Another 555 ?

[zachtheterrible]

How do I get a 555 to make positive pulses? I want .1second of a positive pulse, and then 10 seconds of negative . . .

The reason I need this is because whenever the 555 is turned on, it gives the positive (long) pulse first, and then the negative (short) pulse. If you can give me a way to have a short pulse first (polarity doesn't matter) I will be satisfied.

Thanx :lol:

 

[David Bridgen]

Hi zach.

Take a look at Fig.3 https://www.davidbridgen.com/555.htm

 

[zachtheterrible]

Thanx David, I don't have a chance to fiddle around with it yet (its time for bed), but it looks like that will solve my problem.

Funny thing, I stumbled across your website a couple of days ago. Nicely done.

 

[ljcox]

I would use a counter. eg. a 4017 would give you a 1:9 ratio. You would need an inverter to make the output low for 1 and high for 9.

But if you really want 1:10 then you will need a modulo 11 counter.

The advantage of a counter versus a timer is that the duty cycle is exact, whereas with a timer such as the 555, it won't be due to component tolerances.

Len

 

[zachtheterrible]

David, thanks for that circuit, it does EXACTLY what I want when it works off 3 volts :lol: . Kick it up to 6 volts though and it won't work. Let me explain this circuit that I have posted and you can see what's not working.

First off, let me say that this circuit is going to be used in a wireless security system. Since there will be multiple transmitters on the same frequency, they cannot all be on at once. This circuit is intended to make the transmitter stay on for a very short amount of time, therefore reducing the risk of two transmitters being on at once. This circuit will be implimented in all of the sensor/transmitters.

The signal on Vin is a positive pulse that could last a couple of minutes because it comes from an IR break beam sensor, and somebody could stand in front of it. That signal needs to activate a transmitter.

So, explanation of the circuit: The positive signal turns on Q1, giving power to the 555. The 555 generates very short positive pulses. These positive pulses turn on Q2, allowing the signal from Vin to go through it for very short amounts of time (determined by the positive pulses from the 555). The signal then goes to the base of Q3, which inverts the signal because the transmitter needs a negative pulse.

So why will this circuit only work off of 3 volts and not 6? When I change it to six, I get a negative signal from Vout that just stays negative.

Thank you very much :lol:

EDIT: I forgot to add the value of C1. It is .1uf

 

[audioguru]

Hi Zach,
Aren't you using an ordinary 555? Its minimum supply voltage is 4.5V. The Cmos version is rated to work with a supply voltage down to 1.5V.

Your circuit needs a resistor in series with the base of Q2. The series connected base-emitter diodes of Q2 and Q3 are like a dead short to the 555's output when the supply voltage is high. Then the voltage at pin 2 can't rise high enough for the timing period to end. :lol:

 

[ljcox]

Hi Zach,
Aren't you using an ordinary 555? Its minimum supply voltage is 4.5V. The Cmos version is rated to work with a supply voltage down to 1.5V.

Your circuit needs a resistor in series with the base of Q2. The series connected base-emitter diodes of Q2 and Q3 are like a dead short to the 555's output when the supply voltage is high. Then the voltage at pin 2 can't rise high enough for the timing period to end. :lol:

The resistor needs to be between the Emitter of Q2 and the Base of Q3.

Q2 does not need a Base resistor since it is operating as an Emitter Follower.

There should also be a resistor between B/Q3 and gnd to ensure it is off when it should be off. ie. to soak up Icbo.

Len

 

[audioguru]

Good points Len! :lol:
Your emitter resistor for Q2 protects the 555, the input, Q2 and Q3.

I worked for a company that made a product with two transistors in a darlington configuration like this one. It didn't have a resistor to ground on the base of the lower transistor and some of them didn't work until the resistor was added. :lol:

 

[David Bridgen]

Hi zach.

Only just taken a look at the forums today. Audioguru and Len have taken up the cause, so, as I shall be incommunicado for a few days, I shall leave you in their extremely capable hands.

 

[ljcox]

Good points Len! :lol:
Your emitter resistor for Q2 protects the 555, the input, Q2 and Q3.

I worked for a company that made a product with two transistors in a darlington configuration like this one. It didn't have a resistor to ground on the base of the lower transistor and some of them didn't work until the resistor was added. :lol:

The resistor to gnd also prevents thermal instability.

Len

 

[ljcox]

How do I get a 555 to make positive pulses? I want .1second of a positive pulse, and then 10 seconds of negative . . .

Thanx :lol:

Zack, you mean "10 seconds of 0 Volt (or gnd)" not "10 seconds of negative". There are no negative voltages in your circuit.

The negative end of the battery is connect to gnd which makes its potential 0 Volt.
Len

 

[zachtheterrible]

Zack, you mean "10 seconds of 0 Volt (or gnd)" not "10 seconds of negative". There are no negative voltages in your circuit.

Yea, that's right, I meant to say negative relative to +6v :lol:

I got it working fine now.

Take a look at my diagram and see if the resistor values are ok, thanx :lol:

 

[ljcox]

Zack, you mean "10 seconds of 0 Volt (or gnd)" not "10 seconds of negative". There are no negative voltages in your circuit.

Yea, that's right, I meant to say negative relative to +6v You normally measure voltages with respect to gnd.:lol:

I got it working fine now.

Take a look at my diagram and see if the resistor values are ok, thanx :lol:

The 10k is correct. But what is the output current? The 1k means a base current of about 1 mA so Q3 can sink 10 mA easily.

Len

 

[zachtheterrible]

I didn't measure the ouptput current. Wouldn't the output current depend on the load on Vout? Vout will be connected to an HT12E encoder, the data pin. If it activates the data pin on the encoder, everything will be fine; i won't have to worry about output current.

 

[ljcox]

It should be OK then.

Len