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Scmitt trigger oscillator

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ukeee

New Member
I'm having trouble with a schmitt trigger oscillator. I'm using a 40106 cmos hex inverter. I'm running it of 12V but the output voltage is approximately 2V, I need a voltage of at least 8V. I have tried using different combinations of R and C. Any help would be greatly appreciated thanks.
 

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Nigel Goodwin

Super Moderator
Most Helpful Member
Where are you taking the output from?. The righthand side of the resistor should be a large square wave, and the lefthand side a smaller triangle wave.

The component values look a bit strange as well?, 1K and 1uF don't look a very good combination - I would think that 10K and 0.1uF, or 100K and 0.01uF would be better balanced (and smaller).
 

ukeee

New Member
Yea sorry I was using a 0.1uF and a 10k resistor, I'm taking the output from the right hand side , I've got a little square wave and a little triangle wave on the left hand side. I've also tried using the 100k 0.01uF combination. Thanks for your help
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I'm presuming you are using a scope to check the waveforms?. If so, make sure it's set to be DC coupled, and measure on the output of the gate. Is it going from nearly 0V upto a positive voltage, or is it not going down very low. Also check the HT supply to the chip, on the actual pin itself - the output of a gate "shouldn't" be able to be anything apart from high or low.
 

olly_k

Member
why not add a transistor to the rhs with suitable current limit resistor? That way you will have upto 12v output and nearly anything in between.
 

laroche73

New Member
output

ukeee, do you have anything besides R1 attached to the output, such as a LED or other load? The CD4K series outputs can only sink or source a few mA. Are all of the other inputs tied high or low? (CMOS parts can do funny things when unused inputs are left floating)
 

ukeee

New Member
I'm measuring it using a scope, which is DC coupled, i've tied all the unused inputs low and currently am not loading it. It will be used as the input to another cmos chip.

The output is going from nearly 0V to 2.2V, the scope I'm using is a picoscope ADC 210.

Is it possible that the feedback path is drawing to much current?

Thanks for all your help
 

laroche73

New Member
chip type

One other thing, ukeee. You mentioned the part is a 40106 hex inverter, but the schematic shows a schmitt NAND gate, like the 4093. Is that just a typo?
 

ukeee

New Member
Yea its just because my simulation software doesn't have a 40106 only the 4093, sorry should have mentioned that
 

laroche73

New Member
bad part?

ukeee, you may want to try another part. I wired up the same circuit with a CD40106 (74C14), and it works as expected. The output is a square wave with a 1 ms period, 50% duty cycle, and swings from rail to rail (0 - 11.5V). The input looks like a small triangle wave (since only the early part of the RC charge/discharge cycle is used) with a DC offset, and swings from 3.75V to 7.5V on my breadboard (11.5V Vcc).
 

ukeee

New Member
Thanks for the help at least I know the circuit will work. I'll try using another parts.
 

bgscott

New Member
Hi

Does your 40106 have a BE suffix (buffered outputs) as I have had problems in the past when trying to use UB suffix (unbuffered output) devices. Using BE devices worked eveytime.

Brian
 

ukeee

New Member
Thanks everyone for your help. I was using a 40106 with a UB suffix have now swapped it over with a BE suffix and got it working. Didn't know there were two different versions. Thank a lot, my mystery has been solved!
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I don't like to be critical, but so far the circuit posted has the wrong value resistor, the wrong value capacitor, and the wrong IC as well. It's not really a terribly helpful diagram!.

And on top of all that, you're not even using a real oscilloscope!.
 

laroche73

New Member
To me, it's more of a lesson about making assumptions. Ukeee's post was clear enough to get the idea across. I thought I was asking the right questions, it never occurred to me that someone might be using an unbuffered part in this day and age. Thanks to bgscott for pointing out the obvious (in retrospect).
 
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