# Series Transistor Voltage Regulator

Discussion in 'Electronics Homework Help' started by still-young, Dec 3, 2017.

1. ### still-young

Joined:
Dec 3, 2017
Messages:
2
0
Hi guys, I'm doing an engineering HND atm, there's a question on power supplies where I have to explain how a series transistor voltage regulator works. I understood it in class, but I guess my notes weren't very good because I'm struggling with it now. I know Vbe controls how much current passes through the transistor. Everything I've read online says Vout=Vzener-Vbe, but then says that if Vout increases (increased Vin or decreased RL), Vbe decreases, letting less current through, and therefore Vout decreases again. But that doesn't work with the equation? Because Vzener stays constant, if Vout increases, shouldn't Vbe also increase? So more current would flow through.

I'd appreciate some help guys! Thanks

still-young, Dec 3, 2017

2. ### Externet

Joined:
Aug 24, 2009
Messages:
549
114
Location:
Mideast US
That is a voltage regulator. Vbe can vary from about 0.6V to about 0.7V. Only 0.1V difference, taken as a fixed value.

Your ..."if Vout increases, shouldn't Vbe also increase ?"...
Vout is kept steady, regulated to the value.
Vout=Vzener-Vbe ----> Vz = Vout + Vbe. In other words, Vout will be Vz - 0.65 V

Externet, Dec 3, 2017

3. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
What does "HND atm" mean? Expand your acronyms at least once so folks know what you are talking about. Better yet, if it does not pertain to the problem, don't mention it.

If Vin changes, the constant current operation of the collector won't be affected very much. The collector voltage can change quite a bit before the collector current does, when Q1 is operating in the active mode.

If RL decreases, less voltage will appear across it. Then Vbe will increase to make up the difference between the constant Vzener and RL. The increased Vbe will increase the emitter current, which will raise the voltage across RL to what it was before when RL was higher. Therefore, the voltage across the load will be the same regardless of the load resistance, within the limits of Q1 to supply or reduce the required current.

Ratch

Ratch, Dec 3, 2017
4. ### still-young

Joined:
Dec 3, 2017
Messages:
2
0
Sorry, HND is Higher National Diploma (the qualification I am doing) and atm is at the moment. And that might be where I’ve been going wrong - made a silly error, I had it in my head that decreased load resistance would increase the voltage.

still-young, Dec 3, 2017
5. ### Audioguru

Joined:
Sep 24, 2016
Messages:
2,208
499
Why are you asking about the impossibility of something causing the output voltage to rise?
Instead think about what happens when the input voltage rises. Then the zener diode current increases which increases the output voltage a little.

I disagree that the output voltage is fixed because it does not have any negative feedback so its voltage regulation when the load current increases (decreased load resistance) is BAD:
If the output current increases then the transistor's Vbe increases and the increased transistor's base current steals current from the zener diode reducing its voltage, two things reducing the output voltage.

Audioguru, Dec 4, 2017
6. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
Why impossible? If the load resistance should rise, perhaps from a temperature increase, then the output voltage will rise unless the load current is reduced. I explained how that happens in my last post.

Why would that happen? The input voltage source will just absorb that extra zener current and the zener voltage will be unaffected. That is what zeners do, They keep a constant voltage within a variable current range.

Who said negative feedback is absent from this circuit? Negative feedback occurs because Vbe decreases when the load current increases. We can go over that again if you don't agree on that point.

Two things here. One, stealing current from a zener will not change its voltage unless you starve it by reducing its current below what it needs to regulate the voltage. Two, the zener voltage equals the sum of Vbe and Vout. If the output current tries to rise for whatever reason, the Vout will also try to rise in step. This will cause a decrease in Vbe and lower the current to the load, which will lessen the load voltage.

Ratch

Ratch, Dec 4, 2017
7. ### Audioguru

Joined:
Sep 24, 2016
Messages:
2,208
499
The datasheet for a zener diode shows its dynamic series impedance that lessens its voltage regulation. The dynamic impedance is the lowest for a zener voltage of about 6V that produces the best voltage regulation. Curves show the voltage rising a little when the zener diode current increases which happens when the input voltage increases. The voltage rise is caused by the series dynamic impedance. The curves also show the opposite, the voltage drops a little when the zener diode current decreases. Ratch says the voltage does not change but the datasheet shows that it does change. The voltage regulation of a zener diode is much better when it is fed from a constant current source which is not in this simple circuit.

The transistor is a simple emitter-follower. Its output voltage drops when its output current increases.
Ratch wrongly says that "Vbe decreases when the load current increases" but a datasheet for any transistor shows that the Vbe increases.

Therefore in this simple circuit when the load current increases then the output voltage drops.

One way to ruin the excellent voltage regulation of an IC voltage regulator is to hang an emitter follower on its output without any negative feedback like in this circuit. But when the current boosting transistor is included in the negative feedback loop then the voltage regulation stays excellent.

Audioguru, Dec 4, 2017
8. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
No one says that a single uncompensated zener is a perfect voltage regulator. What are we talking about here, a few millivolts? That circuit is a regulator, not a voltage standard. To a first order, a zener keeps the voltage constant within its current range. Nothing is perfect in this world. If you are going to worry about the zener, the transistor has secondary effects that can vary from the calculations if you measure the results finely enough.

That does not make sense. In any load, the output voltage will increase if the current increases. That is the basic definition of resistance.

Invalid comparison! You are comparing a single device (transistor) with a circuit (regulator). The regulator circuit has feedback which I described in my previous post. This feedback does in fact lower Vbe when the emitter current increases.

The regulator circuit will adjust the Vbe to keep the voltage with a relatively small range.

The submitted circuit is primarily a voltage regulator, not a current booster.

Ratch

Ratch, Dec 4, 2017
9. ### Audioguru

Joined:
Sep 24, 2016
Messages:
2,208
499
But this simple circuit does not feed the zener diode from a constant current source, therefore the zener voltage changes with input voltage changes and with load current changes.

Not with an emitter-follower. The emitter voltage gets dragged down when the load current increases because Vbe increases.

This simple circuit has no negative feedback. It is a zener diode and an emitter-follower buffer.

No it will not.

If the transistor is not a current booster then it reduces the voltage regulation of the zener diode and the transistor should be removed.

#### Attached Files:

• ###### Vbe vs current.png
File size:
38 KB
Views:
124
Audioguru, Dec 4, 2017
10. ### Harald KappModeratorModerator

Joined:
Nov 17, 2011
Messages:
8,382
1,590
Location:
Germany-Europe-Earth-Sol System-Milky Way-Laniakea
@Ratch , @Audioguru :

Gentlemen,
do you really think this in depth discussion will help the op understand the very basic operating principle of this circuit? Which is what this thread was originally about.

Harald Kapp, Dec 4, 2017
11. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
Irrelevant. The zener operates in its breakdown region no matter whether it is fed with a voltage or current source. The Vz vs Iz curve shows an almost vertical slope, denoting a nearly constant voltage despite large current changes. Even batteries or power supplies change voltage somewhat depending on the load, unless they are equipped with voltage sense leads.

The voltage across the load of a emitter follower (EF) is Vbase - Vbe = Vload . If Vbase is held constant by a voltage source, then Vbe must change if Vload changes. If the load current increases, the voltage drop across the load will tend to increase also. That means that Vbe must reduce also for the above equation to balance. That's negative feedback by anyone's definition.

Any EF circuit has negative feedback, just as a common emitter circuit with an emitter resistor does.

No current is being fed into its input, so no current is being boosted. It is simply a regulated voltage source that is able to supply a moderate amount of current, just as any voltage source can do.

Ratch

Ratch, Dec 4, 2017
12. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
I believe the OP's question and concern was addressed and answered. Any further discussion is for the rest of the group's edification and benefit. I wish that others would participate in this dialog and offer opinions based on fact about which of us is correct.

Ratch

Ratch, Dec 4, 2017
13. ### Harald KappModeratorModerator

Joined:
Nov 17, 2011
Messages:
8,382
1,590
Location:
Germany-Europe-Earth-Sol System-Milky Way-Laniakea
This would be true if the load were purely resistive. For other kinds of load this is not necessarily true. Take for example a simple LED (with appropriate current limiting resistor) turned on by a switch. And let us assume another load is parallel, e.g. a second LED. This second LED provides a permanent load current. At turn on of the switched LED the current rises but the load resistance sinks as now 2 LEDs with resistor are connected in parallel. And due to the inevitable internal resistance of any regulator the load voltage will fall.

Due to the non-ideal internal resistance of this voltage regulator, the rising current increases the voltage drop across the regulator (assuming an ideal and constant input voltage). Therefore the voltage across he load will fall, not rise.
A falling output voltage increases Vbe (assuming in 1st approximation Vz = const.). A rise in Vbe in turn provides a rise in Ic which is required to drive the higher load. This is where the negative feedback comes into play.

The question is how much. For a 1N750 (4.7V) with a 1 k series resistor a change in input voltage of 200 mV results in a change in zener voltage of 65 mV, a reduction by 3, at an input voltage to the circuit of 4.8 V.

Harald Kapp, Dec 4, 2017
14. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
Yes, you are correct. I was only considering an ohmic load.

Upon further consideration, I think it is useless to ask how the regulator works if the load resistance changes. If the resistance doubles or is halved, the current is halved or doubled, and the IR output voltage remains the same. As long as the transistor remains in the active mode from the current changes, the output voltage will not change. Now, suppose the input voltage increases. Then the voltage across the transistor Vce and Vout will both increase proportionately. Then, since Vz is constant, Vbe has to shrink because Vbe = Vz - Vout. This will cause the transistor to increase its collector resistance and drop more of the input voltage. This reduces the voltage available to the load to what it was before, and satisfies Vz - Vbe = Vout . That is how negative feedback in this circuit is achieved.

Ratch

Ratch, Dec 4, 2017
15. ### Audioguru

Joined:
Sep 24, 2016
Messages:
2,208
499
Thank you for agreeing with me.

Where are you applying 4.8V? If you apply 9.4V (double the zener diode voltage) to its series resistor and use its 20mA current rating then the resistor value is 4.7V/20mA= 235 ohms. The max dynamic impedance of a 1N750 is 15 ohms from one manufacturer so if the 9.4V drops 1V then the current drops about 1V/235 ohms= 4.26mA and the voltage drop across the 15 ohms of dynamic impedance is 4.26mA x 15 ohms= 64mV. Then the voltage regulation is (4.7V - 65mV)/4.7V= 86%, not good but not too bad. Since the zener diode voltage has reduced then the amount of current drop is also reduced then the voltage regulation is a little better than 86%.

When the load current increases then the Vbe also increases which drops the output voltage a little. When the load current (and emitter current) increase the base current also increases which causes less zener current causing less zener diode voltage.

Audioguru, Dec 5, 2017
16. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
Does that mean that you agree that the circuit contains negative feedback?

You need to specify what causes the load current to increase. I already showed that if the current increase is caused by a reduction in load resistance, then the output voltage does not change. If the increase in load current and load voltage is caused by a increase in input voltage, then Vbe will decrease and more voltage will be dropped by the transistor, and the voltage across the load will decrease. That's negative feedback.

Ratch

Ratch, Dec 5, 2017
17. ### Audioguru

Joined:
Sep 24, 2016
Messages:
2,208
499
No, I agreed that the internal resistance of this voltage regulator causes a voltage loss. "Therefore the voltage across he load will fall, not rise.[/quote]

Not true. The zener diode has some voltage and the emitter follower also has some voltage loss.

When the input voltage rises then the poor voltage regulation of the zener diode causes its output voltage to rise. But the increase of Vbe voltage loss of the emitter follower reduces this effect. That is not negative feedback. Didn't you see my attachment of the Vbe of a transistor increasing when the current increases? But you wrongly say that Vbe decreases.

Audioguru, Dec 5, 2017
18. ### Ratch

Joined:
Mar 10, 2013
Messages:
993
289
That is an indefinite answer. The internal resistance of the voltage regulator (transistor?) is always present and always causes some voltage loss. Specifically, if the input voltage rises, then both the voltage of the transistor and the load will tend to rise proportionately. That will be corrected when only the transistor increases its resistance, and thereby removes the increased voltage from the load.

The common collector configuration is known far and wide as a feedback circuit. See any textbook or this link. http://fourier.eng.hmc.edu/e84/lectures/ch4/node9.html Notice the neat way this site describes the chain of events of the feedback sequence.

As I said before, to a first order, the Vz and Vbe are assumed to be constant when designing this regulator. As an interesting fact, the Vbe will increase only 18 millivolts if the Ic is doubled.

When I said the Vbe decreases, I should have emphasized that it tends to decrease. Of course the feedback will bring it back to its balanced value.

Ratch

Last edited: Dec 5, 2017
Ratch, Dec 5, 2017