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my explaination

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a voltage across a capacitor ,which charges through a resistor, is exponential (It has nothing to do with 555) . Vc=Vo(1-exp(-t/RC)) if i remember correct , Vc voltage across the cap and Vo the voltage applied , t time and RC -> time constant. the comparators just monitor the voltage across the cap with its internal ref and switches on and off the discharge transistor.
 
Yes!! You are right! I got it!
By refering the report i've done before, i know what r u talking about. Thanks a lot :D ;) ;P
 
hi,
i just ask, if the tank circuit is connected to the + and the 100K connected to -ve, what will i get? opposite result? when it detects nothing, low, when detects the field, high..?
instead of the LM393, can i use 741?

thanks..
 
bananasiong said:
if the tank circuit is connected to the + and the 100K connected to -ve, what will i get? opposite result?
No. You would get nothing. The LM393 has PNP transistors at its inputs that need a negative voltage (or ground) at their input for these transistors to work properly.
The circuit will have its output inverted if you swap its inputs.

instead of the LM393, can i use 741?
No. The LM393 comparator works fine with the bases of its PNP transistors at its inputs DC-grounded. A 741 doesn't work if its NPN input transistors are too positive or are too negative.
Also, a 741 is an opamp that is slowed down so that it doesn't oscillate when it has negative feedback applied. So at 10kHz its max gain is only 90. The LM393 is a comparator that never uses negative feedback so it operates as fast as it can, and has a gain of 200,000 at 10kHz.

thanks..[/QUOTE]
 
No. You would get nothing. The LM393 has PNP transistors at its inputs that need a negative voltage (or ground) at their input for these transistors to work properly.
The circuit will have its output inverted if you swap its inputs.
?? I don't get, you said i would get nothing, and the output will be inverted..??

akg said:
there shouldn't be much difference in ur case
Y different answer? I think audioguru is correct with his defination..
 
bananasiong said:
?? I don't get, you said i would get nothing, and the output will be inverted..??
You asked if the tank could be connected to the positive supply. Then the PNP input transistor in the comparator won't work because it would be cutoff.

You asked how to make the output the opposite (inverted). Just swap the input pins of the comparator.
 
You asked if the tank could be connected to the positive supply. Then the PNP input transistor in the comparator won't work because it would be cutoff.

You asked how to make the output the opposite (inverted). Just swap the input pins of the comparator.
I think i didn't state clearly. I mean, the tank connected to the non inverting and the 100K to the inverting, will i get the output inverted (no signal-->o/p=0, with signal-->o/p=Vcc)?
So the output will be inverted, right?
 
bananasiong said:
I think i didn't state clearly. I mean, the tank connected to the non inverting input and the 100K to the inverting input, will i get the output inverted (no signal-->o/p=0, with signal-->o/p=Vcc)?
So the output will be inverted, right?
Yes, that is correct.
 
Thanks... Can i know, what is the input bias, offset current and voltage? All of them from the datasheet, i don't even know how to understand them. I'm going to complete my diploma coming August, but all of them are not in our syllabus. Can i learn them here??
 
bananasiong said:
Can i know, what is the input bias, offset current and voltage? All of them from the datasheet, i don't even know how to understand them.
There are tutorials on the web, but most are many pages long.
If you know how transistors work then opamps are easy to understand.

The spec's change with temperature change but are stated at 25 degrees C for the LM393:
1) Input Bias Current. The base current of each input transistor. The LM393 has PNP input transistors. Typically 25nA, 250nA max.
2) Input Offset Current. The difference between the input bias current of one input transistor and the other input transistor. Typically 5nA, 50nA max.
3) Input Offset Voltage. The amount of input voltage change that is needed to make the input transistors perfectly DC balanced. Typically 1mV, 5mV max.

One input is connected directly to ground through the low resistance input coil. The other input has input bias current through a 100k resistor. Use Ohm's Law to calculate the voltage across the 100k resistor. Typically 25nA x 100k= 2.5mV, 25mV max due to input bias current. The offset current and voltage changes it a little. Since the max input offset voltage spec is higher than the typical voltage difference caused by the 100k resistor then the output voltage will be backwards with some LM393 comparator ICs.
 
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Thanks a lot!!
3) Input Offset Voltage. The amount of input voltage change that is needed to make the input transistors perfectly DC balanced. Typically 1mV, 5mV max.
I don't really understand this. I thought the input offset voltage is simply the multiplication of the input offset current with the resistor??

Use Ohm's Law to calculate the voltage across the 100k resistor. Typically 25nA x 100k= 2.5mV, 25mV max due to input bias current. The offset current and voltage changes it a little.
why 25mV? What do the offset current and voltage change?

Since the max input offset voltage spec is higher than the typical voltage difference caused by the 100k resistor then the output voltage will be backwards with some LM393 comparator ICs.
How is it backward?? How if i replace the 100K with a 200K, then it is 5mV, same as the input offset voltage.

I'm really a newbie.. Thanks for telling me these..
 
bananasiong said:
I don't really understand this. I thought the input offset voltage is simply the multiplication of the input offset current with the resistor??
No. Without any resistance feeding the same voltage to both inputs, the output will be saturated negative or positive due to the input offset voltage. Now give a small voltage difference between the inputs to make the output linear and with a voltage about half-way netween negative supply and the positive supply. The small voltage that you added is the opamp's iput offset voltage.
The voltage across the 100k resistor at one input is an external offset voltage that your circuit makes.

why 25mV?
The input bias current is between the typical amount of 25nA to a max amount of 250nA. The voltage is 25mV across the 100k resistor if the comparator has the max input bias current.

What do the offset current and voltage change?
They change the differential input voltage just like a DC signal at one input.

How is it backward?? How if i replace the 100K with a 200K, then it is 5mV, same as the input offset voltage.
Yes, but not enough to make the output do what you want. Maybe the resistor should be 220k or 270k so that all LM393 ICs will work.
 
Yes, but not enough to make the output do what you want. Maybe the resistor should be 220k or 270k so that all LM393 ICs will work.
U mean 200K is not enough but 220K and 270K will work? But i use 100K, it is still ok.
For the rest, i think i get your point, thanks ;)

For the internal npn at the output. Is it fully saturated at 6mA of Ic? With 5V of supply and the 1K resistor, I get 5mA of Ic right? What will happen if the Ic is too high or too low??
 
audioguru said:
You asked if the tank could be connected to the positive supply. Then the PNP input transistor in the comparator won't work because it would be cutoff.
AG he asked what would happen if the i/p pins of comp: are swapped , it won't do affect anything in his ckt right?(well he would get waveform inverted , but it is no concern here)
 
sheeet, I've done this course twice(don't ask) and what I wouldn't have given for all the tech heads to do the "difficult" questions. As it was the second time around I used to do a month in a weekend.
 
bananasiong said:
U mean 200K is not enough but 220K and 270K will work? But i use 100K, it is still ok.
It works with your LM393 but not with all of them. If the input offset voltage is at the max of 5mV and the input bias current is also at the typical value of 25nA, and add together in the opposite direction to the voltage across the input resistor to ground, then a 200k resistor might or might not work properly because both input voltages would be the same. If the input bias current was lower then it wouldn't work properly.

For the internal npn at the output. Is it fully saturated at 6mA of Ic?
Its min spec is 6mA with a saturation voltage of 1.5V which is very poor. With only 4mA its saturation voltage is 0.4V or less. It doesn't have a high output current because it is "low power".

With 5V of supply and the 1K resistor, I get 5mA of Ic right? What will happen if the Ic is too high or too low??
With a 5V supply and a 1k output resistor, an LM393 at min spec will have a saturation voltage of nearly 1V and a current of nearly 4mA. If the load draws more current then there will be a higher saturation voltage.
 
Yes.. I will replace it with a 220K. Thanks

Currently i'm using this circuit with my microcontroller using different power supply (but they're common grounded), it works well. How if i want them to share a same source. Just connect them in parallel? Anything needs to be changed?
*I'm using Motorola 68HC11E9 microcontroller, 4 AA batteries.
 
Use the same power supply for both circuits. The 555 might stop working when the battery voltage drops to less than 4.5V but a Cmos 555 will work down to 1.5V.
 
Nono.. i mean, i want to use the same supply for the electromagnetic field detector (LM393) and the uController together. Not the 555. i will use one supply to the 555.

what is cmos 555? i'm using NE555
 
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