output of a 555

How are the two resistors connected?

Are they in series to make a 102 ohm resistor (with a total power of about .48 watts dissipated between them) or in parallel to make a

25.5 ohm resistor (with 1.9 watts dissipated between them), when the 555 puts out 7 volts?

The heat from the 555 depends on how much voltage it is dropping (the difference between the supply voltage and the output voltage) while it drives the resistors. That voltage times the resistor current is the biggest component of the 555's power.

they are in series, so 0.48watts. So well under the maximum of 0.2 amps the

555 can take. The other way around would be a sure shot of burning.

ken

What is the supply voltage on the 555?

Have you measured the voltage across each 51 ohm resistor when they are first heated?

Are you sure you are not working with 1/4 watt resistors?

OOOkk, I had a bad wire in my circuit. Its working fine now.

ken

Well finally it still does not work. But it do not heat up anymore. I think I fixed something, the wiring maybe be at fault. I do not know if I should hook up the 555 timer battery ground anywhere.

I want to charge a battery, and control the chargeing with the 555. The 555 timer output is set to the base. The collector to the negative of the battery. The emmitter to the ground of the supply. The positive of the battery to the Supply. Using this setup it do not work. The 555 has its own battery. Should I ground the 555 with the supply ?

ken

Too many words with too little information. Is there any way yo can create a graphic of your schematic, and either email it to me, or post it as an attachment to a message on alt.binaries.schematics.electronics. or to some web host and give us a link to it?

The 555 ground _must_ be connected to the other power supply ground.

Also, you need a resistor between the 555 output and the base of the transistor. The value of the resistor will depend on the base current needed by the transistor, which will in turn depend on the current the transistor is controlling, and its gain.

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of

the

own

I would recomend using a relay with one end of its coil connected to positive of the timer's supply with a limiting resistor and the other end to the transistors collector with emmitter to timer's supply negative. Then connect one side of relay output to the charger's positive supply. Connect other side of relay output to positive of battery to be charged. Negative of battery to be charged can be connected to negative of charger's supply.

*A resistor should be placed in series with battery to be charged so to provide some form of current limiting.* Calculate the value based on charger's supply voltage minus dead battery's lowest voltage and the maximum charging current you want to have. What type of battery are you trying to charge btw? Some types don't like to aggressive of a charging rate. You could use a common ground for both circuits if you wish. I prefer the isolation myself. A diode could also be added for protection against a reverse polarity connection.

Chris

Ok I posted my circuit here:

the software I used did not have have a bridge. and I omitted the circuit for the 555.

ken

You _must_ connect the 555 ground pin to the transistor's emitter. You also need a resistor between the 555 pin 3 and the transistor base, in place of the direct connection you show. The value of the resistor will depend on the current the transistor is controlling, and on its gain.

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Won't work - there's no common ground. Pin 1 of the

555 has to go to both the 6V battery and the 3055 emitter. Are you trying to control the charging of the 12V battery?

Ed

you need a resistor between the two circuit parts.

easiest would be between pin 1 of the 555 and the emitter of the transistor (seing as you already have a wire from pin 3 to the base)

47 ohms 1 watt would probably be about right.

but a better way would be to connect pin to the emitter with a wire and use the resistor between pin 3 and the base.

then it'd be possible get power for the 555 from the ac supply. (using a half bridge another capacitor and a 7805 regulator)

Bye.

And the OP has drawn his bridge wrong. He should either try this with one diode (view in fixed font or M$ Notepad):

| | VCC VCC | + + | | | | .--o---o--. | | | 8 4 | .---||--o-------| --- 25V \_/ | | | .-. | | | | o | | | | | | | | | | | === === === | | 1 5 | '-' GND GND GND | '--o---o--' | | | === | === GND | GND | (created by AACircuit v1.28.6 beta 04/19/05

or this with a bridge rectifier:

| | VCC VCC | + + | | | Dx4 | .--o---o--. |+ | | 8 4 | .---||--o--------o-|

Actually I forgot to put the resistor in the schematic.. sorry there are 2 51 Ohms resistor there, 1/2 watt

ken

yes , but when I do that , the 2, 51 Ohm resistor burns up very rapidly

ken

To clear things up: I am now using a 40v capacitor. before I tresting was using a 12v battery and driving the transistor (no charging battery). I never had this heating problem. I changed the battery for a capacitor at

40v, and now the resistors at the output of the timer going to thebase of the transistor are burning up. To corect this, I used a separate voltage supply for the timer. But again when I connect the grounds together, The resistors burns up again. The only thing that changes is the voltage going from the collector to the emitter. At 12v its ok, but at the 40v it seems to be sucking more current out of the output of the timer. I would imagine my 555 to heat up, but no it does not.

ken

My hunch is that the circuit you have is not the circuit you planned. There has been some error in the construction.

Ok after reading this, I disssembled everything. I found that I had a wire going to the negative of the battery to the ground of the timer... removed that and no heating of my resistors, thanks everyone.

ken