8 channel quiz buzzer circuit using 8051 microcontroller
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
Project: 8 channel quiz buzzer circuit using 8051 microcontroller from this site.
For the first candidate who presses their button, their number will show on the 7-segment display and the buzzer will make a sound.I am using peizo-buzzer in the circuit.
After making all the connections, the display is working but the buzzer is not making any sound. Buzzer delay time in code in 1ms. When we give direct 5V supply to the buzzer it works, but not in the circuit.
Please give me some solution for this.
#include<reg51.h>
#define SEGMENT P2 // PORT2 to Segments of 7-Segment Display
#define SWITCH P1 // Input Switches (buttons) to PORT1
sbit buzz=P3^0; // Buzzer
sbit rst=P3^3; // Reset Switch (Reset the display) - not the microcontroller
sbit digit=P3^7; // 7-Segment Display Common Pin (to enable)
void delay (int); // Delay function
int x=0,y,z;
unsigned char ch={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x98}; // Hexadecimal values from 0 to 9.
void delay (int d)
{
unsigned char i;
for(;d>0;d--)
{
for(i=250;i>0;i--);
for(i=248;i>0;i--);
}
}
void main()
{
SWITCH=0xff;
SEGMENT=0xff;
digit=1;
buzz=0;
rst=1;
while(1)
{
while(SWITCH==0xff); // wait until any button is pressed.
while (SWITCH==0xfe) // Button 1 is pressed.
{
SEGMENT=ch[1];
buzz=1;
delay(1000); // Activate buzzer for 1 second.
buzz=0;
while(rst!=0); // display the digit until the reset is pressed.
}
while (SWITCH==0xfd) // Button 2 is pressed.
{
SEGMENT=ch[2];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xfb) // Button 3 is pressed.
{
SEGMENT=ch[3];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xf7) // Button 4 is pressed.
{
SEGMENT=ch[4];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xef) // Button 5 is pressed.
{
SEGMENT=ch[5];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xdf) // Button 6 is pressed.
{
SEGMENT=ch[6];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xbf) // Button 7 is pressed.
{
SEGMENT=ch[7];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0x7f) // Button 8 is pressed.
{
SEGMENT=ch[8];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
SEGMENT=0xff;
rst=1;
}
}
microcontroller 8051 piezo-buzzer
$endgroup$
|
show 5 more comments
$begingroup$
Project: 8 channel quiz buzzer circuit using 8051 microcontroller from this site.
For the first candidate who presses their button, their number will show on the 7-segment display and the buzzer will make a sound.I am using peizo-buzzer in the circuit.
After making all the connections, the display is working but the buzzer is not making any sound. Buzzer delay time in code in 1ms. When we give direct 5V supply to the buzzer it works, but not in the circuit.
Please give me some solution for this.
#include<reg51.h>
#define SEGMENT P2 // PORT2 to Segments of 7-Segment Display
#define SWITCH P1 // Input Switches (buttons) to PORT1
sbit buzz=P3^0; // Buzzer
sbit rst=P3^3; // Reset Switch (Reset the display) - not the microcontroller
sbit digit=P3^7; // 7-Segment Display Common Pin (to enable)
void delay (int); // Delay function
int x=0,y,z;
unsigned char ch={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x98}; // Hexadecimal values from 0 to 9.
void delay (int d)
{
unsigned char i;
for(;d>0;d--)
{
for(i=250;i>0;i--);
for(i=248;i>0;i--);
}
}
void main()
{
SWITCH=0xff;
SEGMENT=0xff;
digit=1;
buzz=0;
rst=1;
while(1)
{
while(SWITCH==0xff); // wait until any button is pressed.
while (SWITCH==0xfe) // Button 1 is pressed.
{
SEGMENT=ch[1];
buzz=1;
delay(1000); // Activate buzzer for 1 second.
buzz=0;
while(rst!=0); // display the digit until the reset is pressed.
}
while (SWITCH==0xfd) // Button 2 is pressed.
{
SEGMENT=ch[2];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xfb) // Button 3 is pressed.
{
SEGMENT=ch[3];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xf7) // Button 4 is pressed.
{
SEGMENT=ch[4];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xef) // Button 5 is pressed.
{
SEGMENT=ch[5];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xdf) // Button 6 is pressed.
{
SEGMENT=ch[6];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xbf) // Button 7 is pressed.
{
SEGMENT=ch[7];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0x7f) // Button 8 is pressed.
{
SEGMENT=ch[8];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
SEGMENT=0xff;
rst=1;
}
}
microcontroller 8051 piezo-buzzer
$endgroup$
1
$begingroup$
it sounds like the microcontroller output is too weak to drive the buzzer, measure the voltage on the buzzer
$endgroup$
– Jasen
Mar 27 at 8:19
2
$begingroup$
Can you remove disconnect the resistor from pin 10 of the AT89C51 and drive it manually with 5V? (P.S. do use reference names as well for components, it is easier to refer to them that way)
$endgroup$
– Huisman
Mar 27 at 8:58
1
$begingroup$
P3^3
What language is this? It is nonsense in C,^
being the bitwise XOR operator.
$endgroup$
– Lundin
Mar 27 at 9:07
3
$begingroup$
@Lundin It is no nonsense. Please read keil.com/support/man/docs/c51/c51_le_sbit.htm
$endgroup$
– Huisman
Mar 27 at 9:23
1
$begingroup$
@Huisman Of course it is nonsense in C, since the^
operator is already taken and perfectly valid to use inside an initializer. This is clearly not valid C but non-standard extensions. There are of course many other reasons why Keil has such a bad reputation, this is just one of them.
$endgroup$
– Lundin
Mar 27 at 9:39
|
show 5 more comments
$begingroup$
Project: 8 channel quiz buzzer circuit using 8051 microcontroller from this site.
For the first candidate who presses their button, their number will show on the 7-segment display and the buzzer will make a sound.I am using peizo-buzzer in the circuit.
After making all the connections, the display is working but the buzzer is not making any sound. Buzzer delay time in code in 1ms. When we give direct 5V supply to the buzzer it works, but not in the circuit.
Please give me some solution for this.
#include<reg51.h>
#define SEGMENT P2 // PORT2 to Segments of 7-Segment Display
#define SWITCH P1 // Input Switches (buttons) to PORT1
sbit buzz=P3^0; // Buzzer
sbit rst=P3^3; // Reset Switch (Reset the display) - not the microcontroller
sbit digit=P3^7; // 7-Segment Display Common Pin (to enable)
void delay (int); // Delay function
int x=0,y,z;
unsigned char ch={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x98}; // Hexadecimal values from 0 to 9.
void delay (int d)
{
unsigned char i;
for(;d>0;d--)
{
for(i=250;i>0;i--);
for(i=248;i>0;i--);
}
}
void main()
{
SWITCH=0xff;
SEGMENT=0xff;
digit=1;
buzz=0;
rst=1;
while(1)
{
while(SWITCH==0xff); // wait until any button is pressed.
while (SWITCH==0xfe) // Button 1 is pressed.
{
SEGMENT=ch[1];
buzz=1;
delay(1000); // Activate buzzer for 1 second.
buzz=0;
while(rst!=0); // display the digit until the reset is pressed.
}
while (SWITCH==0xfd) // Button 2 is pressed.
{
SEGMENT=ch[2];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xfb) // Button 3 is pressed.
{
SEGMENT=ch[3];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xf7) // Button 4 is pressed.
{
SEGMENT=ch[4];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xef) // Button 5 is pressed.
{
SEGMENT=ch[5];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xdf) // Button 6 is pressed.
{
SEGMENT=ch[6];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xbf) // Button 7 is pressed.
{
SEGMENT=ch[7];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0x7f) // Button 8 is pressed.
{
SEGMENT=ch[8];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
SEGMENT=0xff;
rst=1;
}
}
microcontroller 8051 piezo-buzzer
$endgroup$
Project: 8 channel quiz buzzer circuit using 8051 microcontroller from this site.
For the first candidate who presses their button, their number will show on the 7-segment display and the buzzer will make a sound.I am using peizo-buzzer in the circuit.
After making all the connections, the display is working but the buzzer is not making any sound. Buzzer delay time in code in 1ms. When we give direct 5V supply to the buzzer it works, but not in the circuit.
Please give me some solution for this.
#include<reg51.h>
#define SEGMENT P2 // PORT2 to Segments of 7-Segment Display
#define SWITCH P1 // Input Switches (buttons) to PORT1
sbit buzz=P3^0; // Buzzer
sbit rst=P3^3; // Reset Switch (Reset the display) - not the microcontroller
sbit digit=P3^7; // 7-Segment Display Common Pin (to enable)
void delay (int); // Delay function
int x=0,y,z;
unsigned char ch={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x98}; // Hexadecimal values from 0 to 9.
void delay (int d)
{
unsigned char i;
for(;d>0;d--)
{
for(i=250;i>0;i--);
for(i=248;i>0;i--);
}
}
void main()
{
SWITCH=0xff;
SEGMENT=0xff;
digit=1;
buzz=0;
rst=1;
while(1)
{
while(SWITCH==0xff); // wait until any button is pressed.
while (SWITCH==0xfe) // Button 1 is pressed.
{
SEGMENT=ch[1];
buzz=1;
delay(1000); // Activate buzzer for 1 second.
buzz=0;
while(rst!=0); // display the digit until the reset is pressed.
}
while (SWITCH==0xfd) // Button 2 is pressed.
{
SEGMENT=ch[2];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xfb) // Button 3 is pressed.
{
SEGMENT=ch[3];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xf7) // Button 4 is pressed.
{
SEGMENT=ch[4];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xef) // Button 5 is pressed.
{
SEGMENT=ch[5];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xdf) // Button 6 is pressed.
{
SEGMENT=ch[6];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0xbf) // Button 7 is pressed.
{
SEGMENT=ch[7];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
while (SWITCH==0x7f) // Button 8 is pressed.
{
SEGMENT=ch[8];
buzz=1;
delay(1000);
buzz=0;
while(rst!=0);
}
SEGMENT=0xff;
rst=1;
}
}
microcontroller 8051 piezo-buzzer
microcontroller 8051 piezo-buzzer
edited Mar 27 at 15:51
Shrutika Jagtap
asked Mar 27 at 8:08
Shrutika JagtapShrutika Jagtap
262
262
1
$begingroup$
it sounds like the microcontroller output is too weak to drive the buzzer, measure the voltage on the buzzer
$endgroup$
– Jasen
Mar 27 at 8:19
2
$begingroup$
Can you remove disconnect the resistor from pin 10 of the AT89C51 and drive it manually with 5V? (P.S. do use reference names as well for components, it is easier to refer to them that way)
$endgroup$
– Huisman
Mar 27 at 8:58
1
$begingroup$
P3^3
What language is this? It is nonsense in C,^
being the bitwise XOR operator.
$endgroup$
– Lundin
Mar 27 at 9:07
3
$begingroup$
@Lundin It is no nonsense. Please read keil.com/support/man/docs/c51/c51_le_sbit.htm
$endgroup$
– Huisman
Mar 27 at 9:23
1
$begingroup$
@Huisman Of course it is nonsense in C, since the^
operator is already taken and perfectly valid to use inside an initializer. This is clearly not valid C but non-standard extensions. There are of course many other reasons why Keil has such a bad reputation, this is just one of them.
$endgroup$
– Lundin
Mar 27 at 9:39
|
show 5 more comments
1
$begingroup$
it sounds like the microcontroller output is too weak to drive the buzzer, measure the voltage on the buzzer
$endgroup$
– Jasen
Mar 27 at 8:19
2
$begingroup$
Can you remove disconnect the resistor from pin 10 of the AT89C51 and drive it manually with 5V? (P.S. do use reference names as well for components, it is easier to refer to them that way)
$endgroup$
– Huisman
Mar 27 at 8:58
1
$begingroup$
P3^3
What language is this? It is nonsense in C,^
being the bitwise XOR operator.
$endgroup$
– Lundin
Mar 27 at 9:07
3
$begingroup$
@Lundin It is no nonsense. Please read keil.com/support/man/docs/c51/c51_le_sbit.htm
$endgroup$
– Huisman
Mar 27 at 9:23
1
$begingroup$
@Huisman Of course it is nonsense in C, since the^
operator is already taken and perfectly valid to use inside an initializer. This is clearly not valid C but non-standard extensions. There are of course many other reasons why Keil has such a bad reputation, this is just one of them.
$endgroup$
– Lundin
Mar 27 at 9:39
1
1
$begingroup$
it sounds like the microcontroller output is too weak to drive the buzzer, measure the voltage on the buzzer
$endgroup$
– Jasen
Mar 27 at 8:19
$begingroup$
it sounds like the microcontroller output is too weak to drive the buzzer, measure the voltage on the buzzer
$endgroup$
– Jasen
Mar 27 at 8:19
2
2
$begingroup$
Can you remove disconnect the resistor from pin 10 of the AT89C51 and drive it manually with 5V? (P.S. do use reference names as well for components, it is easier to refer to them that way)
$endgroup$
– Huisman
Mar 27 at 8:58
$begingroup$
Can you remove disconnect the resistor from pin 10 of the AT89C51 and drive it manually with 5V? (P.S. do use reference names as well for components, it is easier to refer to them that way)
$endgroup$
– Huisman
Mar 27 at 8:58
1
1
$begingroup$
P3^3
What language is this? It is nonsense in C, ^
being the bitwise XOR operator.$endgroup$
– Lundin
Mar 27 at 9:07
$begingroup$
P3^3
What language is this? It is nonsense in C, ^
being the bitwise XOR operator.$endgroup$
– Lundin
Mar 27 at 9:07
3
3
$begingroup$
@Lundin It is no nonsense. Please read keil.com/support/man/docs/c51/c51_le_sbit.htm
$endgroup$
– Huisman
Mar 27 at 9:23
$begingroup$
@Lundin It is no nonsense. Please read keil.com/support/man/docs/c51/c51_le_sbit.htm
$endgroup$
– Huisman
Mar 27 at 9:23
1
1
$begingroup$
@Huisman Of course it is nonsense in C, since the
^
operator is already taken and perfectly valid to use inside an initializer. This is clearly not valid C but non-standard extensions. There are of course many other reasons why Keil has such a bad reputation, this is just one of them.$endgroup$
– Lundin
Mar 27 at 9:39
$begingroup$
@Huisman Of course it is nonsense in C, since the
^
operator is already taken and perfectly valid to use inside an initializer. This is clearly not valid C but non-standard extensions. There are of course many other reasons why Keil has such a bad reputation, this is just one of them.$endgroup$
– Lundin
Mar 27 at 9:39
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The 8051 and clones typically use an I/O configuration that is described as "pseudo-bidirectional". They have active pull-down but quasi-passive pull-up (there's a transistor that is turned on briefly to improve the rise time). This means that they can only source a tiny amount of continuous current to an external device. The datasheet (DC Characteristics, page 10) shows that VOH drops to 2.4V at just 60 µA of current.
This is not enough current to drive your NPN transistor. Instead, try putting a logic-level N-channel MOSFET there.
Note also that your other NPN (the one attached to pin 17 — you really do need to use reference designators in your schematics!) should actually be a PNP. This requires inverting the logic in your code that drives pin 17.
$endgroup$
$begingroup$
There's only one digit of display so the transistor attached to pin 17 is not even needed at all. The SEGMENT=0xFF line at the end of the loop blanks the display.
$endgroup$
– Finbarr
Mar 27 at 13:04
$begingroup$
@Dave Tweed any reference for MOSFET ...I want MOSFET number.
$endgroup$
– Shrutika Jagtap
Apr 2 at 10:26
$begingroup$
You could try the 2N7000 or the BSS138.
$endgroup$
– Dave Tweed♦
Apr 2 at 11:04
add a comment |
$begingroup$
You don't say what type of device the buzzer is (electromechanical, piezo, etc.) or how much current it requires to operate, but I don't think either of those is the problem.
One millisecond is too short a time for the buzzer to wake up and make enough noise to be heard. Increase the on-time to 1 second to verify that the code is working, then adjust to taste.
BTW, what type of buzzer is it? Datasheet or link?
$endgroup$
1
$begingroup$
Hi, "One millisecond is too short a time for the buzzer to wake up" FYI the OP isn't claiming to drive the buzzer for 1ms. The code (I added a link to the original source page) is supposed to switch on the buzzer, delay for 1000 x 1ms = 1s (hence thedelay(1000);
), then switch the buzzer off. There are comments in the original code which I linked (not included by the OP) which make this clearer. That doesn't answer all of your questions, but I hope their intent is now clearer.
$endgroup$
– SamGibson
Mar 27 at 12:18
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The 8051 and clones typically use an I/O configuration that is described as "pseudo-bidirectional". They have active pull-down but quasi-passive pull-up (there's a transistor that is turned on briefly to improve the rise time). This means that they can only source a tiny amount of continuous current to an external device. The datasheet (DC Characteristics, page 10) shows that VOH drops to 2.4V at just 60 µA of current.
This is not enough current to drive your NPN transistor. Instead, try putting a logic-level N-channel MOSFET there.
Note also that your other NPN (the one attached to pin 17 — you really do need to use reference designators in your schematics!) should actually be a PNP. This requires inverting the logic in your code that drives pin 17.
$endgroup$
$begingroup$
There's only one digit of display so the transistor attached to pin 17 is not even needed at all. The SEGMENT=0xFF line at the end of the loop blanks the display.
$endgroup$
– Finbarr
Mar 27 at 13:04
$begingroup$
@Dave Tweed any reference for MOSFET ...I want MOSFET number.
$endgroup$
– Shrutika Jagtap
Apr 2 at 10:26
$begingroup$
You could try the 2N7000 or the BSS138.
$endgroup$
– Dave Tweed♦
Apr 2 at 11:04
add a comment |
$begingroup$
The 8051 and clones typically use an I/O configuration that is described as "pseudo-bidirectional". They have active pull-down but quasi-passive pull-up (there's a transistor that is turned on briefly to improve the rise time). This means that they can only source a tiny amount of continuous current to an external device. The datasheet (DC Characteristics, page 10) shows that VOH drops to 2.4V at just 60 µA of current.
This is not enough current to drive your NPN transistor. Instead, try putting a logic-level N-channel MOSFET there.
Note also that your other NPN (the one attached to pin 17 — you really do need to use reference designators in your schematics!) should actually be a PNP. This requires inverting the logic in your code that drives pin 17.
$endgroup$
$begingroup$
There's only one digit of display so the transistor attached to pin 17 is not even needed at all. The SEGMENT=0xFF line at the end of the loop blanks the display.
$endgroup$
– Finbarr
Mar 27 at 13:04
$begingroup$
@Dave Tweed any reference for MOSFET ...I want MOSFET number.
$endgroup$
– Shrutika Jagtap
Apr 2 at 10:26
$begingroup$
You could try the 2N7000 or the BSS138.
$endgroup$
– Dave Tweed♦
Apr 2 at 11:04
add a comment |
$begingroup$
The 8051 and clones typically use an I/O configuration that is described as "pseudo-bidirectional". They have active pull-down but quasi-passive pull-up (there's a transistor that is turned on briefly to improve the rise time). This means that they can only source a tiny amount of continuous current to an external device. The datasheet (DC Characteristics, page 10) shows that VOH drops to 2.4V at just 60 µA of current.
This is not enough current to drive your NPN transistor. Instead, try putting a logic-level N-channel MOSFET there.
Note also that your other NPN (the one attached to pin 17 — you really do need to use reference designators in your schematics!) should actually be a PNP. This requires inverting the logic in your code that drives pin 17.
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The 8051 and clones typically use an I/O configuration that is described as "pseudo-bidirectional". They have active pull-down but quasi-passive pull-up (there's a transistor that is turned on briefly to improve the rise time). This means that they can only source a tiny amount of continuous current to an external device. The datasheet (DC Characteristics, page 10) shows that VOH drops to 2.4V at just 60 µA of current.
This is not enough current to drive your NPN transistor. Instead, try putting a logic-level N-channel MOSFET there.
Note also that your other NPN (the one attached to pin 17 — you really do need to use reference designators in your schematics!) should actually be a PNP. This requires inverting the logic in your code that drives pin 17.
edited Mar 27 at 11:33
answered Mar 27 at 11:24
Dave Tweed♦Dave Tweed
123k10153267
123k10153267
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There's only one digit of display so the transistor attached to pin 17 is not even needed at all. The SEGMENT=0xFF line at the end of the loop blanks the display.
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– Finbarr
Mar 27 at 13:04
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@Dave Tweed any reference for MOSFET ...I want MOSFET number.
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– Shrutika Jagtap
Apr 2 at 10:26
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You could try the 2N7000 or the BSS138.
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– Dave Tweed♦
Apr 2 at 11:04
add a comment |
$begingroup$
There's only one digit of display so the transistor attached to pin 17 is not even needed at all. The SEGMENT=0xFF line at the end of the loop blanks the display.
$endgroup$
– Finbarr
Mar 27 at 13:04
$begingroup$
@Dave Tweed any reference for MOSFET ...I want MOSFET number.
$endgroup$
– Shrutika Jagtap
Apr 2 at 10:26
$begingroup$
You could try the 2N7000 or the BSS138.
$endgroup$
– Dave Tweed♦
Apr 2 at 11:04
$begingroup$
There's only one digit of display so the transistor attached to pin 17 is not even needed at all. The SEGMENT=0xFF line at the end of the loop blanks the display.
$endgroup$
– Finbarr
Mar 27 at 13:04
$begingroup$
There's only one digit of display so the transistor attached to pin 17 is not even needed at all. The SEGMENT=0xFF line at the end of the loop blanks the display.
$endgroup$
– Finbarr
Mar 27 at 13:04
$begingroup$
@Dave Tweed any reference for MOSFET ...I want MOSFET number.
$endgroup$
– Shrutika Jagtap
Apr 2 at 10:26
$begingroup$
@Dave Tweed any reference for MOSFET ...I want MOSFET number.
$endgroup$
– Shrutika Jagtap
Apr 2 at 10:26
$begingroup$
You could try the 2N7000 or the BSS138.
$endgroup$
– Dave Tweed♦
Apr 2 at 11:04
$begingroup$
You could try the 2N7000 or the BSS138.
$endgroup$
– Dave Tweed♦
Apr 2 at 11:04
add a comment |
$begingroup$
You don't say what type of device the buzzer is (electromechanical, piezo, etc.) or how much current it requires to operate, but I don't think either of those is the problem.
One millisecond is too short a time for the buzzer to wake up and make enough noise to be heard. Increase the on-time to 1 second to verify that the code is working, then adjust to taste.
BTW, what type of buzzer is it? Datasheet or link?
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1
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Hi, "One millisecond is too short a time for the buzzer to wake up" FYI the OP isn't claiming to drive the buzzer for 1ms. The code (I added a link to the original source page) is supposed to switch on the buzzer, delay for 1000 x 1ms = 1s (hence thedelay(1000);
), then switch the buzzer off. There are comments in the original code which I linked (not included by the OP) which make this clearer. That doesn't answer all of your questions, but I hope their intent is now clearer.
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– SamGibson
Mar 27 at 12:18
add a comment |
$begingroup$
You don't say what type of device the buzzer is (electromechanical, piezo, etc.) or how much current it requires to operate, but I don't think either of those is the problem.
One millisecond is too short a time for the buzzer to wake up and make enough noise to be heard. Increase the on-time to 1 second to verify that the code is working, then adjust to taste.
BTW, what type of buzzer is it? Datasheet or link?
$endgroup$
1
$begingroup$
Hi, "One millisecond is too short a time for the buzzer to wake up" FYI the OP isn't claiming to drive the buzzer for 1ms. The code (I added a link to the original source page) is supposed to switch on the buzzer, delay for 1000 x 1ms = 1s (hence thedelay(1000);
), then switch the buzzer off. There are comments in the original code which I linked (not included by the OP) which make this clearer. That doesn't answer all of your questions, but I hope their intent is now clearer.
$endgroup$
– SamGibson
Mar 27 at 12:18
add a comment |
$begingroup$
You don't say what type of device the buzzer is (electromechanical, piezo, etc.) or how much current it requires to operate, but I don't think either of those is the problem.
One millisecond is too short a time for the buzzer to wake up and make enough noise to be heard. Increase the on-time to 1 second to verify that the code is working, then adjust to taste.
BTW, what type of buzzer is it? Datasheet or link?
$endgroup$
You don't say what type of device the buzzer is (electromechanical, piezo, etc.) or how much current it requires to operate, but I don't think either of those is the problem.
One millisecond is too short a time for the buzzer to wake up and make enough noise to be heard. Increase the on-time to 1 second to verify that the code is working, then adjust to taste.
BTW, what type of buzzer is it? Datasheet or link?
answered Mar 27 at 11:42
AnalogKidAnalogKid
2,78837
2,78837
1
$begingroup$
Hi, "One millisecond is too short a time for the buzzer to wake up" FYI the OP isn't claiming to drive the buzzer for 1ms. The code (I added a link to the original source page) is supposed to switch on the buzzer, delay for 1000 x 1ms = 1s (hence thedelay(1000);
), then switch the buzzer off. There are comments in the original code which I linked (not included by the OP) which make this clearer. That doesn't answer all of your questions, but I hope their intent is now clearer.
$endgroup$
– SamGibson
Mar 27 at 12:18
add a comment |
1
$begingroup$
Hi, "One millisecond is too short a time for the buzzer to wake up" FYI the OP isn't claiming to drive the buzzer for 1ms. The code (I added a link to the original source page) is supposed to switch on the buzzer, delay for 1000 x 1ms = 1s (hence thedelay(1000);
), then switch the buzzer off. There are comments in the original code which I linked (not included by the OP) which make this clearer. That doesn't answer all of your questions, but I hope their intent is now clearer.
$endgroup$
– SamGibson
Mar 27 at 12:18
1
1
$begingroup$
Hi, "One millisecond is too short a time for the buzzer to wake up" FYI the OP isn't claiming to drive the buzzer for 1ms. The code (I added a link to the original source page) is supposed to switch on the buzzer, delay for 1000 x 1ms = 1s (hence the
delay(1000);
), then switch the buzzer off. There are comments in the original code which I linked (not included by the OP) which make this clearer. That doesn't answer all of your questions, but I hope their intent is now clearer.$endgroup$
– SamGibson
Mar 27 at 12:18
$begingroup$
Hi, "One millisecond is too short a time for the buzzer to wake up" FYI the OP isn't claiming to drive the buzzer for 1ms. The code (I added a link to the original source page) is supposed to switch on the buzzer, delay for 1000 x 1ms = 1s (hence the
delay(1000);
), then switch the buzzer off. There are comments in the original code which I linked (not included by the OP) which make this clearer. That doesn't answer all of your questions, but I hope their intent is now clearer.$endgroup$
– SamGibson
Mar 27 at 12:18
add a comment |
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$begingroup$
it sounds like the microcontroller output is too weak to drive the buzzer, measure the voltage on the buzzer
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– Jasen
Mar 27 at 8:19
2
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Can you remove disconnect the resistor from pin 10 of the AT89C51 and drive it manually with 5V? (P.S. do use reference names as well for components, it is easier to refer to them that way)
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– Huisman
Mar 27 at 8:58
1
$begingroup$
P3^3
What language is this? It is nonsense in C,^
being the bitwise XOR operator.$endgroup$
– Lundin
Mar 27 at 9:07
3
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@Lundin It is no nonsense. Please read keil.com/support/man/docs/c51/c51_le_sbit.htm
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– Huisman
Mar 27 at 9:23
1
$begingroup$
@Huisman Of course it is nonsense in C, since the
^
operator is already taken and perfectly valid to use inside an initializer. This is clearly not valid C but non-standard extensions. There are of course many other reasons why Keil has such a bad reputation, this is just one of them.$endgroup$
– Lundin
Mar 27 at 9:39