Oscillator output clock has less peak to peak

Thread Starter

to PVR

Joined Jan 5, 2024
2
we are using oscillator AX3DAF1-156.2500T -156.25MHZ(LVDS), we are given 3.3V VCC, enable and Resistor termination with 100 ohms, output clock has less than 200mV PK-PK. and offsetting to 3.3V amplitude.
 

BobTPH

Joined Jun 5, 2013
9,110
datasheet?

What current is the device specified to source and sink? A 100Ω resistor would pull 33mA. Is that within spec?
 

WBahn

Joined Mar 31, 2012
30,230
Your 200 mV is a little low, but not terribly so. I don't know what you mean by "offsetting to 3.3 V", but if the common-mode value of the output is that high, then it is probably starving the current source in the part. How are you measuring the signal?

Also, you might put additional bypass capacitors on the supply, perhaps 1 uF and 10 nF, given the frequency you are operating at.
 

Lo_volt

Joined Apr 3, 2014
321
How's your bandwidth on your oscilloscope? Is it possible that the scope can't handle frequencies that high? If that were the case, it would show up as a heavily attenuated signal.
 

MisterBill2

Joined Jan 23, 2018
18,926
What is the reason for that 100ohm resistor connected across the output?? THAT is certainly causing the low output voltage. Remove the 100 ohm resistor and the output voltage will probably be in spec.
 

Reloadron

Joined Jan 15, 2015
7,543
What is the reason for that 100ohm resistor connected across the output?? THAT is certainly causing the low output voltage. Remove the 100 ohm resistor and the output voltage will probably be in spec.
This is the data sheet(s).

This is a test circuit example:
LVDS.png
Used on the LVDS differential output test. I am not sure exactly which version the thread starter has. I assume based on the image posted by thread starter it's the differential output flavor I posted above with the 100 ohm resistor across the output.

Ron
 

MisterBill2

Joined Jan 23, 2018
18,926
I suggest remove the resistor and see what they have. To develop 3.3 volts across 100 ohms will take 33 milliamps if V=IR
Most clock modules do not deliver that much current, do they???
 

Reloadron

Joined Jan 15, 2015
7,543
I suggest remove the resistor and see what they have. To develop 3.3 volts across 100 ohms will take 33 milliamps if V=IR
Most clock modules do not deliver that much current, do they???
None I am aware of? Just looking at the data sheet and saw their test setups. I also have no idea how the thread starter is measuring?

Ron
 

michael8

Joined Jan 11, 2015
421
we are using oscillator AX3DAF1-156.2500T -156.25MHZ(LVDS), we are given 3.3V VCC, enable and Resistor termination with 100 ohms, output clock has less than 200mV PK-PK. and offsetting to 3.3V amplitude.

What are you expecting? The LVDS output swing minimum spec from the data sheet is 250 mV.

Also what are you connecting to measure the value. 10 pF is about 100 ohms at 156 Mhz. You would need
a differential input to measure it. I'd guess that AC grounding one side might cut the output swing in half.
 

MisterBill2

Joined Jan 23, 2018
18,926
I am wondering if that differential output is isiolated from the source that they are using to offset the output. And I am thinking that the "100 ohms" is the impedance of the link to whatever the load would be. I learned about the low value load resistors many years ago, overheating an LM733. The fact that it can drive a 75 ohm video input does not mean that it can drive a 75 ohm resistor.
 
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