Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
Annu007,
I assume that you are referring to an asymmetrical square wave, since the DC value of a symmetrical square wave is zero. A simple single pole low pass filter will do nicely. Connect a DC voltmeter to the filter output. The time constant should be at least 10 X the period of the square wave, which means in your case, >(10X.0125) microseconds.
Regards,
Kral
Annu007,
The formula is correct for the case where the square wave goes only from 0 to a positive value (Vcc). In this case, the square wave is not symmetrical (it only goes positive). However, if the square wave goes both negative and positive, you must take into acount the signs of the voltages. For example consider the case where the voltage goes to +1 volt for 1/4 the period, and goes to -2 volt for 1/2 the period, and is zero for 1/4 of the period.
Let T = the period
Net area = (+1)X(1/4)T + (-2) X (1/2)T + (0) X (1/4)T = (1/4)T -1T +0T = -(3/4)T
.
The DC component = Net Area/T = (-3/4T)/T = -3/4
.
In all cases, the filter-DC voltmeter combination works.
Regards,
Kral
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
