# Voltage Regulating Circuit

#### b0mbst3r

Joined Oct 13, 2016
5

Hey guys,

I'd like to introduce this Voltage Regulator circuit. As you can see it incorporates x2 NPN Transistors and a 5.6V Zener Diode (ZD).

This circuit is a part of one of my electronic module assignments, where I am expected to explain the following:

a) Explain the purpose of the 5.6V ZD
b) Explain the purpose of the Q2 NPN Transistor
c) Suggest the effects on the Output Voltage if the 5.6V ZD was replaced by a 4.7V ZD

The following points are how I have interpreted this circuit, and partly what I have been told is happening (please feel free to correct me if I am incorrect at any point). I will ask questions at the end of each point and if possible I’d appreciate if someone could answer them:

If we were to use VR2 as our "Load";

1. If the resistance at VR2 decreased, the current would increase, ultimately decreasing the Voltage. This decreased Voltage would also be found at the Base of Q2, which if the Voltage is below 0.7V (transistor operating threshold) would decrease its conductivity and "switch off". This in turn means that the Voltage at the Base of Q1 is greater than Q2, so it will conduct more, meaning that the current will amplify freely to the output, where this current value is the same value as the demand current.

a) What is the current flow direction from VR2? Would it try to flow up through R3 and into the Base of Q2 until realizing it is switched off, and then having to continue along the bottom of the circuit until going into the negative terminal of the Bridge Rectifier?

b) From the positive terminal of the Bridge Rectifier would there be 12V, until splitting at split point 1 for R2, and at split point 2 for R1 (would the Voltage now be low at the Base of Q1?), but because there is a capacitor in the circuit (C2) would then then increase the voltage back up so that Q1 conducts more?

2. Point 2 is pretty much the opposite of Point 1 where if the resistance at VR2 increased, the current would decrease, ultimately increasing the Voltage. This increased voltage can be seen at the Base of Q2, where it is amplifying the current as it continues down the wire to the ZD. Now because this high Voltage reaches the ZD, it would be clamped down at 5.6V until continuing its route to the negative terminal of the Bridge Rectifier.

Now this is where I don’t understand what is happening:

Following on from the end of Point 2; the same again is happening here as in point 1 where 12V exits the Positive terminal of the Bridge rectifier until a High(?) Voltage is seen at the Base of Q1? Because of the same issue again with the Capacitor? Or is the capacitor simply there to even out the “DC” output and NOT to boost the circuit voltage? The reason for my confusion is that I told these transistors would work together in opposition where one receives a high Base Voltage and amplifies; and the other receives a Low Voltage and switches off, is this correct?

I was also told that Q1 has a Common Base, meaning that it would amplify current, and Q2 has a common Emitter, meaning it would amplify voltage, is this true?

Any help would be appreciated guys.

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#### AnalogKid

Joined Aug 1, 2013
8,533

Without C1 in the circuit, the voltage at the output of the bridge is a full-wave rectified sinewave, going from 0 V (GND) to approximately 15 V. This is( 12 V x square root of 2) - 2 power diode Vf drops. While diodes usually are assumed to have Vf of approx 0.6 to 0.7 V, power diodes easily can have Vf = 1.0 V or more depending on their construction and the current flowing through them. So to keep the math easy we'll assume that each diode in the bridge drops 0.985 V.

Connecting C1 filters the full-wave rectified sinewave into a DC level with ripple. The amount of ripple depends on the size of C1 and the output current into the load. To keep things simple, we are going to assume that C1 is perfect and there is no ripple. So everything connected to the bridge (C1, R1, R2, and Q1 collector) all see a nice clean 15 Vdc.

R1's main function is to bias zener diode D1 into reverse conduction. We'll assume that D1 is perfect, so it's reverse (zener) voltage is exactly 5.6 V no matter how much current is added by R2 and Q2.

Q1 is not a common base amplifier. It is an emitter follower, also called a common collector amplifier.
https://en.wikipedia.org/wiki/Common_collector

Q2 is a form of a common emitter amplifier, but in this case it is acting as a differential amplifier with an inverting output.

To answer one of your questions, no - current does not "up" through VR2 backwards into the Q1 emitter.

C2 is a noise filter on the base of Q1 to reduce the amount of ripple that appears there. It has no other effect on circuit operation, and does not have anything to do with how the circuit regulates the output voltage.

If Q2 is not installed, then all of the current through R2 flows into the base of Q2, turning it on. With Q1 installed, the current from R1 now has two paths to GND. One is through Q2 base-to-emitter diode, R4, A, and VR2 to GND, and the other is through Q1 collector-to-emitter and D1 to GND. As Q1 conducts more or less, the amount of current it steals away from the Q2 base varies. This is how Q1 controls Q2, by stealing away some of its base current.

Digest that part. I'll do more later.

ak

#### BR-549

Joined Sep 22, 2013
4,938
I would describe the circuit as a current controlled constant voltage electronic capacitor.

#### AnalogKid

Joined Aug 1, 2013
8,533
It is a variable gain voltage controlled voltage source. The zener plus Q1 form a 6.2 V constant voltage source. This goes through a non-inverting amplifier (Q1 and Q2) with the gain controlled by varying the amount of negative feedback (through VR1).

ak

#### b0mbst3r

Joined Oct 13, 2016
5
Thank you for your reply AnalogKid, the way you have explained some points; I'm starting to better understand the circuit.

If the voltage drops at VR2, can you explain how the voltage also drops at the base of Q2 and not at the base of Q1? Is it because the base of Q1 has less "voltage splits" from the supply (BR)? Which would ultimately mean Q1 base is not receiving more voltage and conducting more, amplifying the current to match the demand at VR2.

So, can you explain the opposite for me, if the voltage were to increase at VR2, could you explain the journey of the current detailing voltage and current levels (up/down) at each component in the circuit please?

Or in fact do the same for both increase and decrease in voltage?

#### Veracohr

Joined Jan 3, 2011
720
Higher voltage at the top of VR2 implies higher voltage at Q1 emitter. This means higher voltage at Q2 base, turning it on more, which pulls current away from the base of Q1, thus decreasing the emitter current, and lowering the output voltage.

#### b0mbst3r

Joined Oct 13, 2016
5

Can anyone explain HOW lowering the ZD to 4.7V for example would lower the overall voltage output (at VR2)?

#### AnalogKid

Joined Aug 1, 2013
8,533
Q2 acts as a differential amplifier, with the base as the inverting input and the emitter as the non-inverting amplifier. VR1 and R3 form the negative feedback loop that sets the gain. Any change in the voltage at the emitter shows up as a proportional change in the output voltage. This is complicated a little by the addition of the Q2 Vbe as mentioned above.

When the Q2 emitter goes down, more of the current coming through R2 is shunted away from the base of Q1, decreasing it's conduction, decreasing its emitter current, decreasing the voltage developed across R4 and VR2.

ak