[Vinnie R.]

Has anyone bought an aftermarket clock (ex. LC AUDIO XO2) that outputs a 5V signal, but wanted to run it into a chip that runs on 3.3V or lower?  Was a simple resistor divider good enough to scale down the voltage?  What was the magnitude of the resistors (ohm, Kohm, Mohm)?

I want to preserve the clock waveform integrity and just remove the excess voltage.  

Thanks for all your input,

Vinnie

[randytsuch]

Has anyone bought an aftermarket clock (ex. LC AUDIO XO2) that outputs a 5V signal, but wanted to run it into a chip that runs on 3.3V or lower?  Was a simple resistor divider good enough to scale down the voltage?  What was the magnitude of the resistors (ohm, Kohm, Mohm)?

I want to preserve the clock waveform integrity and just remove the excess voltage.  

Thanks for all your input,

Vinnie

Hi Vinnie,
I put in a superclock into my SACD 1000 a while ago.  I was worried about the same issue, but I looked at all the chips the clock went to, and although they were 3.3V chips, they all handled a 5V input signal, so I did not have to worry about connecting a superclock.

If you build a resistor divider, I would think something in the low Kohm range maybe.  You don't want to go too low, because then the resistor to ground would load down the signal.  If you go too high, you're putting a lot of resistance in the signal path, and limiting the drive current for the clock signal.  Most new IC's these days don't require much input current, so drive current is not that big a concern, and long as you don't make your resistors too big.

Hope this helps.  Good luck.

Randy

[Vinnie R.]

Hi Vinnie,

If you build a resistor divider, I would think something in the low Kohm range maybe.  You don't want to go too low, because then the resistor to ground would load down the signal.  If you go too high, you're putting a lot of res ...

Hi Randy,

Thanks for your reponse.  Well, I tried a 1K and 2K resistor divider to make the output 3.3V.  The output went from 0-5V to about 2-4V, and it became a triangle on the scope instead of a square-wave.    Maybe an impedance mismatch issue, or maybe I'm loading down the output putting the 2K to ground?  This is strange indeed  

Should I try a 10K and 20K to see if this improves, or maybe a 100K and 200K?  Also, right before the output of the X02, the signal goes through a 56ohm resistor  

It seems like everyone who puts clocks in don't look into the 3.3V vs. 5V issue.  It seems important to get a correct waveform into the chip.  If the chip ends up clipping the 5V waveform because it takes 3.3V, that is BAD     And if the chip runs off of 3.3V and you feed it 5V, that could cause a 'fight' inside the chip between the two mismatched voltages, right?  Arrrggghhhh!  Well, I'll just have to keep trying and scoping the signal.  I want this baby to be PERFECT, with a 50% duty cycle as the SM8707H datasheet recommends!  

-Vinnie

[randytsuch]

Hi Randy,

Thanks for your reponse.  Well, I tried a 1K and 2K resistor divider to make the output 3.3V.  The output went from 0-5V to about 2-4V, and it became a triangle on the scope instead of a square-wave.    Maybe an impedance mismatch issue, or maybe I'm loading down the output putting the 2K to ground?  This is strange indeed  

Should I try a 10K and 20K to see if this improves, or maybe a 100K and 200K?  Also, right before the output of the X02, the signal goes through a 56ohm resistor   ...

With the 2K to ground, sounds like you loaded it down, maybe.  See what happens with 10K and 20K, I think 100K is kind of big.  One thing is you are building a RC network, because the IC inputs always have some capacitance.  If your resistor is too big, the C comes into play, and you will have a sine wave instead of a square wave.

BTW, I always figured if the chip was made to handle a 5V input, it should be OK, at least it was not damaging the chip, but I see your point, it could have been less than ideal.

Randy

[Vinnie R.]

With the 2K to ground, sounds like you loaded it down, maybe.  See what happens with 10K and 20K, I think 100K is kind of big.  One thing is you are building a RC network, because the IC inputs always have some capacitance.  If your resistor is too big, the C comes into play, and you will have a sine wave instead of a square wave.

BTW, I always figured if the chip was made to handle a 5V input, it should be OK, at least it was not damaging the chip, but I see your point, it could have been less than ideal.

Randy

Randy,

Lars from LC Audio returned my email (rather quickly) last night and mentioned that I should just connect a 220ohm resistor across the output and that should give me an impedance matched 3.3V  
I didn't think I could do that, but he's the designer.  

Thanks for your help, I'll post back my results to share with anyone else who may be in the same situation.  

-Vinnie

[Tinker]

Has anyone bought an aftermarket clock (ex. LC AUDIO XO2) that outputs a 5V signal, but wanted to run it into a chip that runs on 3.3V or lower?  Was a simple resistor divider good enough to scale down the voltage?  What was the magnitude of the resistors (ohm, Kohm, Mohm)?

I want to preserve the clock waveform integrity and just remove the excess voltage.  

This is a common problem now that there are so many logic standards, particularly new ones designed for low power or high-speed applications.

See if you can find out more about your chip. As Randy has found, overvoltage is safe in many cases. Most new logic devices are tolerant of over-voltage conditions. Older ones have diode input clamps which can become reversed biased and pass current back to the source through the V+ pin.

At low speeds the resistors should work OK. Otherwise (if interested) have a look at level translation switches, which are used in data-bus matching problems.

T