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CMOS 555 Output Voltage drop

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trevor261

New Member
Hi all. I am using an LMC555 for a monostable circuit in a LV project. The chip is set up to operate a buzzer via pin 3 every time I trigger it. I have been forced to use a transistor to drive the buzzer because the voltage level at the output pin is too low. The supply voltage is about 3.2V (2 X AAA) and the datasheet specifies only a small voltage drop at the output. My buzzer can sound at 1V but when I connect it between pin-3 and Ground I get nothing. Measuring the voltage across it when the LMC555 is triggered I get about 1.5V. If I replace the CMOS 555 with a normal 555 timer (~NE555) the circuit works fine. I don't want to use a normal 555 timer though because it won't work when the batteries fall below 3V, draws too much current, and requires two more capacitors in my circuit.

So my issues are: - Why am I getting such a small output voltage in comparison to the supply?
- Why isn't my buzzer working with the LMC555 even though it can sound at 1.5V?

Thanks, Trevor G.
 

BrownOut

Banned
Your buzzer is probably loading down the timer's output. Using a transistor to buffer the signal is correct.
 

audioguru

Well-Known Member
Most Helpful Member
Your battery is 3.2V when it is brand new then its voltage quickly drops to 2.4V.

The datasheet of the Intersil ICM7555 shows a graph of its typical output current and voltage when the supply is 2.0V, 5.0V and 18.0V.
With a 2.0V supply its output high current is only 1mA when it has a 1V loss. Its current is only 1.1mA into a dead short.
With a 5.0V supply its output high current is only 5mA when it has a 2.5V loss. Its current is 7mA into a dead short.

The output low current of a Cmos 555 is much higher.

The output high or low current of a normal 555 exceeds 100mA when it has a 2V loss, a huge difference.

Your buzzer probably needs a low driving impedance but a Cmos 555 has a fairly high output impedance.
 

BrownOut

Banned
I knew you'd come up with the numbers and a graph. I tried to beat you to it, but couldn't find the info. :)
 

audioguru

Well-Known Member
Most Helpful Member
I didn't bother posting the graph because many people do not understand that the Cmos 555 has a fairly high output resistance (which results in a voltage drop and low output current) when its output is high.
 

trevor261

New Member
I guess this would be the disadvantage of the CMOS 555 over the traditional 555. My buzzer draws about 1.5mA at 3V, I expected the chip to be able to handle that much at least. I was going off the National LMC7555 datasheet and it states that the chup can source up to 50mA no worries and at Vs = 5V and Isrc = 2mA, the typical output voltage is 4.7V. Looks like I'll have to stick with the transistor man for my little buzz.
 

BrownOut

Banned
No. The datasheet says Voh is 4.7V (typ) at only -2mA when Vs is 5V.

Oh wait, you are using a different part. You said LMC555.
 
Last edited:

audioguru

Well-Known Member
Most Helpful Member
I guess this would be the disadvantage of the CMOS 555 over the traditional 555. My buzzer draws about 1.5mA at 3V, I expected the chip to be able to handle that much at least. I was going off the National LMC7555 datasheet and it states that the chup can source up to 50mA no worries and at Vs = 5V and Isrc = 2mA, the typical output voltage is 4.7V. Looks like I'll have to stick with the transistor man for my little buzz.
There is an LMC555, a TLC555 and an ICM7555 and they are all Cmos 555s with the same spec's.
National's datasheet says that it "is tested to -10mA and +50ma output levels" but the spec's show that its output high current is 10mA when its supply is 12V and its output voltage loss is a max of 1.5V. Its output low current is 50mA when its supply is 12V and its output voltage loss is a max of 2V.

I suspect that your "buzzer' is actually a piezo beeper with a built-in oscillator that uses 1.5mA at 3V but needs to be driven from a low impedance. The LMC555 will have an output of 1.5mA at 3V when its supply is about 4.5V. Try it with a 100uF capacitor on its output to ground to make it a low impedance.

They show much less current at lower supply voltages.
 
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