Hey all, first off, i`m still a newbie to electronics so bear with me.

i`m trying to build the EMG circuit found on this instructable: http://www.instructables.com/id/Muscle-EMG-Sensor-for-a-Microcontroller/?ALLSTEPS

And i was wondering if i could use 1uF Electrolytic Capacitors instead of the 1uF Tantalum capacitors and the 1uF Ceramic capacitors on this circuit
since i couldn`t find any tantalum or ceramic anywhere in my country.

 

A bit hard to see that schematic but I see some 1M resistors. You're likely to run into leakage problems with aluminum electrolytics.
What country are you in?

 

It certainly won't damage anything if you do and I suspect you wouldn't notice any difference at all.
Two of the capacitors are across the supply so leakage there won't be a problem and the other one is across an 80.6k resistor so i don't think that will be either.

 

Sudan, any other alternatives ?

 

Okay .. any tips about handling the polarity ?

 

I also read some where that i should use a higher voltage cpacitor when using alternatives, for example this circuit`s suppy is 18v so should i use a 100v+ capacitor ?

 

100V is a bit excessive, 50V would be good.
On the circuit, the curved side of the capacitor is the negative connection.

 

Thank you

 

35 or 50 volt parts would be fine. The caps across the power supply are marked as to polarity. The cap near the output should have the + on the left and - on the right.

 

I agree completely about C1 and C2.

Are you guys sure about C3 and C4? Just because whoever drew that showed them polarized one way doesn't mean (to me) that it's a fact.

 

I agree completely about C1 and C2.

Are you guys sure about C3 and C4? Just because whoever drew that showed them polarized one way doesn't mean (to me) that it's a fact.

The TS was asking about the 1uF caps. C3 is 0.01uF and should be non-polar - the DC voltage across it is not predictable.
C1, C2 and C4 are 1uF and the polarity across them will be as marked.

 

C4 is used in the second to last stage. Prior to that stage, there is a rectifier circuit, and the polarity of that rectifier circuit is such that the voltage across C4 will be + on the left (actually virtual ground on the left) and - on the right. The final stage is an inverter, so the final output will be positive going.