How to double PCI clock?

An FPGA with an internal PLL and an external Schottky diode pair to power and ground should do the job. What is the energy content of the

11 volt spikes on the PCI bus?

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Rick

Or if you don't need an FPGA, a sufficiently fast 744046 and a flip-flop would do.

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Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

This is from xilinx' xapp311: ``` The PCI specification requires that a 5V PCI compatible device be able to withstand repeated application of 11V pulses sourced through a 22 ohm resistor without damage. For 5V tolerant pins, this mandates the use of an external clamp diode. The application of 11V to an unprotected I/O structure will cause damage. ``` that's all I know... I'm thinking playing with an atmel PLD not one of xilinx ones, but google turned up that pdf among others... Just adding an FPGA for clock gen seems excessive... I was hoping there were integrated devices for that...

I think you over estimate FPGAs.

1.4 mm square. Is that small enough? Xilinx is into the big. When you need something that isn't big and power hungry look elsewhere. Xilinx FPGA.

If you add a 22 ohm resistor in series and a dual Schottky diode clamp, you should have no trouble with 11 volt spikes on any part. If you want a device that will take 11 volts without separate protection, you will need to find a part designed specifically for that.

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Rick

Yes, I've seen the icestorm stuff. The part that scares me off is the `buy 3 units, coz you'll brick two debugging the software' as I will be perfectly able to brick two parts all by myself, then end up with -1 working unit....

I'll try that.

I sell FPGA boards and have shipped around 2000 units. Out of those I have had 1 FPGA failure in the factory where the device would not program. I have never "bricked" an FPGA and I don't even know what that means in an FPGA. An MCU can be bricked by overwritting a flash bootloader. Flash FPGAs don't have anything in Flash to be overwritten. If the icestorm tools are so bad that they damage FPGAs then you can always stick with the vendor's tools. They work well and are free.

I seriously doubt you will ever see 11 volt spikes on the bus. I'm not certain what this spec is for. I seem to recall that PCI allows for hot insertion, but that is probably compact PCI I am thinking of.

I guess my point is that FPGAs have many I/O modes that can operate compatibly with TTL and CMOS as well as many other standards. Due to advanced process technology and a lack of need for 5 volt tolerance they have dropped that feature, but it is easy enough to add back. There are also level converter chips of many flavors, but they seldom will withstand 11 volts on the I/O out of the box. I expect everyone just adds the series resistors and goes with the flow.

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Rick

11V.

The PCI bus is obviously multi-drop and therefore needs to be unterminated with worst case 5v signals reflecting to get your 11 (ish). However the PCI clocks (plural) are not on the bus they are generated on th e motherboard and distributed point to point by a clock buffer circuit. If I recall, they are simply series terminated and each PCI device just sees 5 V. Being plug in boards, the clocks might need significant electrostatic pr otection but it is a very long time since I designed such a board and I don 't recall.

Colin

Apologies to the purists for using google. I'm sitting behind a firewall co ntrolled in India and I'm not even going to wonder whether I could install a decent newsreader.

Thank you, that's good news.. a CY2303 will do, then...

Even if a PCI trace were driven by a 5 volt driver, it would have some output impedance and the trace (transmission line) would have a characteristic impedance acting as a voltage divider, reducing the incident wave. When the driver matches the load the incident wave is reduced to half the unloaded voltage. The reflected wave then brings the transmission line up to the full voltage of the unloaded driver. With a lower driver impedance than the transmission line the incident wave is higher than half and the reflected wave then brings the transmission line up to higher than the source voltage, but it won't be twice the voltage. That would only be the case of a true voltage source with no output impedance.

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Rick