syee10

Joined Mar 26, 2010
59
Hi,

I have a problem here everyone. This is really make me going to crazy. I have an exact connection just like the connection in Figure 11.
Suppose i have a 3V input to the X1 and 3V input to the Y1. X2 and Y2 is grounded. Z is also grounded. In this case i should get a 0.9V because 3X3/10 it will be 0.9V. I am sure my there is no wrong in my connection because i had check through it many times. The answer i get from the output W is around 12.2V. What happen to my circuit actually? Is my AD633 IC spoil or broken and give me this result? Please assist me =(

Ron H

Joined Apr 14, 2005
7,014
Have you checked the voltage on every pin (and I mean all 8 of them), right at the point where the pin enters the body of the IC? If you have, then your IC is probably bad.

syee10

Joined Mar 26, 2010
59
How to confirm the IC is spoil or broken? Do you have any methods to check?

eblc1388

Joined Nov 28, 2008
1,542
Ron just told you exactly how to check for that.

syee10

Joined Mar 26, 2010
59
Ok now i found what is the problem already. This is funny problem and i really dun know why this happen. Forget about what i post in the previous post, regardless which type of square wave i put into my input X1 and Y1. Square wave with positive peak and negative peak or square wave with positive peak and 0 is working fine with the AD633.
The funny thing is, when i connect a probe (connect to oscilloscope) to either my input X1 or Y1, the output is working fine! And when i disconnect the probe again from my input X1 or Y1, the output is always show me 12.3V!! Why is this happen? Why must i connect a probe to either X1 or Y1 in order to get a correct output?! Please comment..

beenthere

Joined Apr 20, 2004
15,819
What does the circuit look like (schematic)?

ifixit

Joined Nov 20, 2008
652

syee10

Joined Mar 26, 2010
59
The circuit is exactly same as the Figure 11 in the datasheet. The input to X1 and Y1 is a square waveform of 50Hz positive 1-5 high level voltage and 0 low voltage. What's wrong with the connection? Is the connection in Figure 11 in the datasheet miss out some components? Why must i connect a probe at either X1 or Y1 in order to get me a correct output?!

Ron H

Joined Apr 14, 2005
7,014
It may be due to oscillation. Do you have good power supply decoupling capacitors (typically 100nF) connected with short leads from pins 3 and 6 to ground?

syee10

Joined Mar 26, 2010
59
Yes i have connected the capacitor 0.1uF to the ground at pins 3 and 6. What you mean due to 'oscillation'?

Ron H

Joined Apr 14, 2005
7,014
Yes i have connected the capacitor 0.1uF to the ground at pins 3 and 6. What you mean due to 'oscillation'?
Many analog circuits will oscillate if they do not have low power supply impedances. The capacitors provide this. See this article for more information.
http://electronicdesign.com/article...ign-techniques-solve-high-speed-op-amp-s.aspx

If you don't know what oscillation is, you are an electronics beginner, and analog multipliers will probably be a challenge for you.

As others have said, your problem may be a loose connection.