Battery getting hot

Thread Starter

Katherine1

Joined Mar 1, 2013
24
I did a circuit using 9V battery. However, the battery becomes hot very quickly. May I know what;s the problem?

Attached is the schematic circuit. Is there a possibility that I have drawn wrongly which leads to short circuit?
 

Attachments

DerStrom8

Joined Feb 20, 2011
2,390
There are actually a few errors in that schematic, but #12 pointed out the main one. The battery is backwards. Also, pin 4 should be connected directly to +. Right now you're "resetting" the chip at the same time that you "set" it. You can connect it directly to positive in order for the chip to time out on its own, or you can connect it via a separate pull-up resistor to a second switch, that when pressed would ground it and reset the chip before the timing period ends.
 

Thread Starter

Katherine1

Joined Mar 1, 2013
24
There are actually a few errors in that schematic, but #12 pointed out the main one. The battery is backwards. Also, pin 4 should be connected directly to +. Right now you're "resetting" the chip at the same time that you "set" it. You can connect it directly to positive in order for the chip to time out on its own, or you can connect it via a separate pull-up resistor to a second switch, that when pressed would ground it and reset the chip before the timing period ends.
Oh really, thx for pointing out for me. This is my first time doing a circuit. I'll try to reverse the battery and connects pin 4 to the positive. One more problem, the switch i use is a 4 leg push button (https://www.sparkfun.com/products/9190) . when i press it, it pops back. Does that mean it cant be used in my case?
 

DerStrom8

Joined Feb 20, 2011
2,390
Oh really, thx for pointing out for me. This is my first time doing a circuit. I'll try to reverse the battery and connects pin 4 to the positive. One more problem, the switch i use is a 4 leg push button (https://www.sparkfun.com/products/9190) . when i press it, it pops back. Does that mean it cant be used in my case?
I see you're using that pushbutton as the trigger. That is exactly the type of button you want. The 555 will be set with only a single input pulse. Pressing and releasing the button will be perfect for the trigger.
 

LDC3

Joined Apr 27, 2013
924
Another problem is that you are putting about 8 V on the speaker (which is usually 8 ohms). This would result in a 1 amp current. Much more than the battery can supply. I think a typical 9 V battery can supply 0.1 amps for a very short time.
 

Wendy

Joined Mar 24, 2008
22,789
The transistor has no current limiting for the transistor. That is a dead short to the output of the 555.

A transistor BE junction looks like a diode, a forward biased diode. Put a 1KΩ resistor between pin 3 and the Base of the transistor.
 

DerStrom8

Joined Feb 20, 2011
2,390
The transistor has no current limiting for the transistor. That is a dead short to the output of the 555.

A transistor BE junction looks like a diode, a forward biased diode. Put a 1KΩ resistor between pin 3 and the Base of the transistor.
Oops, forgot to put that in my post listing the problems. Thanks Bill :p
 

DerStrom8

Joined Feb 20, 2011
2,390
Another problem is that you are putting about 8 V on the speaker (which is usually 8 ohms). This would result in a 1 amp current. Much more than the battery can supply. I think a typical 9 V battery can supply 0.1 amps for a very short time.
It should still work though, it just won't be as powerful.
 

timescope

Joined Dec 14, 2011
298
BPR1 could be one of those piezo buzzers with a built in oscillator that only requires a DC supply to produce a sound (a cylindrical black thing with a hole on top and two pins on the base)

The circuit is configured as a re-triggerable monostable by connecting the reset and trigger together: when SW2 is pressed, the capacitor is discharged and a new timing cycle begins when it is released.

One possible problem would be the leakage current of the 1000uF capacitor : if it is too large, pin 6 will never reach the threshold of 2/3 of the supply voltage.

Timescope
 

LDC3

Joined Apr 27, 2013
924
BPR1 could be one of those piezo buzzers with a built in oscillator that only requires a DC supply to produce a sound (a cylindrical black thing with a hole on top and two pins on the base)
Well if this is the case, most piezo buzzers have a current limit under 50 mA. With no other resistance, the current would be way to much for it.
 

timescope

Joined Dec 14, 2011
298
Use the "Estimated Service" information in the datasheet.

First you must implement the corrections suggested and get the circuit working properly then measure the current consumption.

1. Correct the battery polarity
2. Add resistor in series with the transistor base
3. Replace the 555 timer as it is probably damaged

Timescope
 

LDC3

Joined Apr 27, 2013
924
From the datasheet, if the load is 620 ohms with an average voltage of 7 volts, the average current would be about 11.3 mA. A load of 180 ohms has an average current of about 38.9 mA.
 
Top