Hi
Could you please help me with this circuit
It is very simple and i have done it several times before but i don't know what is wrong with it this time

The problem is that the led stays on all the time even when the switch is open
The buzzer is working correctly : when the switch is open it is on and when the switch is closed it is off
But the led is on all the time either the switch is open or closed
I tried to change the values of the resistor R2 but it didn't help

 

Put a 10kΩ resistor (or lower) across Q1 base and emitter.

 

Assuming you have wired it correctly. . .
Either Q1 is faulty, or the 4069 is faulty.
You have missed out the base current limiting resistor of Q2, which could have damaged IC1.
What is the voltage on Q1 base when the switch is open?
Does the LED go off if you remove IC1?

 

Presumably, the leakage current on IC1 4069 is turning on Q1.

 

The "base limiting resistor" to which Ian0 referred is a resistor in series with the base to limit the base current. If you want 1 ma of base current (for example) 200 microamps, the resistor R2 in the circuit below would be about 20k. (5V-0.7V)/200e-6.

 

Or, you can configure the circuit this way, and not use an inverter IC.
There will be about 1mA flowing thru LED when switch is open but the 2v@20mA LED will be extremely dim and not visible.

 

There will be about 1mA flowing thru LED when switch is open but the 2v@20mA LED will be extremely dim and not visible.

Don't be too sure about that. Modern LEDs are quite bright at 1mA. I never run them more than 5mA.

 

Install a resistor in parallel with the LED if need be

 

The problem is that the led stays on all the time even when the switch is open
The buzzer is working correctly : when the switch is open it is on and when the switch is closed it is off
But the led is on all the time either the switch is open or closed
I tried to change the values of the resistor R2 but it didn't help

Ignoring IC1, the first thing I'd suspect is a faulty Q1. What transistor did you use? IS it a transistor? Or is it a FET? If it's a FET then once the gate is charged it can stay energized for a very long time. Adding a resistor from the gate to the negative rail will drain the capacitive charge the gate can hold, and D1 will shut off.

Before we can do further diagnostics on your circuit we need to know what transistors you used. Otherwise we're just assuming the transistor is a common NPN BJT.

 

I also notice you have approximately 20mA feeding the base of Q1. Not knowing what transistor you used, you may have shorted the transistor into conduction all the time, thus explaining why the LED is constantly lit.

It is very simple and i have done it several times before but i don't know what is wrong with it this time

It's the simple circuits that most often fool us because we might overlook some small detail. Also, assuming 2Vf for D1 and 0.6Vf for Q1, 5V-2.6=2.4V. Under that assumption; 2.4V÷220Ω= approximately 11mA. Those seem like small currents, but again, we don't know what transistor you used.

 

There will be about 1mA flowing thru LED when switch is open

Install a resistor in parallel with the LED if need be

Yes, 1-2kΩ across the LED should keep it fully off when the transistor is off.

 

Don't be too sure about that. Modern LEDs are quite bright at 1mA. I never run them more than 5mA.

Your correct the designer should be mindful of the LED specs.
This LED would look dark (intensity is practically zero) at IFwd of 1mA.

 

Your correct the designer should be mindful of the LED specs.
This LED would look dark (intensity is practically zero) at IFwd of 1mA.

View attachment 352040

Intensity would be 5% of the 20mA intensity, but don't forget that the eye has a logarithmic response.

 

Intensity would be 5% of the 20mA intensity, but don't forget that the eye has a logarithmic response.

but the eye would not see a logarithmic intensity curve as in LED dimming, the LED is either on or off.

 

Or, you can configure the circuit this way, and not use an inverter IC.
There will be about 1mA flowing thru LED when switch is open but the 2v@20mA LED will be extremely dim and not visible.

If a milliamp is acceptable, then you can leave quite a lot of components out of it:

 

Why do you even need a transistor?

 

Why do you even need a transistor?

Nice try, but LED is OFF when buzzer is ON - the transistor is a needed inverter.

 

Nice try, but LED is OFF when buzzer is ON - the transistor is a needed inverter.

I did not know that is what the TS wanted.
Here is the revised circuit.

 

How about:


The (primary) purpose of R2 is to bleed some (or even most) of the current through the LED and R1 so that the current into the base of Q1 is not too high. As long as it turns hard on it will ground the base of Q2.

 

but the eye would not see a logarithmic intensity curve as in LED dimming, the LED is either on or off.

A logarithmic response has nothing particularly to due with a dimming curve.
It means that a 5% amount of light will look much brighter than if the eye had a linear response to light.