Basic N555 Square Wave Generator Circuit

Thread Starter

sab201

Joined Nov 18, 2023
64
Good Day,

I have made an N555 square wave generator based on the circuit below and it seems to work just fine.
NE555 Square wave generator circuit~3.jpg
I have learnt that the 1 M Ohm potentiometer is used to increase or decrease the horizontal pulse width between the waves. That is the frequency.

But what I could not figure out is whether a higher resistance value of the pot causes increase in the pulse width or it happens the other way around which is lower resistance increases it.

Thanks.
 
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ScottWang

Joined Aug 23, 2012
7,411
But what I could not figure out is whether a higher resistance value of the pot causes increase in the pulse width or it happens the other way around which is lower resistance increases it.
A higher resistance value of the pot causes an increase in the pulse width and the frequency is going low.
 

BobTPH

Joined Jun 5, 2013
9,129
Does a capacitor charge faster or slower when you increase the series resistance? Answer that and you can probably figure it out.
 

dl324

Joined Mar 30, 2015
16,989
I have made an N555 square wave generator based on the circuit below and it seems to work just fine.
The symbol you used is useless for conveying circuit intent. If you draw it as shown below, it's easier to discern functionality:
1713635719342.png
I have learnt that the 1 M Ohm potentiometer is used to increase or decrease the horizontal pulse width between the waves. That is the frequency.
Your terminology is awkward. Changing R3 changes the frequency while maintaining a duty cycle of about 50%.

When you use a potentiometer, you should indicate the clockwise direction. The way I've drawn it, frequency increases with clockwise rotation.
But what I could not figure out is whether a higher resistance value of the pot causes increase in the pulse width or it happens the other way around which is lower resistance increases it.
Until the pot is set to less than around 100k, the duty cycle is going to be about 50%. Below 100k, it'll increase. At 10k it would be 67%. It's never going to be less than 50%.
 
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Thread Starter

sab201

Joined Nov 18, 2023
64
When you use a potentiometer, you should indicate the clockwise direction. The way I've drawn it, frequency increases with clockwise rotation.

Until the pot is set to less than around 100k, the duty cycle is going to be about 50%. Below 100k, it'll increase. At 10k it would be 67%. It's never going to be less than 50%.
Scottwang and dl324,

Both have opposing posts. According to Scottwang - higher resistance decreases the frequency increasing the pulse width.

dl324, thanks for explaining about the duty cycle this is also important when considering the frequency and pulsewidth. So, clockwise rotation in my pot increases the resistance and anticlockwise decreases it. The pin connections are same in both circuits. If that is the case, then the frequency increases with increase in resistance as per your post?

What I notice is that when I decrease the resistance the Voltage output from the 555 timer increases to max to 3 Volts at 0 ohms. So when I set the pot to 0 ohms will the frequency be at its maximum or minimum.

Thanks for all the replies.
 
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Thread Starter

sab201

Joined Nov 18, 2023
64
You can check the internal structure of the potentiometer to adjust the resistance.
Yes based on the pin connections of the potentiometer in the drawing I have marked it on the circuit.sketch-1713766390654.jpg
I suppose what matters is the resistance value connected between terminals 7 and 6 of the 555 timer IC. Now I have set this value to zero. 7 and 6 are shorted which inturn shorts it to pin 2 of the IC. So will the frequency of the wave be at its maximum or what will be the outcome.

Thanks.
 
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Thread Starter

sab201

Joined Nov 18, 2023
64
Does a capacitor charge faster or slower when you increase the series resistance? Answer that and you can probably figure it out.
Bob TPH,

When the series resistance is increased the capacitor charging time increases since the voltage drop across the resistor increases thereby reducing the voltage available to charge it.

So when there is no resistor, the capacitor should charge faster. That should increase the frequency and reduce the pulse width.

Edit: With zero resistance in the circuit, there is no frequency and no pulsed output.
 
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ScottWang

Joined Aug 23, 2012
7,411
Yes based on the pin connections of the potentiometer in the drawing I have marked it on the circuit.View attachment 320455
I suppose what matters is the resistance value connected between terminals 7 and 6 of the 555 timer IC. Now I have set this value to zero. 7 and 6 are shorted which inturn shorts it to pin 2 of the IC. So will the frequency of the wave be at its maximum or what will be the outcome.

Thanks.
If you want to get the highest frequency then you can set R2 = 1K and cut off the Pin1 of R3 and then in series with a 1K between Pin1 of R3(potentiometer ) and pin7, if you don't care about the width of the discharge pulse then you can choose the in series resister smaller than 1K.
 

Thread Starter

sab201

Joined Nov 18, 2023
64
If you want to get the highest frequency then you can set R2 = 1K and cut off the Pin1 of R3 and then in series with a 1K between Pin1 of R3(potentiometer ) and pin7, if you don't care about the width of the discharge pulse then you can choose the in series resister smaller than 1K.
Ok in order to get the maximum frequency, I set R3 to zero and connect a 1k resistor as shown below...
sketch-1713772066601.jpg
Edit: Could you please tell me what will be the frequency in the original case when R3 is set to zero. Thanks.
 
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ScottWang

Joined Aug 23, 2012
7,411
Thanks...This formula is really helpful. So according to this formula I find that when R3 is zero, then there will be no tL and there is no pulsed output from the circuit at all.
Your original circuit was just like that, but when you added the 1K[call R4] then the condition is different, you will get the output pulse High and Low.
 

ScottWang

Joined Aug 23, 2012
7,411
According to the formula, the Highest Frequency as below --

RA = R2 = 1K

RB = R1 + 1K; R1=[0~1M], 1K [Call R4]

C3 = 0.1uF

T(High) = 0.693(1K+1K)*C

T(Low) = 0.693(1K)*C

Period = T(High) + T(Low) = 0.693(RA+2RB)*C

Frequency ≈ 1.44/(RA+2RB)*C

F ≈ 1.44/(1K+2*1K)*1uF
 

dl324

Joined Mar 30, 2015
16,989
Both have opposing posts. According to Scottwang - higher resistance decreases the frequency increasing the pulse width.
I don't see where our posts are in conflict. If you increase resistance, frequency will decrease. Duty cycle is close to 50% until pot resistance is sufficiently low.

If you're confused by direction information, that's because your schematic doesn't indicate clockwise direction. My schematic has an arrow that's used to indicate clockwise rotation.
 

AnalogKid

Joined Aug 1, 2013
11,128
Yes based on the pin connections of the potentiometer in the drawing I have marked it on the circuit.
The pin numbering is incorrect. The standard for both trim pots and panel mount is that pin 3 is the clockwise end of rotation, as viewed from the front of the shaft. See post #9.

ak
 
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