hi eetech00
I got the IC's.... Now am staring to work on them... but I am a little confused with that 50k pot... grounding two legs of pot will make two resistance paths ...one complete 50k and the other adjusted resistance path form pin 3 through R1... Is it correct ??? what I mean to say is that with this configuration 50k path is always available independent of the position of potentiometer adjusted leg. like current has to take two paths ... should'nt it be like it has to only pass thro adjusted leg of potView attachment 75025

Once the current gets to the wiper on the pot, it will all flow through the wiper. That is because the wiper is wired as a short across the remainder of the pot. The current does NOT have to take two paths in this case of a short.

 

but the wiper need to be adjusted... it will be short if the wiper is on the extreme end.... may the wiper is some where at the middle then how is it considered short??

 

It shorts the resistance of the pot which is connected to ground in parallel with the wiper.

Yes, it will be a complete short if its at one extreme end, but that's the purpose of R1 in the circuit - to prevent a complete short to the pin.

You say it's a 50k pot. Lets look at the case where the wiper is in the middle. The portion of resistance on either side of the wiper is 25k. But since the wiper and one end of the pot is connected to ground, the remaining portion of resistance (the other 25k) is bypassed or shorted. Hence, it will not have any effect on the circuit; no current will flow through it.

 

Hi Ahoo,

djsfantasi is correct. The potentiometer is wired to function as a 0 -50k ohm variable resistor. It is wired correctly.
R1 will prevent Pin 3 from seeing anything less than 100k ohms.

When R2 is turned fully one direction, its value will be 50k ohms. When turned fully the opposite direction, its value will be zero ohms.
Anywhere in-between will be a value between zero and 50k ohms. Preset it to the middle of the range (25k) before adjusting the circuit.
It won't hurt anything if you don't, but it will make the circuit easier to setup.

eT

 

ok got it....... another question...... this freq-to-volt IC LM2907 need the signal to cross zero for it to function properly and give a linear voltage proportional to frequency... Is it the case with LM2917 too??? if it is then I guess the signal I will be getting from my hall sensor may not be crossing zero. Am I right????

 

Hi

No...the input to the 2917 does not need to cross zero.
The frequency of the sensor output will follow the spark frequency.

eT

 

thank you very much.....

 

hey guys can you tell me what will be the magnitude of the output voltage of hall sensor....I cannot understand the datasheet properly... Like what I am trying to say is that "will the hall sensor be able to cross the minimum value of voltage magnitude that can be detected by LM2917... By the way I just got my hall sensor today and am starting working on it.....

and one more thing can I replace LM317L with 7809 IC????????

What is the purpose of electrolytic capacitors (having polarity) in the circuit. Can I use the other one having no polarity??

 

Hi

No...the input to the 2917 does not need to cross zero.
The frequency of the sensor output will follow the spark frequency.

eT

hi eetech00
hope you will be fine..
I ordered the hall sensor online and I received them... but while you were drawing that circuit you missed one important point that is the strength of magnetic field around the spark plug wire.......... I think it is in the range of 26 Guass I guess... and the magnetic flux around the spark plug wire wont be that much strong.... just look at the formula and you will see that even for larger current values and shorter distances from wire the value of magnetic field strength is quite low....some where like for 20A and 1 mm distance it says like 40 Guass.. and I dont think that current in spark plug wire will be 20A range..

Am I right???

 

Any one there ................. Plzzzzzzzzzzzz reply

 

Any one there ................. Plzzzzzzzzzzzz reply

hi

I'm making some guesses, but...
Plug wire is about 7mm in diameter and using about .5m long wire is about 5k ohms. If voltage is about 10kv, current is 10k/5k=2amps
so..
r= 0.0035
I=2
Uo=1

1*2/2*pi*0.0035=90.94

So...I think its ok...

eT

 

[QUOTE="Ahoo, post: 780236, member: 249622
and one more thing can I replace LM317L with 7809 IC????????
/QUOTE]

If the supply voltage stays about 3 volts above the 9v supply, then yes, you can substitute, but if you need to adjust the supply
voltage later, you won't have the convenience of the adjustable regulator. If it were me, I would leave the adjustable regulator in..

eT