# DC value

• posted

Hello, I need to measure the average DC value of a square wave of 80MHz frequency (going from 0V to 5V, 50% duty). When I measure using a Digital Multi-meter I get 5V. Shouldn't I be reading 2.5V for an average value?

I was told by somebody to use a filter...can you explain why I need a filter and what kind of filter?

Regards

• posted

Chances are, whatever produces the square wave was designed to have a

50 ohm load on it. Without the 50 ohm load, the voltage is 2 times greater than with the load. It's 0V to 10V, or 5V on average, and your measurement is correct.

A 50 ohm load is required for impedance matching, and will pull the voltage down to the designed values of 0-to-5V.

Mark

• posted

** Huh ?

Sure about that 80 megahertz ??

........ Phil

• posted

Mark, So how do you measure the average value?

• posted

Yes, its 80MHz.

• posted

"ABC"

** Generated by what ?

Kindly supply the make and model of your square wave gene.

Plus - how do you KNOW the output is 5 volts above zero ?

...... Phil

• posted

You measured it correctly, but try again with 50 ohms between the output and ground, and your meter reading across the 50 ohms.

Mark

p.s. there's a small chance 75 ohms is the appropriate load.

• posted

high frequencies can confuse the digital to analogue converter in many meters. 80MHz is a very high frequency :).

add a low pass filter in front of the meter ---[10K]----+----[10K]---------+------------ | | source ===== .1u ceramic ===== to meter | | ------------+------------------+------------ (if that looks like gibberish paste it into notepad)

that should give results closer to what you expect, if your meter has an input resistance of 10Mohms this circuild will cause it to read low by about 0.2%

```--
Bye.
Jasen```

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