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Why drop in voltage in this circuit ?

Angy15

Nov 18, 2015
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In this simple circuit. When Transistor is ON ideally we should get 10V due to voltage divider for the buzzer to operate.But I am getting 6.45V.So I changed R1 to 470 E and I am getting 9V.

So my doubt is why am I getting 6.45 V instead of 10V . Is this due to limitng current due to resistance R1?
 

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Sadlercomfort

Ash
Feb 9, 2013
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Perhaps the resistance of the buzzer is effecting the voltage divider.

Try and find the resistance of the buzzer, then work out R2 and the 'R3' in parallel.

You can work out the resistance of a buzzer, for example a 9V buzzer pulling 10mA:

9/10mA =900Ω
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
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This is happening because you're you've failed to include the resistance of the Buzzer in your voltage divider calculations.
I suggest that you use the transistor itself to produce the 9V to drive the buzzer directly off T1's Cathode. You can do this by moving your voltage divider to the base of T1. T1's Cathode voltage will be ~ 10V - .7V= 9.3V.

Note: The above assumes that the input of the voltage divider =20V. ;)

Chris
 

AnalogKid

Jun 10, 2015
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What is the voltage on the base of T1 when the buzzer is on?

ak
 

davenn

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Perhaps the resistance of the buzzer is effecting the voltage divider.

Try and find the resistance of the buzzer, then work out R2 and the 'R3' in parallel.

You can work out the resistance of a buzzer, for example a 9V buzzer pulling 10mA:

9/10mA =900Ω


Yup, nice one, there's always a good reason for these things and you realised what the Op didn't ;)


Dave
 

CDRIVE

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Yup, nice one, there's always a good reason for these things and you realised what the Op didn't ;)


Dave
Yes my post alluded to that. That's why I told him to move the divider to the base. The transistor's current gain will enable him to use the very common 1K resistors he's now using. It will also have the advantage of handling a greater range buzzer resistances without effecting the required values of the voltage divider.

Chris
 

CDRIVE

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Here is what Chris is suggesting:

View attachment 24816
Yes Bob except, that although Angy15's schematic doesn't indicate it, I'm assuming he's gating the base of Q1 with a switch or from a previous un-shown stage. So if it's a N/O switch it would be placed between the Vcc node of R1 and the Vcc respectively. If it's from a previous stage then R1 would be the collector resistor of the previous stage. ;)

Chris
 

BobK

Jan 5, 2010
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Yep, I thought of that, but didn't know which so I just drew the always on schematic.

Bob
 

dorke

Jun 20, 2015
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Angy,
Why are you using 20V ?
...that is a very "strange" value to use.
Isn't there a 9v or 12V in your "circuit".
 

CDRIVE

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Yup, I knew that you knew what I was referring to. I just wanted everyone else, especially Angy15 to understand. Thanks for posting it.

Chris
 

eetech00

Nov 17, 2014
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Move the buzzer to the collector side of Q1.

20V supply is a little odd though.

Also, add a diode across the coil of the buzzer
 

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