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I am currently studying an intro course in digital logic and had a small project I needed to create. I had to use a NOT gate(74HC04) to set off a buzzer when a simulated front door of a house opens(so when a switch opens, the buzzer goes off).

It was a really simple circuit and I got full marks but my question is this, the output voltage of my NOT gate(74HC04) is +5V and is enough to make the buzzer sound but not very loud at all. Is there any way to use the NOT gate setup but somehow also have the buzzer connected to a 9V battery to make the sound louder? I read that the maximum voltage for the IC(74HC04) is +7.0V so I'm worried I will break the IC connecting it straight to the 9V battery.

So to sum up my problem, I need the logic function of the NOT gate but need the buzzer to have +9V across it when a switch opens/breaks. If you cannot do this with a logic gate is there any other simple way this circuit can be set up?

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    \$\begingroup\$ There are some members of the 74 family designed for this. They have open-collector outputs that can be pulled up to 30V. Check out 7406 and 7407 for a start. \$\endgroup\$ – Brian Drummond Apr 25 '16 at 22:05
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One way you could go about it is by using a logic level MOSFET. It is called logic level because you can fully turn it on (as in turn on a switch) with logic levels (~ 4.5 to 5V) at its gate.

schematic

simulate this circuit – Schematic created using CircuitLab

So what this does is the following: When there is 5 volts at the output of the NOT gate, that voltage will turn the MOSFET on since it is working as a switch here. Turning on the MOSFET means that there is a conducting path between the source and drain terminals and therefore the buzzer circuit is now complete (It will sound). When there is 0 volts coming from the NOT gate, then the MOSFET will be off and the buzzer won't have a ground connection and hence won't sound.

The 10K resistor has to do with some of the characteristics of the MOSFET such as large input impedance and capacitance. It just ensures that the MOSFET won't stay on after you once turned it on or will randomly turn on by means of static charges etc.

Since it is an intro course, this may be out of the scope but that is how these issues are typically addressed. Hope it helps!

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  • \$\begingroup\$ Thanks so much for this response, this is the circuit I was looking for! Am I right is thinking that all 3 for those triangles are ground and they can all be connected together? \$\endgroup\$ – Joey Apr 26 '16 at 1:02
  • \$\begingroup\$ Yes, you are correct! All of those are ground connections, you can connect them up together. \$\endgroup\$ – Big6 Apr 26 '16 at 2:08
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This is easier than you think . . .

schematic

simulate this circuit – Schematic created using CircuitLab

You can drive it with close to 10 volts (p-p) without a 9 volt battery.

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    \$\begingroup\$ This still only goes 0-vcc; logical not flips between high<->low; low is almost always 0v, high, is 5v, 3v3, or whatever. One cannot get more voltage out than high-low! \$\endgroup\$ – esoterik Apr 25 '16 at 23:34
  • \$\begingroup\$ We are swinging a total of (nearly) 10 volts across the buzzer's terminals. A 5 volt bipolar drive is more power than a 9 volt uniploar drive. \$\endgroup\$ – Mark Apr 25 '16 at 23:40
  • \$\begingroup\$ Another way to look at it: The unipolar method is swinging from 0 to 9 volts. The bipolar method is swinging from -5 volts to +5 volts. \$\endgroup\$ – Mark Apr 25 '16 at 23:50
  • \$\begingroup\$ Where are you getting -5 volts, logical low is 0 v, logical high is VCC (nominally 5v or 3v3); not gates convert 5v to 0v and 0v to 5v, there is on -5v! N.b. one can not create more voltage than your supply; to get 10 volts one would need two supplies a +5v and a -5V. Standard logic is, however, High 5v, Low 0v (aka ground); n.b. high can be other voltages eg. 3v3, 1v5 or 1v8.) If you are trying to describe a non-standard setup (e.g. setting GND to -5v) you should indicate that in your schematic! \$\endgroup\$ – esoterik Apr 26 '16 at 0:14
  • \$\begingroup\$ If you look at the voltage across the buzzer terminals with a voltmeter (regardless of polarity), then you will measure +5 volts across the buzzer when the input is in one state, and -5 volts across the buzzer when the the input is in the opposite state. Can't you see that? I suspect that you just aren't looking. You need to stop thinking of everything in the circuit being referenced to ground, because, in this configuration, the buzzer is not. \$\endgroup\$ – Mark Apr 26 '16 at 0:35
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Check out the 7406. It has an "open collector" output, which means the output transistor isn't connected to the 5v supply. Place your buzzer between the 9v and this output pin, and it will go to ground when the input is high. Note that there are limits--too much current, or voltage spikes from a mechanical buzzer can fry the chip. But in principle, or with a small enough buzzer, this will work.

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  • \$\begingroup\$ But I'm voting for @Mark's answer :) \$\endgroup\$ – gbarry Apr 26 '16 at 0:01
  • \$\begingroup\$ you mean the one that is wrong? \$\endgroup\$ – esoterik Apr 26 '16 at 0:18
  • \$\begingroup\$ Try the simulation, if you think it is wrong. \$\endgroup\$ – Mark Apr 26 '16 at 0:37
  • \$\begingroup\$ @mark I did! there is never more than 5 volts across the buzzer! \$\endgroup\$ – esoterik Apr 26 '16 at 0:49
  • \$\begingroup\$ I never said there would be. There is not supposed to be. \$\endgroup\$ – Mark Apr 26 '16 at 1:08

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