Unknown type of wavefarm...!!

Thread Starter

bhargav shankhalpara

Joined May 9, 2010
20
Hello everyone...!!

i am working on one old circuit. Now i want to make same circuit with another microcontroller but i want same output as now old circuit is giving.

My old circuit contain ATMEGA16A Microcontroller. form its datasheet i come to know at its pin no.1 have three functions like XCK for USART Purpose, T0 for timer/counter external pulse for counting and third is Simple port I/O Pin.

I am getting squarewave in nature. which i have seen in my DSO NANO. see attached photo for wavefarm...!!

My question is that...

i can obtain this square type wave on pin no. 1 (positive terminal of DSO) only when i connect ground terminal of DSO with Positive supply which i have given to microcontroller. but when i connect ground terminal of DSO with supply ground i get straight line like dc of 0 volts.

so why this signal i can see only with respect to positive terminal...why not with ground.(as it should be show in inverted form but it should be....).

which type of this wave...?

some attributes i have read on DSO ( With DSO + terminal at pin no.1 of microcontroller and - terminal of DSO with positive supply of microcontroller) are as bellow...

1. Vrms = 3.04V
2. Vavg = -2.28V
3. Vp-p = 4.60V4. Vmax = +320mV
5. Vmin = -4.28V

Wavefarm image is attached.

http://obrazki.elektroda.pl/1832621300_1409857001.jpg

Thank you...!!
 

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Thread Starter

bhargav shankhalpara

Joined May 9, 2010
20
Looks like an open collector to me, i.e. high impedance for the rising edge, low impedance for falling edge.
Hi kubeek...

Thanks for your quick answer...!!

as your suggestion i connect external pull up of 22k with pin no.1 and than i can measure signal with both positive and ground reference...!! its open collector output.

next i want to ask that how i can generate that same kind of wave using that pin. because when i try to generate simple PWM signal of 50% duty cycle. it give output directly without connecting resistor.

i want to obtain same output because next part of circuit is working fine with this output.

Thanks...!!
 

kubeek

Joined Sep 20, 2005
5,794
You can do this by setting PORTx on that pin to 0, and then setting DDRx to 1 to pull the pin low, and to 0 to leave the pin in high impedance state.
 
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