AC square wave from 555

Thread Starter

JW_Bryncelynnog

Joined Mar 25, 2020
12
I have a circuit running from a +/- dual rail supply. I want to produce a square wave signal voltage (driving 10k resistance) about 22 kHz which gives me a positive +/- signal relative to ground.

I can build the 555 with an almost 50% duty, but it is the negative part of the output I am struggling with.

I have searched for ages for a way to do this with a 555, but not come up with anything that seems to do it.

Is there an easy way to achieve this? Any suggestions most welcome!
 

ericgibbs

Joined Jan 29, 2010
10,214
hi JW,
I assume you are using a dual power supply +/- Vout for the project.?
If yes, then a simple unity gain OPA circuit could be used to give a +/-Vout square wave.
E
 

Hymie

Joined Mar 30, 2018
837
The simplest solution would be to power the 555 timer from the +/- rails, bear in mind that the maximum supply voltage for a 555 timer is 18V.
 

Thread Starter

JW_Bryncelynnog

Joined Mar 25, 2020
12
The simplest solution would be to power the 555 timer from the +/- rails, bear in mind that the maximum supply voltage for a 555 timer is 18V.
I thought about that. I could turn the voltage regulators to +/- 6 or 7 volts no problem.

I was just wondering about whether the 555 will also need the ground as a reference? If not, happy days!
 

Hymie

Joined Mar 30, 2018
837
If the 555 timer is wired across the rails at +/-6V or +/-7V then the timer will behave as powered by 12V or 14V with its output swinging rail to rail (or thereabouts).
Remember that anywhere shown on your 555 timer circuit as connected to 0V, must be connected to the minus volt rail in your application.
 

DickCappels

Joined Aug 21, 2008
6,469
The NE555 is referenced to ground through the power supplies, so no further referencing is needed.

The trick might be getting the +peak equal but opposite to the - peak. If your duty cycle is truly 50% then the best way to achieve this is to do as Alec_t suggests in post #4, that of capacitive coupling the NE555 output to a resistor, the other end of which is grounded.
1588516109546.png

Except the NE555 should be represented as a voltage source rather and a current source. Just make sure that Xc is much lower than R at the fundamental frequency.
 

crutschow

Joined Mar 14, 2008
25,135
The output of the LMC555 CMOS version goes from rail to rail (as long as the load is high impedance).
Its maximum voltage is 15V so will operate from± ±6V or ±7V supplies.
 

Thread Starter

JW_Bryncelynnog

Joined Mar 25, 2020
12
The NE555 is referenced to ground through the power supplies, so no further referencing is needed.

The trick might be getting the +peak equal but opposite to the - peak. If your duty cycle is truly 50% then the best way to achieve this is to do as Alec_t suggests in post #4, that of capacitive coupling the NE555 output to a resistor, the other end of which is grounded.
View attachment 206119

Except the NE555 should be represented as a voltage source rather and a current source. Just make sure that Xc is much lower than R at the fundamental frequency.
Thanks all for your help.By resistor to ground you mean my 0v rail? That would make sense.
 

Hymie

Joined Mar 30, 2018
837
It is very easy to configure a 555 timer to produce a 50% duty cycle.

If using a capacitor to couple the 10kΩ load; as DickCappels says the capacitance value must be small compared to the resistance at the fundamental frequency. At 22kHz a 1nF capacitor will have an impedance of around 10KΩ, making such a technique impractical at 22kHz into a 10kΩ load.
 

MisterBill2

Joined Jan 23, 2018
6,383
... or you could just use a capacitor to couple the 555 output to your 10k load.
INDEED IT WOULD!! A capacitor sized to have low reactance at your chosen frequency will split the amplitude of the square wave fairly close to 50%between positive going and negative going.
 

MisterBill2

Joined Jan 23, 2018
6,383
The RC time constant, to come close to a squarish wave, should be quite a few cycle periods. The "R" is the sum of the source and load impedance.
 

sparky 1

Joined Nov 3, 2018
265
A basic Schmitt trigger RC oscillator the scope shows 0.4V positive. How do I add offset control so positive and negative are equal ?
Is it better to move to an op amp circuit ?
7414 22 kHz square wave.JPG
 
Last edited:

MisterBill2

Joined Jan 23, 2018
6,383
Try the same simulation circuit with a 0.1 MFD capacitor in series with the output. and then try it with another 10K resistor across that point to the common side, (=supply negative).
Ground is a mythical zero impedance plane connected to the supply common. But it is a quite useful concept unless creating a PCB layout. And using a ground symbol makes circuit drawings easier to follow.
 

Alec_t

Joined Sep 17, 2013
11,419
A basic Schmitt trigger RC oscillator the scope shows 0.4V positive. How do I add offset control so positive and negative are equal ?
Try a CMOS gate (CD40106, or CD4093), instead of a 74xx gate. A high value resistor between the gate input and a supply rail can be used to adjust the duty-cycle.
 
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