# Help with VOL VOH question...Voltage output on oscialltor

Discussion in 'Electronic Basics' started by MauiBlue, Feb 15, 2004.

1. ### MauiBlueGuest

I have this component:
http://rocky.digikey.com/WebLib/ECS/Web Data/ECS300C Dual Output Oscillator.pdf

The component I'm trying to interface with requires a 5v clock at 158.3khz.
I've got the proper clock signal by setting this chip with the proper
dividers, I'm getting a clean 5v signal to it as well.
Now, my problem is this: The volatge reading on the clock pin is 2.4volts.
According to the datasheet at VOH it should put out .9 x input voltage or
roughly 4.5volts.
Can anyone here explain what the VOL and VOH on the datasheet mean, and more
importantly what I might need to do to get the correct output voltage?

Thanks!

Chris

2. ### Jim ThompsonGuest

What kind of load do you have on the pin?

...Jim Thompson

3. ### Tim DicusGuest

Hi Chris,

Are you using an o-scope to measure the voltage? What does the waveform look like? Correct frequency?

Tim

4. ### John LarkinGuest

Are you using a voltmeter? A square wave is +5 half the time, 0 half
the time, and that averages 2.5, which the voltmeter reports.

Voh is high voltage output from the gadget, under some specified load.
Vol is low ditto. The clock should be a square wave that flipflops
between the two at 158,300 cycles per second.

Funny, I tend to type 'voltage' as 'volatge' a lot myself. Have a lot
of trouble with 'system' and 'state' a few other specific words.

John

5. ### Tim ShoppaGuest

Average DC voltage or peak voltage?
Since they do not specify the load current, I have to assume that it's
open-circuit voltages in "low" and "high". Average 0V and 5V and you get
2.5V, which your 2.4V isn't too far from.

2.4V isn't too far out of line for a TTL high. Is it possible that you
have a similar programmable oscillator, but a "TTL-out" instead of "CMOS-out"?
Any TTL gate ought to make a good buffer... if you really need it. I'm
not sure that anything is really wrong, especially if 2.4V is the average
voltage on the oscillator output.

Tim.

6. ### JamieGuest

yes.
the .9 acounts for the lost in the chip to the ratio of your
supply source for the chip.
what it means is when the VOH (voltage ouput in High State) it will
be ~souceVoltage * 0.90...
seeing that your getting 2.4 v and i can assume that your using a
ac type meter to measure this i would say that your working just fine
due to the meters averaging response.
look at that on a scope or put your meter in DC mode and put the
peak hold on. i think you will then see a much higher reading.