Bus Frequencies and Ohms Law

In summary: So by increasing the frequency of the bus, you're really just increasing the voltage on the processor side since the bus is effectively just a carrying medium for the data.
  • #1
Brown399
16
0
What does increasing the frequency of a computers main bus [the bus that connects the CPU to the chipset] imply about the voltage ["information"] traveling on a bus? (what IS the frequency of a bus effectively measuring, data units per second?)

The way i see it (please correct me if I'm wrong, which i will presume i am), increasing the number of 'ups and downs' in any given second will result in more voltage going through the bus (which means more will be denoted a logical 0 or 1, so effectively more data is passing through the bus).

If it was 2 cycles per second, and was increased to 4 per second, would that not imply that more voltage passes through the bus?

And if this is implied ( that more voltage passes through the bus at any given second) how can this be, without violation Ohms law...? How can a change in frequency change the amount of data per second that can be processed in any given second (I'm talking about overclocking, increasing a processor from 3.6->4.0ghz for example)? How can this be explained by physics without breaking fundamental laws (i presume this is because frequency has a more complex nature than the basics of Ohms law).
 
Computer science news on Phys.org
  • #2
Brown399 said:
What does increasing the frequency of a computers main bus [the bus that connects the CPU to the chipset] imply about the voltage ["information"] traveling on a bus? (what IS the frequency of a bus effectively measuring, data units per second?)

The way i see it (please correct me if I'm wrong, which i will presume i am), increasing the number of 'ups and downs' in any given second will result in more voltage going through the bus (which means more will be denoted a logical 0 or 1, so effectively more data is passing through the bus).

If it was 2 cycles per second, and was increased to 4 per second, would that not imply that more voltage passes through the bus?

And if this is implied ( that more voltage passes through the bus at any given second) how can this be, without violation Ohms law...? How can a change in frequency change the amount of data per second that can be processed in any given second (I'm talking about overclocking, increasing a processor from 3.6->4.0ghz for example)? How can this be explained by physics without breaking fundamental laws (i presume this is because frequency has a more complex nature than the basics of Ohms law).

It's best not to say "voltage through". Voltage is a potential difference across something, and that voltage difference causes current to flow through the thing.

Increasing the clock frequency of a bus indeed allows more data to pass through the bus per second. Since the data in a digital system consists of rectangular waveforms, there is a lot of harmonic content in the signals. Even if you only have a bus operating at 10MHz, the edges of the waveforms will be in the few nanosecond range, which means that the signal harmonic bandwidth goes up to about 0.35/rise time, or a few hundred MHz. This can cause signal integrity issues if the bus length gets to be longer than about 1/10 of a wavelength for the highest harmonics. Once you get to high enough frequencies, you need to treat the bus like a transmission line, and be sure to properly terminate it.
 
  • #3
berkeman said:
Increasing the clock frequency of a bus indeed allows more data to pass through the bus per second.

Data is represented by voltage in a computer, and therefor for more data to transfer through the same bus, more CURRENT (v/r) has to flow through that bus.

I've increased my processor about .7GHz, how is this possible without violating Ohms law?

I = V/RSilicon is has a constant resistance, so that obviously cannot reduce in magnitude and result in more current flow.

That only leaves voltage.

When i increased the frequency of my bus to 250 from 200mhz, the number of up-down cycles was increased. How does this increased frequency actually allow more data to transfer from the north bridge (on the main bus) to the CPU?
 
Last edited:
  • #4
Brown399 said:
Data is represented by voltage in a computer, and therefor for more data to transfer through the same bus, more CURRENT (v/r) has to flow through that bus.

I've increased my processor about .7GHz, how is this possible without violating Ohms law?

I = V/R


Silicon is has a constant resistance, so that obviously cannot reduce in magnitude and result in more current flow.

That only leaves voltage.

When i increased the frequency of my bus to 250 from 200mhz, the number of up-down cycles was increased. How does this increased frequency actually allow more data to transfer from the north bridge (on the main bus) to the CPU?

The digital signals drive basically rail-to-rail, so there is trivial voltage drop across the resistance of the bus traces. You are really mostly driving the capacitance of the bus (formed between the traces and the inner layer ground plane beneath them).

As you drive to higher frequencies, it takes more power, because I ~ CVf (Capacitance * Voltage swing * frequency). The current pulses take place when you change the voltage on a bus trace, either high-->low or low-->high. Holding the trace high or low takes very little current.

Hope that helps to clear up your doubt. You can read more about the basics of computers and digital signals at HowStuffWorks.com
 
  • #5
Brown399 said:
When i increased the frequency of my bus to 250 from 200mhz, the number of up-down cycles was increased. How does this increased frequency actually allow more data to transfer from the north bridge (on the main bus) to the CPU?

Bits of information are transferred at the system clock rate. If you increase the system clock rate, you increase the number of bits transferred per second. So you can read variables from RAM at a faster rate, for example.
 

1. What are bus frequencies?

Bus frequencies refer to the number of times a bus or computer component can send and receive data within a given time frame. It is measured in Hertz (Hz) and is an important factor in determining the speed and performance of a computer system.

2. Why is Ohm's Law important in understanding bus frequencies?

Ohm's Law, which states that the current through a conductor is directly proportional to the voltage and inversely proportional to the resistance, is important in understanding bus frequencies because it helps us calculate the amount of current that can flow through a bus without damaging it. This is crucial in designing computer systems and ensuring their proper functioning.

3. How do bus frequencies affect computer performance?

Bus frequencies have a direct impact on the performance of a computer system. A higher bus frequency means that data can be transferred at a faster rate, resulting in faster processing and better overall performance. On the other hand, a lower bus frequency can lead to slower data transfer and slower performance.

4. Can bus frequencies be overclocked?

Yes, bus frequencies can be overclocked, but it is not recommended unless you have proper knowledge and experience in doing so. Overclocking can result in an unstable system and can also damage computer components if not done correctly.

5. How can we improve bus frequencies?

There are several ways to improve bus frequencies, such as using faster and more efficient components, optimizing the system's BIOS settings, and making sure that the system is properly cooled. It is also important to regularly update drivers and firmware to ensure that the system is running at its optimal frequency.

Similar threads

Replies
9
Views
1K
Replies
10
Views
483
  • Classical Physics
Replies
6
Views
2K
  • Electromagnetism
Replies
16
Views
675
  • Introductory Physics Homework Help
Replies
2
Views
748
Replies
19
Views
3K
  • Computing and Technology
Replies
1
Views
3K
Replies
1
Views
828
  • Electrical Engineering
Replies
5
Views
750
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Back
Top