hello friends i seek ur guidence in understanding this simple looking circuit

 

heellllllo

 

There is a circuit simulator here:
Maybe playing around with it will help you understand how transistors work.
http://www.falstad.com/circuit/index.html

 

Y is R2 used.i undstood R1 which limits d input volt..but what is the function of R2.y is it grounding d voltage.what z reason behind that...

 

I observed the same thing in many circuits that I have come across.
Usually while applying grounding in working with transistors, we usually do depict a resistance just before the ground terminal. Maybe this is to indicate that ALL the current does not get grounded and probably takes into consideration the internal resistance of the wire leading to the ground terminal.

 

By looking at the circuit i gather that there z no need of that base ground resistance.so what would hapen if we remove it.what is it doing there?

 

To give the transistor a hard ground and turn it off if required. It would work without it, but the chance for catastrophic failure goes way up if something goes wrong.

 

Hmm u mean its usd as a circuit protector?
In which condition it would work and how

 

Actually I was wrong. Looking at it again it is meant to turn the transistor on, I'm used to seeing the emitter facing down.

Without it the circuit would/could never turn on.

There is something important missing in the schematic, where is the sensor or input? All I see is a power supply, as it is drawn the circuit makes no sense.

 

Turn on...ok...plz tel me it its working principle

 

Turn on...ok...plz tel me its working principle

 

hello friends i seek ur guidence in understanding this simple looking circuit

The transistor is a PNP type, so the base must be more negative than the emitter, for it to work.

Since a silicon diode, is sitting at the emitter, then a voltage of 0.7v. will be dropped across it when conducting.

so looking at the base emitter loop, you can see that as you start from the positive side of the battery, you have a -0.7v across the diode, and another -0.7v. across the emitter base junction, so that requires the base voltage to be at -1.4v. min, in order for the transistor to start conducting.

The R1 and R2 form a voltage divider, as the variable resistor, (VR1) is made larger, then less positive voltage is at the divider, and the divider voltage becomes more neg, to cause the base voltage to become more negative, to the point it reaches -1.4v. with respect to the +supply, and then conducts.

If the buzzer is a DC buzzer, then the capacitor in the right side is used as a intermitten to turn the buzzer on and off with a possible damping effect, in a exponential manner.

if the buzzer is just a piezo crystal, then the capacitor is used to cause the piezo crystal to turn on and off intermittenly at a frequency to make it work as a buzzer.

 

R2 is providing the bias that turns on the transistor. It is PNP and that might be confusing you - because you asked why is it grounded. Base of a PNP needs to be more negative than the Emitter and in this positive supply circuit that means the base has to have a bias path towards ground.

It is R1 and VR1 that are actually tricky to understand.

They will always have only 1.4 volts across them. the rest of the supply will be across R2.

Best I can understand is R1 is to set a maximum current that is allowed from that 1.4 volts.
The VR1 is meant to allow this current to be decreased.
The main use of these resistors will be to steal current that would otherwise flow through the Base Emitter path of the transistor.

Because of how they compare to the size of R2 and the limit of about 1.3 to 1.6 Volts across them R1 and VR1 will only have a small fraction of the R2 current with most going to the base of Q1.

That fraction is largest at the 5V supply and decreases at the higher levels of the supply voltage range.

Interesting - and I hope somebody can confirm this for me.

A piezo on other side of a diode from the supply like this circuit might try to act as a voltage booster. It alternates drawing current to reach full voltage then tends to kick back voltage and current.

The capacitor is a large .0047 Farad storage/filter size to prevent the spiking from the kickback.

The buzzer as drawn is a full package including the exciter circuit?

 

My first interpretation of the circuit would indicate that this is some sort of audible alarm circuit that is there to sound when the voltage +ve drops below its minimum voltage level. The problem I see with the circuit as drawn is that the alarm would tend to sound continuously as long as the voltage +ve is applied. I am guessing that the circuit has been drawn wrong. I believe that the terminal of VR1 that is tied to +ve in the drawing should actually be connected to T1's emitter instead. In that case VR1 would be adjusted with power applied until the piezo stopped sounding. Thereafter, if the voltage at +ve dropped below its minimum voltage the piezo would sound briefly. The length of time that the piezo sounded would be dictated by the time it took to discharge C1 below the voltage level needed to activate the piezo alarm. If I am correct, R1 would play a critical role in both establishing the piezo alarm off setting when power was present and it would play an equally important role in setting the amount of current delivered to the piezo when the power drops below the minimum level as set by VR1.

hgmjr

 

The transistor is a PNP type, so the base must be more negative than the emitter, for it to work.

Since a silicon diode, is sitting at the emitter, then a voltage of 0.7v. will be dropped across it when conducting.

so looking at the base emitter loop, you can see that as you start from the positive side of the battery, you have a -0.7v across the diode, and another -0.7v. across the emitter base junction, so that requires the base voltage to be at -1.4v. min, in order for the transistor to start conducting.

The R1 and R2 form a voltage divider, as the variable resistor, (VR1) is made larger, then less positive voltage is at the divider, and the divider voltage becomes more neg, to cause the base voltage to become more negative, to the point it reaches -1.4v. with respect to the +supply, and then conducts.

If the buzzer is a DC buzzer, then the capacitor in the right side is used as a intermitten to turn the buzzer on and off with a possible damping effect, in a exponential manner.

if the buzzer is just a piezo crystal, then the capacitor is used to cause the piezo crystal to turn on and off intermittenly at a frequency to make it work as a buzzer.

beautiful answer this helped me lot to understand thanks..........now last thing that i did not understand about this circuit is why is that diode used at emiter........there is necessary supply at the base of transistor which makes it to conduct then what is the use of diode?

 

beautiful answer this helped me lot to understand thanks..........now last thing that i did not understand about this circuit is why is that diode used at emiter........there is necessary supply at the base of transistor which makes it to conduct then what is the use of diode?

As I stated in my earlier reply I still think the circuit is misdrawn. I believe the diode is intended to provide a charge path for capacitor C1 when +ve is present and is intended to go into its reverse bias mode so that all of the charge in C1 will be forced to power the piezo through transistor T1 when the power drops below the minimum voltage level as set by potentiometer VR1.

I don't have the part number for the piezo but I suspect that it is one of the type that has the oscillator circuit built into it so that it does not rely on an external oscillator circuit to get it to oscillate. Of course, this is mere speculation until the part number for the piezo device is known.

hgmjr

 

The Diode increases the voltage difference from the +Supply and the transistor base - doubling it - by adding a second diode drop.

That makes the R1 and VR1 more effective.

The way the voltage is described as +5 to +15 makes me wonder.

I think that the supply is meant to be 5VAC riding on a 10VDC level.
At least it could be. It might be 60 Hz or from transformer reduced mains AC or 120 Hz from full wave rectified AC but any low frequency would do between about .25Hz and 1.2kHz would be interesting.

AC in the supply would turn the Piezo Buzzer into a siren.

If it is only about turning on the Piezo then why bother with the transistor. Just connect the piezo to the power.

Look at the currents.

Lets say R1 and VR1 toal 14kiloOhms and they have a voltage drop of about 1.4V.

They have a current of about 100 microAmps. That current will not change much with changes in the supply because those two diode drops will try to keep the share of voltage across R1 + VR1 at about 1.2 to 1.6 Volts.

Lets look at supply voltages of 6.4 Volts and 11.4 Volts.

R2 will drop about 5 volts and 10 volts in those siutuations.

500 uA and 1000uA through R2.

The current through the base of the transistor is the difference.
Subtract the 100uA through the R1 path.

So 400uA to 900uA is the base current at 6.4 and 11.4 Volts supply.

Assume a transistor current gain of 100 (maybe more). The collector current will vary from 40milliAmps to 90milliAmps.

That is what will drive the Piezo buzzer.

 

I hate that when something obvious appears after I have made what I thought was a good effort to describe what is happening.

But I am here to learn too.

The HUGE Capacitor is the power supply.

It sounds the buzzer alarm when the Power Supply voltage drops.

R1+VR1 are now negative to the transistor Base and Emitter and increase current to the base of the transistor.

A Piezo can run for long enough time on that Capacitor and create a good beep signal.

That Diode is there to prevent the Capacitor from discharging back into your supply, except through the emitter and base of the PNP.

It is a low voltage alarm.

DC operation is not the same as transient response. That is something a person needs to keep learning sometimes.

Assume that the Supply shuts down and drops all the way to ground. In reverse bias that diode can drop 5 to 15 volts.

So that means that that R1 and VR1 are now in parallel to R2 and they add enough to as much as double the base current on the transistor.

 

That is what I have been trying to convey to the OP. I don't think I have been successful so far. I still think the circuit is misdrawn as I described.

hgmjr

 

Yes thank you. I remember you were saying that it is a power failure alarm before and I just needed to catch up.

I think it works as an audio monitor with a less intense sound when the power is good and that chages to at least a brief high power scream when the power dies.

It would also be good for warning of ripple and and AC on the DC level.