Signal Bridge
Ethernet Bridge with a Data Rate of 10Mbps
Toronto, Canada -GAO Tek Inc. (www.GAOTek.com) is offering its Ethernet Bridge which is designed for broadband access of Ethernet and widely used to transfer data point-to-point from E1 channels to RJ-45 interfaces. By using the resources of existing E1 circuits in telecommunication networks, the Ether Bridge extends the Ethernet transmission distance and the scope of application while minimizing investment.
GAO Tek's optimized Ethernet Bridge, model RB 10M, complies with IEEE802.3 protocol standards and is fully compatible with Ethernet v.2.It provides two selectable timing modes: an internal oscillator mode and a line clock mode. In line clock mode the clock is derived from the received signal.The Ethernet Bridge also supports both full duplex and half duplex operation. The rate of percolation and transmission is 30,000 frames per second.Furthermore,this Ethernet Bridge offers a WAN interface compliant with ITU-Y G.703 standard and a 10M Ethernet port compliant with IEEE802.3 and IEEE502.1Q standards.
Visit http://www.GAOTek.com for more information or to purchase this product online.
For any sales inquires please contact:
1-877 585-9555 ext. 601 - Toll Free (USA & Canada)
1-416 292-0038 ext. 601 - All Other Areas
sales@gaotek.com
About GAO Tek Inc.
GAO Tek Inc. (www.GAOTek.com) is a global leader in research, development and manufacturing of high performance telecommunication testers,electronic measurement instruments, embedded development tools and other electronic products that serve the needs of electronic professionals internationally.
Toronto, Canada -GAO Tek Inc. (www.GAOTek.com) is offering its Ethernet Bridge which is designed for broadband access of Ethernet and widely used to transfer data point-to-point from E1 channels to RJ-45 interfaces. By using the resources of existing E1 circuits in telecommunication networks, the Ether Bridge extends the Ethernet transmission distance and the scope of application while minimizing investment.
GAO Tek's optimized Ethernet Bridge, model RB 10M, complies with IEEE802.3 protocol standards and is fully compatible with Ethernet v.2.It provides two selectable timing modes: an internal oscillator mode and a line clock mode. In line clock mode the clock is derived from the received signal.The Ethernet Bridge also supports both full duplex and half duplex operation. The rate of percolation and transmission is 30,000 frames per second.Furthermore,this Ethernet Bridge offers a WAN interface compliant with ITU-Y G.703 standard and a 10M Ethernet port compliant with IEEE802.3 and IEEE502.1Q standards.
Visit http://www.GAOTek.com for more information or to purchase this product online.
For any sales inquires please contact:
1-877 585-9555 ext. 601 - Toll Free (USA & Canada)
1-416 292-0038 ext. 601 - All Other Areas
sales@gaotek.com
About GAO Tek Inc.
GAO Tek Inc. (www.GAOTek.com) is a global leader in research, development and manufacturing of high performance telecommunication testers,electronic measurement instruments, embedded development tools and other electronic products that serve the needs of electronic professionals internationally.
About the Author
About GAO Tek Inc.
GAO Tek Inc. (www.GAOTek.com) is a global leader in research, development and manufacturing of high performance telecommunication testers,electronic measurement instruments, embedded development tools and other electronic products that serve the needs of electronic professionals internationally.
Signal Bridge
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if the AC signal wave is 10V peak to peak.....?
if the AC signal wave is 10V peak to peak.....then the peak voltage will be 5V but when i convert the 10V peak to peak signal with a bridge rectifier,what will be the output DC
If the rectifier is perfect (no forward voltage drop) and there is no filter at the output to store energy, then the bridge rectifier will convert the AC wave into its absolute value. That is a connected series of half cycles of a sine wave, all 5 volts peak. The average of that waveform is its DC component and that is the same as the average of a single half cycle. The average of a half sine cycle is 0.637 times the peak value, which is 5 volts, in this case, so that means the average (DC component) of the full wave rectified 10 volt peak to peak sine wave is 5 volts*0.637=3.185 volts.
If you are building a real circuit, not calculating for ideal ones, then you have to subtract the forward drop of two rectifiers that are in series with each half cycle. And if you are adding a capacitor filter to the output, the average will be closer to the peak voltage, because the capacitor will store extra energy at the peaks that supplies the voltage during the valleys of the rectified wave.
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Regards,
John Popelish