# 555 PWM: Load Shedding PLC input. Frequency difficulties.

Discussion in 'The Projects Forum' started by 1rspn, Feb 12, 2013.

1. ### 1rspn Thread Starter New Member

Feb 12, 2013
3
0
Hello there,
I am building a 555 oscillator circuit for a college project and come across a small difficulty. It is a input circuit for a PLC which will implement Load Shedding.
The 555 circuit is to simulate the pulse output from a Digital Supply Meter, where each pulse were to equal 1kWh say (this value is not important at this stage). So I have an astable 555 circuit arrangement, with a potentiometer to vary the frequency output. This output is connected to a counter within a PLC. The PLC will give an output, once the counter reaches a set value, that will cause the frequency of the 555 circuit to reduce by a set amount. (eg. 5Hz)
I have tried configurations of increasing the values of the R1, R2, or C1 to create a decrease in the frequency by so much. But this reduction of frequency ends up proportional to the Potentiometer.

2. ### 1rspn Thread Starter New Member

Feb 12, 2013
3
0
For example:
if R1 was the potentiometer of 10KΩ,
R2 = 10KΩ
C1 = 100μF
T = Time period

T = ln(2)x0.0001x(10000+2x10000) = 2.08s

Over 30mins: # of pulses would be 865

When R1 is increased to 100kΩ,

T = ln(2)x0.0001x(100000+2x10000) = 8.32s

Over 30mins: # of pulses would be 216

Now if I was to switch in an added resistor in series with R2, I would increase both the Time High and the Time Low, hence decreasing the frequency. (this is the effect the output of the PLC would have)

New Outputs:
R1 = 10kΩ
R2 = 11kΩ
C1 = 100μF

T = ln(2)x0.0001x(10000+2x11000) = 2.22s

Over 30mins: # of pulses would be 811

When R1 is increased to 100kΩ,

T = ln(2)x0.0001x(100000+2x11000) = 8.46s

Over 30mins: # of pulses would be 213

This means that when R1 = 10kΩ, the change of frequency, due to the increase of R2, is 865-811 = 54 Pulses. or 0.03 Hz

When R1 = 100kΩ, the change of frequency, due to the increase of R2, is 216-213 = 3 pulses. or 0.0017Hz

Could anyone suggest ideas or methods to change the circuit to ensure the same # of pulses drop, no matter how I simmulate the frequency of the Digital Supply Meter?

Btw I find that Voltage Control Oscillator method, or trying to use the control pin, sill doesn't rectify my problem.

Feb 12, 2013
3
0
4. ### tracecom AAC Fanatic!

Apr 16, 2010
3,935
1,430
Have a look at this post.