
UBRRL = BAUD_PRESCALE;
Look at the page 135 of...
I think you forgot to read the button position in the main loop.
void main(){
int l;
while(1){
read_button(); // read the button to know if is pressed or not
if (button==0) // if switch pressed execute this part{
...
}else{ //if not pressed execute this part
...
}
}
}
You need to read the...
It is not very clear, but it seems the VREF/2 pin in the ADC is left unconnected, so according the datasheet, the ADC value goes from 0 (0h) to 5V (FFh).
This is why the code first do:
a=5*adc_val;
b=a/255;
(Now I realize that this is the part of the code you actually understands)
The rest of...
This code implements (some kind of) interpolation/value adjust:
I assume you get the value of the temp sensor in "adc_val".
After the execution of this piece of code you get the temperature value from...
Re: problem
Have you tried some solution? I don't think somebody will give you a complete solution, if you don't make some effort at first. As you say it is YOUR project, and you must do something before.
Best Regards.
The problem is the amplification factor, and the voltage of the source, and/or the power supply.
The gain of the differential amplifier is: \[\frac{V_{out}}{V_{2}-V_{1}} = (1+\frac{2*R_{2}}{R_{1}})*\frac{R_{6}}{R_{4}}\], with the resistor values of your circuit you get...
I agree with you, but sometimes if somebody gives you a piece of code, with a good explanation, you learn a lot more (and is alot less frustating that search over internet for a code that barely applies in your situation), don't you think?
Best Regards.
Re: help needed
I don't have EWB (Electronics WorkBenck right?), but here you have good start for an Instrumentation Amplifier with Operational Amplifiers. You can see in that site an explanation of the purpose of each resistor.
This means that the amplifier, amplifies the difference betwen...
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