another quickie - Electronic Circuits Projects Diagrams Free

andrew2022 · 12th July 2004, 07:51 PM

i need help with a buzzer circuit. i have 4 inputs (mayb more, but once i get the 4 workin it shudnt be too hard for me to expand it) what i need connected to a buzzer. what i need is whan an input goes high for the buzzer to sound until a button is pressed, but i need the buzzer to then stay off whilst the input is high, and not sound until the output goes low then high again. the harder part is i need it to sound once another input goes high, even if another input is high.

does any1 know how this can be done (pref using logic, not microcontrollers) or or can give me any advice on how it can be done?

thanx

laroche73 · 12th July 2004, 09:15 PM

You need edge-triggered (as opposed to level-triggered) logic to do this. You'll probably also need to debounce the inputs and buffer the output. Feed each (debounced) input to the clock pin of a positive-edge triggered D-type flip-flop (1/2 of a 74XX74, tie the "D" and "Preset" inputs high). Feed the outputs to an OR gate (a 74XX32 wired properly will substitute for a four-input OR gate). Tie the "Clear" lines of the flip-flops together with a pullup to VCC, and place the pushbutton switch between this node and ground. Two 74HC74's and a 74HC32 would be enough for the logic function.

On second thought, the inputs probably don't need to be debounced. Multiple high-to-low transitions at the inputs due to switch bounce won't change the flip-flops' state after the first low-to-high edge. Only pulling the "Clear" line low will.

andrew2022 · 13th July 2004, 04:56 PM

er... could someone please repeat that in english? (or make a lil bit more simple to understand). any1 have any diagrams?

laroche73 · 13th July 2004, 05:42 PM

Sorry about that, Andrew. Didn't really have time to draw it up last night. The attached schematic should help. You could also substitute a 4-input NAND gate like the 74HC20 for the multiple 74HC32 OR gates, as long as you use the Q' (not) outputs from the flip-flops instead of the Q outputs to drive the NAND gate inputs (A NAND gate is equivalent to a negative-logic OR gate). It depends what parts you have on hand, either way it's the same package count.

andrew2022 · 13th July 2004, 06:06 PM

thanx for the help. will make that circuit now

andrew2022 · 13th July 2004, 06:15 PM

yet another question. for the input, i need power to flow back to ground, allowing the LED to light. could the input be connected to ground via a resistor allowing the input to the buzzer circuit to read the 'high'?

laroche73 · 13th July 2004, 08:16 PM

Quote:

could the input be connected to ground via a resistor allowing the input to the buzzer circuit to read the 'high'?

It depends what you want the LEDs to do. If you want the LED to only come on while the input's held high, then yes. The R1/D1 connection in the schematic below shows how to do this.

If you want the LED to come on and stay on until the reset button is pushed, you can use the flip-flops Q' output to drive the LED, as shown with R2/D2 (You'll need to calculate the resistor values for your LEDs).

There's a potential problem to be aware of when using LEDs on input circuits like this. The forward voltage across a LED is typically around 1.6 - 2.2V (depends on LED color and other factors). If resistor R1 was connected to VCC and LED D1s' cathode to ground, the voltage across the LED may not be enough to guarantee a high TTL level. That's why the schematic shows R1/D1 swapped from their usual positions. Connected this way, the input sees the voltage drop across the resistor (VCC - Vf) instead of the LED's forward voltage. This should be over 3V with red, green, orange or yellow LEDs, and is enough for a TTL high level.

Since the first method depends to some extent on the forward voltage of the LEDs, the second arrangement (driving the LEDs from the flip-flop outputs) is preferred. In fact, if you were using a blue LED, you'd probably have to switch the positions of R1 and D1 back again, due to the high Vf of blue LEDs (~ 3.7 - 4V). Not great from a design standpoint.

andrew2022 · 13th July 2004, 08:42 PM

thanx for the help. i have not yet been able to make the circuit, due to the fact i dont have the parts (altho al b orderin them ASAP)

also, is there a CMOS version of the 74hc74? (i have looked thru a few catalogues but cudnt find 1, i found a CMOS alternative to the 74hc32) (the circuit will be running from 12V)

one last thing: the TTL versions are rated to 6V, but what can the maximum high input voltage be (not the supply). would i have to lower this to lower than 6V?

laroche73 · 13th July 2004, 09:04 PM

The 74HC series is a CMOS family, the HC stands for "high-speed CMOS". The supply voltage max is 6V (Max input voltage is the same as the supply voltage). The older CD4000 series is the only CMOS family that will operate at 12V.

An equivalent (almost) to the 74HC74 is the CD4013B dual D-type flip flop. However, the polarities of the Set/Reset inputs are reversed (active-high instead of active-low). A substitute for the 74HC32 is the CD4071B quad 2-input OR gate. Or you could use 1/2 of a CD4072B dual 4-input OR gate. You'll almost certainly need an additional driver after the OR gate for the buzzer, CD4000 parts have very limited drive capability.

The warning about LED's and input levels still applies, though the nominal switching point for CMOS inputs is 1/2 Vcc. 'HC logic behaves this way as well, 'HCT logic is a CMOS family with TTL-type input switching thresholds.

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12th July 2004, 07:51 PM	(permalink)
andrew2022	another quickie i need help with a buzzer circuit. i have 4 inputs (mayb more, but once i get the 4 workin it shudnt be too hard for me to expand it) what i need connected to a buzzer. what i need is whan an input goes high for the buzzer to sound until a button is pressed, but i need the buzzer to then stay off whilst the input is high, and not sound until the output goes low then high again. the harder part is i need it to sound once another input goes high, even if another input is high. does any1 know how this can be done (pref using logic, not microcontrollers) or or can give me any advice on how it can be done? thanx

12th July 2004, 09:15 PM	(permalink)
laroche73	buzzer circuit You need edge-triggered (as opposed to level-triggered) logic to do this. You'll probably also need to debounce the inputs and buffer the output. Feed each (debounced) input to the clock pin of a positive-edge triggered D-type flip-flop (1/2 of a 74XX74, tie the "D" and "Preset" inputs high). Feed the outputs to an OR gate (a 74XX32 wired properly will substitute for a four-input OR gate). Tie the "Clear" lines of the flip-flops together with a pullup to VCC, and place the pushbutton switch between this node and ground. Two 74HC74's and a 74HC32 would be enough for the logic function. On second thought, the inputs probably don't need to be debounced. Multiple high-to-low transitions at the inputs due to switch bounce won't change the flip-flops' state after the first low-to-high edge. Only pulling the "Clear" line low will.

13th July 2004, 04:56 PM	(permalink)
andrew2022	er... could someone please repeat that in english? (or make a lil bit more simple to understand). any1 have any diagrams?

13th July 2004, 06:06 PM	(permalink)
andrew2022	thanx for the help. will make that circuit now

13th July 2004, 06:15 PM	(permalink)
andrew2022	yet another question. for the input, i need power to flow back to ground, allowing the LED to light. could the input be connected to ground via a resistor allowing the input to the buzzer circuit to read the 'high'?

13th July 2004, 08:42 PM	(permalink)
andrew2022	thanx for the help. i have not yet been able to make the circuit, due to the fact i dont have the parts (altho al b orderin them ASAP) also, is there a CMOS version of the 74hc74? (i have looked thru a few catalogues but cudnt find 1, i found a CMOS alternative to the 74hc32) (the circuit will be running from 12V) one last thing: the TTL versions are rated to 6V, but what can the maximum high input voltage be (not the supply). would i have to lower this to lower than 6V?

13th July 2004, 09:04 PM	(permalink)
laroche73	CMOS versions The 74HC series is a CMOS family, the HC stands for "high-speed CMOS". The supply voltage max is 6V (Max input voltage is the same as the supply voltage). The older CD4000 series is the only CMOS family that will operate at 12V. An equivalent (almost) to the 74HC74 is the CD4013B dual D-type flip flop. However, the polarities of the Set/Reset inputs are reversed (active-high instead of active-low). A substitute for the 74HC32 is the CD4071B quad 2-input OR gate. Or you could use 1/2 of a CD4072B dual 4-input OR gate. You'll almost certainly need an additional driver after the OR gate for the buzzer, CD4000 parts have very limited drive capability. The warning about LED's and input levels still applies, though the nominal switching point for CMOS inputs is 1/2 Vcc. 'HC logic behaves this way as well, 'HCT logic is a CMOS family with TTL-type input switching thresholds.