Describe whats going on in this circuit

I am trying to make a voltage regulator that converts voltage from 12v to 9v, at first i wanted to use a resistor, but someone told me that, it wouldnt work so great b/c I am trying to feed those 9v to a transmitter and the current will vary depending on how far the transmitter is from the receiver. My friend drew this circuit for me and told me that this would take care of it. Can you please describe step by step how this circuit convert and regulates voltage. I do now how a transistor and resistors work but I am not sure how the diode works and how all these three devices are working together in order to regulate voltage and current. Thank you (PLEASE BE DETAIL IN YOUR DESCRIPTIONS, I REALLY WANT TO GET DIFFERENT DESCRIPTIONS FROM DIFFERENT PEOPLE, SO I CAN UNDERSTAND IT BETTER)

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PS. I am making this circuit for educational purposes, I know I could easily buy a chip or a presicion diode that would regulate the voltage and current. Thank YOU

...transmitter operates at 9v and consumes up to 100mw at its maximum, but that is at its maximum, it can probably operate at less.

****UPDATE****

ALL THE PICTURES ARE FROM THE SAME CIRCUIT BUT FROM DIFFERENT ANGLES, HOW DOES IT LOOK?


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IS A 1.2K RESISTOR AND A 9.1 ZENER DIODE, TRANSISTOR I AM NOT SURE ABOUT THE DETAILS, ILL FIND OUT SOON
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It is too bad that you didn't attach your schematic here.

The zener diode limits the voltage across it to 9.5V. The resistor has 12V-9.5V= 2.5V across it so its current is 2.5V/1k= 2.5mA.
The 2.5mA drives the zener diode, transistor or both.
The transistor is an emitter-follower so its output voltage is 9.5V - 0.7V= 8.8V if the output current of it can be supplied with a base current of only about 1.5mA.

it is there now, sorry i didnt check very good before it posted it

Life might be easier if you use a 7809 regulator. That, two caps, and you're there.

the resistor is connected to the diode and the base of the transistor right?

Correct but its good if you can replace the resistor to 220 ohms to 470 ohms.

220 hm:, because his transmitter dissipates 100 mW

The current limiting resistor (R1) and Zener diode provide a constant reference voltage for the base-collector junction of transistor.

For a small power dissipation it doesn't matter much.even 1K works well.

For tru_cutru you can get more informations on this site.

https://en.wikipedia.org/wiki/Current_source

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Begin by assuming that the 12V dc power supply is able to provide a lot of current at 12V while suffering no voltage drop. In other words, the power supply has a very low source resistance. This assumption helps establish how to analyze the circuit using ohm's law. In ohms law there are three variables, Voltage, Current and Resistance. If you know two of these, you can calculate the third for any circuit path. And we know that the 12V terminal always has 12V no matter how much current is drawn from it.

Next, we consider how much current will flow through the resistor. One side of the resistor is connected to the 12V supply, so that is fixed. On the other side of the resistor is the zener diode connected to ground, and the base of the transistor. At the moment, there is nothing connected to the emitter of the transistor, and we know that it is a NPN transistor, so we can say that no base current will flow since there is no path for it to flow out of the emitter. So all of the current that flows through the resistor will also flow through the zener diode. Now, look at the zener diode current vs voltage curve. It tells us that as the current starts to flow through the resistor and then through the diode the voltage across the diode will flatten out and remain at the "zener" voltage once the current is high enough. The current will then come to rest at a specific value which is easily calculated. Since the voltage across the diode is now known, we calculate the voltage across the resistor to be 12V minus the diode voltage, or 2.5 volts. So now we can calculate the current through the resistor to be 2.5 divided by 1Kohms, or 2.5 mA. This is also the current that flows through the zener diode. The voltage at the junction of resistor and zener diode is thus "stabilized" and remains at one value even if, for some reason, the current amount flowing through the diode happens to change. But how could this happen?

So far, we do not have anything connected to the transistor's emitter. To make some use of this circuit, you attach a load to the emitter, a load with some value of resistance from emitter to ground. Let's start by attaching a load of, say, 800 ohms (just to pick a value out of the blue).

At the instant that we apply power to the circuit with the load attached, there is now a place for emitter current to flow out of the transistor. As the voltage ramps up, the transistor base voltage increases, and thanks to the action of the transistor, its emitter voltage also ramps up, at roughly 0.7 volts lower than the base voltage. The current that flows into the base stabilizes at the amount required for the emitter to achieve 8.8 volts across 800 ohms, divided by the beta of the transistor (the DC current gain). So we can calculate the amount of base current that flows, or rather, we can estimate it pretty closely anyway. It would be 0.11 mA.

This calculation is based on the assumption that the base voltage remains steady. This happens to be true even though now we are taking some base current. The reason for this is that even though it would seem like we should be taking a bit more current through the resistor now, in fact we are not. What is happening is that we are stealing a tiny bit of current from the diode branch and feeding it into the base instead. Since the diode voltage drop remains stable at 9.5 volts for such a small reduction in current through it, the diode voltage does indeed remain steady.

Now, imagine that we change the load resistance from 800 ohms to some other amount. What happens is that the transistor then pulls more or less current into its base, and is stealing a little bit more or a little bit less current from the zener diode. But the zener has this handy characteristic that the voltage across it doesn't change when this happens. And since the emitter of the transistor remains 0.7 volts lower in voltage than the base regardless of small changes in current low through it, the transistor emitter voltage also remains steady.

That's how this circuit "regulates" the voltage across the load.

Can I replace the Resistor value something like in ohms scale?

Is there any dropout in the circuit if I use a higher value resistor?

Thanks

The dropout is equal to the base voltage plus the saturation voltage at the current being drawn. You can find these values from the transistor's datasheet.

What current are you planning to draw?

You need to check that the resistor is low enough to allow enought current to flow into the transistor for the current being drawn.

As previously mentioned it is far more easy to use a voltage regulator IC like the LM7809, LM7808 or LM317 in an application like this.

why don't you simulate this circuit? You can play with different values and see what happens.

I WILL POst a pic of the simulated circuit to check if i did it right or not

bananasiong said:

220 hm:, because his transmitter dissipates 100 mW

Click to expand...

that is what i had at first when i tried to lower the voltage with a resistor, but would you have any idea why he (my friend) put 1k - 1.5k resistor in that circuit?

audioguru said:

It is too bad that you didn't attach your schematic here.

The zener diode limits the voltage across it to 0.5V. The resistor has 12V-9.5V= 2.5V across it so its current is 2.5V/1k= 2.5mA.
The 2.5mA drives the zener diode, transistor or both.
The transistor is an emitter-follower so its output voltage is 9.5V - 0.7V= 8.8V if the output current of it can be supplied with a base current of only about 1.5mA.

Click to expand...

he said the transistor would only consume .3v???

I simulated the schematic diagram, how does it look?

after the schematic image they are all the same simulation of the schematic but from different angles

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after the schematic image they are all the same simulation of the schematic but from different angles

I can't see the transistor's part number. If it is an NPN like in your schematic then the zener diode is connected backwards.

The transistor also might be connected backwards.
The power supply also might be connected backwards.

2n3904
w67
I am surprised it doesnt say A B C or D anywhere, but my friend told me that it wouldnt matter which transistor I use because the amperage consumed by the transmitter is so little. However, it does matter if is npn or pnp because....?

You still don't know how much current the transmitter uses. It might be more than the 200mA max of the little 2N3904 which would melt with such a high power dissipation.

The 1k resistor feeding it and the zener diode has a value that is way too high because the minimum current gain at a collector current of 100mA is only 30 and is much less at higher currents. So at 200mA the base current might be 13mA and the base resistor would need to be about 200 ohms.

Your zener diode is definitely backwards. Its end with the black bar should connect to the base of the transistor.

Why doesn't the schematic match with the connection on the breadboard?
As shown the yellow wire is the collector, so the red wire (assumed as Vcc) should be connected to the collector, and the other end of the resistor as well. The emitter should be grounded.
The pins of 2N3904 are EBC, the flat surface is facing you and the pins are pointing down.