# BJT acting as an adj. resistor in a linear regulator

Discussion in 'General Electronics Chat' started by TimNJ, Dec 31, 2013.

1. ### TimNJ Thread Starter New Member

Dec 31, 2013
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1
Hello everyone,

I have a question regarding how a linear regulator works internally.

I understand the basic topology with a feedback loop from the output being compared to a reference voltage (using an op amp) and then this op amp tries to make its inverting input equal its noninverting input.

In the simplified schematics that I have looked at, an NPN BJT is used. It is commonly said that the BJT is being used as a "variable resistor" dropping a certain voltage across it to maintain a specified voltage on the output, as the op-amp wishes.

But how exactly is this working? I initially thought that the BJT was controlling the current that could flow from its collector to emitter. But apparently this is not the case...because it seems that if the load demands a greater current, a greater current will pass through the BJT, regardless of the extent of saturation. With a greater current passing through that junction, a greater voltage drop is created, and the op amp adjusts accordingly.

Now it may seem that I have some idea what I am talking about, but in reality, I'm not too sure how the transistor is creating that voltage drop. How is it functioning as a so called "adjustable resistor". I need to brush up on my transistor theory, but could anyone perhaps put this in their own words that might make more sense to me (and others who are reading)?

Thank you.

2. ### #12 Expert

Nov 30, 2010
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7,354
The transistor isn't doing any adjusting, the op-amp is. If the output voltage tends to drop, the op-amp simply sends more current to the base of the pass transistor, etc.

Op-amps being rather fast and output capacitors supporting the increased load for less than a millisecond, and you don't see any changes without using an oscilloscope.

3. ### TimNJ Thread Starter New Member

Dec 31, 2013
9
1
Thank you. Yes, I understand the op amp is what is adjusting the drive to the base.

Here are the parts that I don't understand. They are basic but I'm having a hard time with them.

Why does the output voltage drop if the current increases? Is it because applying a base current creates a certain collector to emitter resistance? And if the load increases, the drop, Vce , increases. In turn, the op amp would drive the the base harder, and the resistance would drop allowing for a smaller voltage drop across the device, increasing the voltage across the load..

Does that make sense?

4. ### Brownout Well-Known Member

Jan 10, 2012
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In the case of a series transistor regulator, it is acting as a voltage controlled current source. The controlling voltage is the load voltage, and there is a reciprocal relationship between the controlling voltage and supplied current. And so, as the load draws more current, load voltage drops due to circuit resistance. As the voltage drops, more current is supplied by the transistor, which is thus the additional current required by the load.

5. ### #12 Expert

Nov 30, 2010
16,705
7,354
I think you're right, but rather than sort out your post, I am going to put the answer in my words. That way there won't be any questions in the middle.

When the load current increases, the output voltage starts to drop because the op-amp is not a mind reader. It didn't know you were going to increase the load current. A few microseconds later, the op-amp detects the dropping voltage and increases the current to the base of the output transistor.

Now the output transistor passes more current with about the same voltage collector to emitter it had a millisecond ago. You can say the output transistor is representing a lower amount of resistance.

In practical matters, I don't think in terms of equivalent resistance that is represented by an output transistor. That is because bipolar transistors are current driven devices and thinking in terms of current is sufficient to design with them. Mosfets, on the other hand, are voltage driven and do seem to represent a resistance, but you very quickly convert that activity to resume thinking in terms of current.

It's like, "How many ohms are in a diode?" It depends on the current through the diode. Even if you knew the answer, it is much more convenient to work with the voltage drop across a diode to design a circuit. The relevant question is, "How much voltage am I going to lose in this diode?"

6. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Perhaps your misunderstanding was right there in your original post:

The pass element transistor is NOT in saturation, it is in it's active state, and there is lots of voltage across it's collector to emitter terminals.

How much voltage? Well, the input voltage minus the output voltage, of course.

7. ### TimNJ Thread Starter New Member

Dec 31, 2013
9
1
Wonderful explanation. I appreciate it.

Everything makes sense now, except one basic concept. When the load increases, the output voltage drops (momentarily). Why exactly does the output voltage drop with an increased load. Wheres is the voltage being lost (before the op amp detects the change)?

8. ### TimNJ Thread Starter New Member

Dec 31, 2013
9
1
I think I used the wrong terminology. I didn't mean saturation. That's why I wrote the extent of saturation, but I guess that doesn't really make sense. Thanks.

9. ### TimNJ Thread Starter New Member

Dec 31, 2013
9
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Right that makes lots of sense. The one thing I'm just not understanding is why the voltage across the load decreases with an increased load if we are not treating the pass transistor as a resistive device. Where is the voltage lost so that the output voltage is less? (right before the op-amp picks up on it).

10. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
Load voltage is just load resistance times load current (the ever familiar: I*R) When the load resistance decreases, then I*R also decreases, for constant current, I. But when the pass transistor senses this drop in load voltage, it increases load current, bringing I*R back to the prescribed value.

11. ### #12 Expert

Nov 30, 2010
16,705
7,354
The load can be seen as a resistor. When the load changes to having less resistance, the same amount of current (that was there a millisecond ago) through the load resistor creates less voltage according to Ohm's Law. This decreased voltage is what the op-amp senses. Then it reduces the apparent resistance of the output transistor properly.

Brownout was faster than I was.

12. ### #12 Expert

Nov 30, 2010
16,705
7,354
This sentence is valid if you see the voltage on the base of the transistor as not changing. In the first instant of load increase, it is true. Merely the increased voltage from base to emitter tends to increase current flow through the transistor. This usually is not completely sufficient to hold the output voltage constant, but the op-amp detects how much it is insufficient and sends more current to the transistor's base. This, in turn, increases the base to emitter voltage enough to correct the amount of current the transistor passes to the load.

TimNJ likes this.
13. ### TimNJ Thread Starter New Member

Dec 31, 2013
9
1
Much thanks to both of you. As you can see my understanding of electronics is a little scattered.

Happy new years.

14. ### Brownout Well-Known Member

Jan 10, 2012
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998
We'll help you to get it into focus.

15. ### #12 Expert

Nov 30, 2010
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We usually get it done, even if it takes 5 or 6 of us.

16. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Did you notice that the pass transistor was described as a controlled current source? A current source has a high output impedance. The voltage across the load is E = IR, where I is the current supplied by the current source, and R is the effective resistance of the load. If something suddenly reduces the load resistance, then the output voltage is proportional to the new value of R.

Look at the attached simulation. Suppose that initially the load is just R1=5Ω, and the current source is supplying 1A, to make the load voltage V(out)=5V. Now, at time=1sec, suddenly, R2=45Ω is switched in parallel with R1, making the new load resistance = 5*45/(5+45) = 4.5Ω. The current hasn't changed, so the new output voltage V(out) [Green Trace] jumps from 5V to 4.5V because E=IR=1A*4.5Ω = 4.5V.

Now, in the real IC regulator, the Current Source is not a constant 1A.

After the dust settles, the current source would have to increase from 1A to I=E/R= 5V/4.5Ω = 1.111A in order to bring the output voltage back up to 5V.

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17. ### LvW Active Member

Jun 13, 2013
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Just a small comment: The output characteristics of a FET Id=f(Vds) is rather similar to Ic=f(Vce) of a BJT. Thus, you may consider both transistor types as a dynamic resistance - or not. However, this view is independent on the question how the device is controlled at its input node (gate or base, respectively).