# Pressure Circuit

Discussion in 'The Projects Forum' started by n0nesense, Jan 11, 2013.

1. ### n0nesense Thread Starter New Member

Jan 11, 2013
8
0
Hi everyone. Thank you for viewing.

I have some problems with my pressure circuit. I am using a Force Sensing Resistor that act as a basic resistor. When there is pressure, the resistance will drop. It is somewhat like a LDR.
https://www.sparkfun.com/products/9376

The 1M ohm pot is supposed to be a dummy that act as the sensor. In simulation when both POT at max, the buzzer is not working. After i reduced the resistance of the 1M ohm pot, then only there is output. Overall, the expected result is obtained (in simulation).

But when i try it on a breadboard, the output is approx to 8V at all time, regardless of the variation of pot. So, i wonder how can this problem occur when it does not in simulation. Is it the 555 or the transistor? or the overall circuit?

This is the actual circuit that i refer to create my version
http://electroschematics.com/6122/pressure-sensor-alarm/

2. ### wayneh Expert

Sep 9, 2010
11,927
2,858
What output is 8v, the voltage on pin 3?

I think the sum of R1 and R2 may be too large, and not enough current flows to pull the gate voltage down. Try leaving one of them out.

And try to isolate the problem. If you apply a voltage to the gate of the MOSFET, does the buzzer work? Why are you using only 3V? That's not enough to switch the MOSFET.

3. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
655
BS170 gate threshold voltage (Vgs(th) can be anywhere between 0.8V and 3V, according to the datasheet. In your circuit, the gate-to-source voltage will vary between 0V and ≈2.5V, depending on the settings of the two variable resistors. Your 3V source needs to be increased to at least 6V to ensure that the circuit will work.
You have learned that a simulator is only as good as the spice models of the devices you are using. The models are generally "typical". Your hardware part will not be exactly typical. The simulator is also only as good as the person using it.
Also - you don't need D1.

EDIT: It looks to me like your circuit, when you get it working, will turn the buzzer off when pressure is applied to your sensor. With no pressure applied, the buzzer will be on. This is opposite of the circuit you linked to, which has a piezo sensor.

Last edited: Jan 11, 2013
4. ### n0nesense Thread Starter New Member

Jan 11, 2013
8
0
yes, the output at pin 3. Thank you for the suggestion. I will be sure to try it out.

5. ### n0nesense Thread Starter New Member

Jan 11, 2013
8
0
Thank you for the modification suggestion, I will try it out when i go to lab this monday and let you know what happens.
Do you mean that the way i edited it (from the linked circuit) has turned the circuit the opposite way of how it is supposed to work? If so, then this is bad, very bad

6. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
655
When the piezo device is subjected to pressure, it delivers an impulse of voltage. This charges up the cap on the gate of the MOSFET, turning it on for a short amount of time. This causes the 555 output to go high for a short period of time, turning on the buzzer.

When pressure is applied to your circuit, the voltage on the gate of the MOSFET goes low. If it was previously on, it will be turned off. While the MOSFET was on, the buzzer would have been sounding. Pressure would turn the buzzer off momentarily.
You might be able to get the action you want by connecting the buzzer between the output and +9V instead of to ground.