Hi, just joined the forum. I am trying to build a timer, that after 5 minutes will turn on a buzzer.

My plan was to build a monstable 555 that when the output on pin 3 is high it would apply its voltage to the base of the pnp tranasistor which would turn the buzzer off, then when the 5 minutes are up (i have found the correct capacitor and resistor combination to do so) the output goes low and the pnp would have no power at the base and turn the buzzer back on.

I have both a functioning PNP circuit and a functioing monostable circuit but when i attach them together the 555 does not turn the pnp off... any ideas why, I am stumped.

 

The 555 has a darlington transistor output that has a high voltage that is 1.2V less than its supply voltage. The PNP transistor needs to have a base voltage that is at least 0.4V less than the supply voltage. Use a voltage divider to drive the base of the transistor or add two series diodes to its emitter.

A Cmos 555 like an LMC555, TLC555 and ICM7555 have outputs that go up to the supply voltage.

 

thanks for the reply. First i would like to say that I am new to electronics although I learn quickly. My supply voltage is 9V and the output on the 555 is 6.4 volts if I remember correctly. I an unsure of what a voltage divider is or two series diodes to its emitter means... could you explain, perhaps post a schematic, I can understand a schematic very well.

 

If you're only getting 6.4V out of your 555, there is a problem. Do you have a resistor between your 555 supply pin and the battery supply?

It would help if you posted a schematic of exactly what you have so far.

 

Why don't you just connect the buzzer between +9V and the output of the 555?

 

If the buzzer draws more than ~200mA at 5V, don't connect it directly. Try using the discharge out (pin-7) as in the attached. Or, if discharge is being used for the timer, you can also do the same thing with the pin-3 output.

 

Why don't you just connect the buzzer between +9V and the output of the 555?

ive tried that, when the output on pin 3 goes low and sinks to ground the buzzer comes on, when it goes high the buzzer is on but at a lower tone so that doesn't work

 

If you're only getting 6.4V out of your 555, there is a problem. Do you have a resistor between your 555 supply pin and the battery supply?

It would help if you posted a schematic of exactly what you have so far.

No resistor between pin 8 and 9v+ I will be posting a schematic soon

 

ive tried that, when the output on pin 3 goes low and sinks to ground the buzzer comes on, when it goes high the buzzer is on but at a lower tone so that doesn't work

So, do any of you guys understand how the 555 output can sink current when it's high? According to the schematic, it can't.
Kevin, are you using the same power supply for the 555 and the buzzer?

 

Ron,

The only way it could sink current when it's high is the potential difference between Vcc and whatever the other potential the buzzer is connected.

I'd wait for kevin's schematic before trying to decipher anything else.

 

Thanks for the replys, I appologize for not schematic, I am very busy and I know its impossible to do much electonic diagnosis without one thou I hoped it was somthing simple. Please find below the Monostable 555 I have setup and its connection from pin 3 to the pnp transistor. The monostable works properly, but the pnp will not turn off when the outputs high on pin 3

 

One obvious missing element is a base current limiting resistor.

You need around a 20K resistor in between the output of the 555 and the base of the pnp transistor.

hgmjr

 

thanks for the reply. I thought that was what the 1k resistor is doing. If not do I replace the 1k with a 20k or do I put an additional 20k resistor right in front of the pnp? or right after the 555... does it matter, to me it doesnt... other than the 1k mabe being uncessesary? Here, Ive updated it like this?

 

That's better. Now you should be able to eliminate the 1K resistor altogether.


hgmjr

 

thanks. so why would a 20k resistor work when a 1k would not?

 

thanks. so why would a 20k resistor work when a 1k would not?

The issue with the 1K is twofold. First, it is not in a position to act as a current limiting resistor for drive current from the 555 into the base of the PNP. Second, 1K is too low in value and thus would not limit the base current adequately.

hgmjr

 

How the PNP can be switched off, but the buzzer in the same configuration (between 555 output and +9V) couldn't, is a mystery to me, especially if the 1k resistor was in place also. Kevin, have you tested the circuit you posted? And when you tested it with the buzzer connected between the 555 output and +9V, did you also have a resistor connected from the 555 output to ground?
We really need to know how much current the buzzer requires before we can recommend a drive circuit for it.

 

Yes i have tested the circuit as posted. and with the 20k resistor just causes the 555 to heat up to the point where I cant hold my finger on it so that does not work. The output voltage to the pnp after the 20k is 2.3 volts.

The buzzers specs are

Min. sound pressure level : 85dB at 12VDC/30cm
Operating voltage : 12VDC
Rated voltage : 3 - 30VDC
Resonant frequency : 5mA at 12VDC
Max. current consumption : 3.8±0.5
Operating temperature (°C) : -20 ~ +70

When I hook the buzzer to 9v and the output when it sinks the buzzer is loud and clear which I want, when the output on 3 is high the buzzer still sounds but it is alot quieter. I tried a 220 to 20k resistor there and still no go

 

A 20k resistor on the output of the 555 will not cause it to get hot. You have something wired wrong.

 

The 555 is probably oscillating at a very high frequency which is causing it to get hot.
Because the circuit is missing the two supply bypass capacitors recommended in the datasheet for the LM555.

Oh yeah. The transistor and 555 are blown up since there was no current-limiting resistor from the output of the 555 and the base of the transistor.

Add the supply bypass capacitors and replace the 555 and the transistor and with the 20k current-limiting resistor the circuit should work properly.