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# Linear Power Supplies

#### engr_joni_ee

I would like to understand how do the linear power supplies actually work.

I know that there are two main types of lab power supplies.

1- Linear PSU (Power Supply Unit)
2- SMPS (Switch Mode Power Supply)

The linear power supply can have various options and features available for example.

1.1 Voltage Limit - also known as Constant Voltage (CV)
1.2 Current Limit - also known as Constant Current (CC)
1.5 Remote sensing
1.6 Remote programming

I have attached a circuit diagram that represents the basic design of linear power supplies.

The electronic components in the attached circuit diagram are:

1.1.1 Input power transformer, T1
1.1.2 Bridge rectifier, D1-D4
1.1.3 Reservoir (or smoothing) capacitor, C1
1.1.4 Pass transistor, Q1
1.1.5 Output voltage sense and feedback, Q2
1.1.6 Voltage limit setting, VR1
1.1.7 Voltage reference, ZD1

There are few things I know regarding the attached circuit.

- The input is 220 V AC which is connected through a fuse to the input of the Transformer T1.
- The ratio of number of turns of primary and secondary coils in Transformer T1 decide the output voltage level at the secondary coil of the Transformer T1. For example if the ratio is 22:1 then the 220 V AC will be down to 10 V AC.
- The bridge rectifier consisting of diodes D1 to D4 will rectify. The ripples will be smoothen by the filter circuit or a large capacitor C1 (1000 uF) as mentioned in the attached diagram.

I am wondering what is the formula to calculate Vout provided the values of resistors R2, R3, R4, Potentiometer, and reference voltage 5.1 V are given ?

I understand that Q1 is pass transistor and Q2 is feedback transistor in the attached circuit.

Something more I know is how it work in linear regulator in which we also has a pass element (a transistor) and error amplifier (an OpAmp).

As the output voltage of the linear regulator drops the voltage divider's mid point voltage will also drop that will be compared with the reference voltage, making the error amplifier output low which will turn on the pass transistor, making the output voltage level increase.

As the output voltage level increase beyond the set value. The mid point voltage of the voltage divider will also increase that will be compared with the reference voltage, making the error amplifier output high which will turn off the pass transistor, making the output voltage level drop. This is how the linear regulator work.

Now I need to understand how the linear power supply's attached circuit work. I guess the operation of the attached circuit diagram is also similar but I am not getting it completely.

Can someone please explain how the circuit works ? and how do we calculate the Vout with the values of resistors R2, R3, R4, Potentiometer, and reference voltage 5.1 V ?

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To understand the working of that circuit you need to understand how a transistor works; no base current -> no collector current and also since emitter current is a sum of these two there is no emitter current.. As the transistors are npn type, you need to raise base voltage about 0,6V higher than the emitter to get the current flowing..

-Before You plug in the circuit, emitter and base voltages are zero..
-You connect the mains.. c1 voltage starts to rise..so will q1 base and collector voltage..
-after some microseconds base crosses 0,6v..
-after that the emitter of q1 starts rising but it stays that 0,6v below the base voltage (as emitter follower does)..
-After some more microseconds the output crosses the voltage that is zener voltage added with 0,6v base threshold of q2 and multiplied by a factor of the voltage divider r3, r4 and vr1 (..r2 is also kinda part of that network, but i count it as parallelling the vr1..that makes the compound resistance of the two logarithmic in regards with the mechanical turn..) the transistor q2 starts conducting and pulling q1 base voltage down to regulation point..
-If the q1 base voltage has a tendency to go too low, the q2 starts conducting less due to it's base current going down since the output voltage is fed back via the voltage divider against the zener voltage causing q2 base current to adjust accordingly..

I don't have any ready made math for You, but i think You can conclude it from these known voltages and voltage divider values..(i think you only need ohm's law and kirchoff's law and some very basic semiconductor knowledge for the math..)

Just for You to know; These types of simple regulators are not very accurate keeping their set voltage levels if/when the semiconductor temperatures change.. I think it looks like this circuit has the advantage of the transistors compensating each other out at least somewhat, but zeners still are not the most accurate..although i think 5.1v zener voltage is also in the region that is typically most temperature stable..

Hope this helps You at least a bit..

Transformer (T1): Steps down the AC voltage from 220V to a lower AC voltage. The ratio of the primary to secondary windings determines this step-down ratio.
Bridge Rectifier (D1-D4): Converts the AC voltage from the transformer to pulsating DC voltage.
Smoothing Capacitor (C1): Smoothens out the pulsating DC from the rectifier to provide a relatively steady DC voltage.
Pass Transistor (Q1): Acts as a variable resistor, controlled to maintain a steady output voltage.
Output Voltage Sense and Feedback (Q2): Monitors the output voltage and provides feedback to control Q1.
Voltage Limit Setting (VR1): A potentiometer that sets the desired output voltage.
Voltage Reference (ZD1): Provides a stable reference voltage for the feedback circuit.

The 12Vrms AC will rectify to > 15.5 no-load.
The TIP31 is a Darlington drops with a minimum of 1.5V with a load.
The error amplifier is just a single NPN current amplifier so the voltage gain will be the ratio of Rc/Re which depends on the Zener knee resistance. My estimate is the Av ~ 100 for Q2 which corrects the error of the tapped output minus Vbe with the 5.1V.

I prefer to test any regulator with a stepped load to see the dynamic error and load regulation error simultaneously.

You will see the performance at 1A is much worse than the old LM317 for this test. The reason is the lack of error gain in Q2. The next error is the stability of the Zener with high ripple voltage from the unregulated output through 1k.

The pot has resistors on either side to improve adjustment sensitivity may be improved and must change for a different voltage. I decided to show you a 10x ripple reduction using a Darlington also for Q2 if you swap the parts.

The followings points are clear.

- In the start, the output voltage is zero. This also means that base of the Q2 is also zero that comes from resistance network.
- As the power supply turn on. The voltage across C1 increases.
- The base of Q1 crosses 0.6 V.
- The emitter of Q1 starts rising but it stays that 0.6 V below the base voltage.
- The increasing output voltage will also make the base of the Q2 increasing.
- As the base of the Q2 reaches at Zener voltage added with 0.6 V, the transistor Q2 starts conducting. Note before Q2 start conducting, the Q1 was conducting.

What happend next ?
The Q1 will turned off ?

What happenS next ?
Linear error feedback modulates Q1b according to feedback gain
Will Q1 be turned off?
No

If you examined my simulation, you might understand better.
Here with more plots.

Notice the output Q1e (emitter) is 9.022V corrected with 15k below pot.
Then when active pulsed load is connected the voltage drops but never exceeds the no-load voltage because Q2 only regulates in one direction.
Again using the mouse if replace your Q2 with my floating Darlington NPN below the circuit above, you will see ripple error drop 10-fold.

D.A.(Tony) Stewart nicely shows the same with the simulation; As this is linear regulator and not switched mode, the transistors (hopefully ;D) will not switch totally on (saturate) or totally off but instead stay somewhere in the middle.. As i wrote, q2 pulls q1 base voltage downwards *towards* "regulation point" which would be output voltage +0,6v..if it goes too far the base current of q2 goes down due to voltage difference getting smaller accros q2 emitter and divider resistor network..so it should set itself to that point determined by abovementioned components and voltages..
As D.A. (Tony) Stewart showed, the transistor gains play crucial role of how well the circuit regulates..either load or input variations..this is so simple circuit that it's not easy to get this to oscillate, but you can play a mind game where a feedback over-compensates and is a bit slow..then it would overshoot the output at start, but fix it a bit late..undershoot..fix it a little late..overshoot..etc..

With only 1 large cap it is a 1st order RC loop circuit like a compensated Op Amp but lacking the high gain at DC.

Adding caps anywhere in the loop or output delays feedback and will not reduce output ripple and that also makes it less stable meaning possibly increasing ripple.

This design is a good example of how NOT to design a linear regulator.

Uh..oh..i forgot something from the end of the explanation in my last post..as the base current of q2 goes down also the collector current goes down..so according to ohm's law the voltage on base of q1 (the resistor r1 does the dirty work here with the voltage when q1 collector current changes) starts to rise and that will affect output voltage as the q1 emitter voltage will follow the base voltage with beforementioned 0,6v difference..thus the circuit should 'lock in'..

What happend next ?
The Q1 will turned off ?
It stays in "regulated mode", i.e. "linear operation", that why it is called "linear power supply".

It does NOT turn OFF, it does not switch OFF ... because it is not a SWITCH_Mode_PS

***
Btw: this circuit is not only mature, it also is unprecise (will drift) and suffers from high powr dissipation (a lot of heat to be dissipated by a huge heatsink, mabe needs a fan) and it lacks a current limiter.

Nowadays one use switch mode technology.
For sure linear power supplies have their advantages, especially: regulation speed, low noise, high precision .... but fur this you need a more advanced design.

Klaus

You should know how to analyze the plot voltages in my response then figure out how it works. You know “a picture worth a thousand words” just by knowing discrete components. If not review Vbe controlled collector currents.

If you remember when Vbe = 0.6V Ic will be ~ 1mA. But above this voltage like Vbe2 =639mV from Q2 that Ic must be a bit more than 1mA. In fact the plots show the voltage drop is about 5 V across the 1k collector of Q2, which is good result. But 0.7V is too much and 0.6V is too little. Above Vbe = 0.6V or Ic = 1mA bulk Rbe starts show the variations between small & large current devices, which causes variations in Vbe found in datasheet plots.

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I understand that when Q2 turns on, this will lower the base current of Q1, causing the Vout to decrease. The decreasing Vout will cause the base voltage of Q2. If the base voltage of Q2 is below 5.1 V+0.6 V, the Q2 will turn off. This will increase the base current of Q1, causing the Vout to increase.

The Q1 will not completely turned on and will not completely turned off but it stays in the middle and keep regulating the Vout. This is how we call it linear power supply. Is that correct ?

How do we calculate the value of Vout using the resistors R2, R3, R4, and VR1 is not known to me.

Hi,

R2 .. in my eyes is just for safety. The pot wiper .. when wiping ... may get high impedance. The wiper and the resistive sureface is not perfect thus it may lose connection.
Just imagin there is a little spot of dirt on the resistor surface that prevents the wiper to keep contact with the surface.
In this case (without R2) the feedback path would be OPEN and thus the output voltate would go higher than wanted.
R2 simply prevents the feedback path to go completely OPEN and thus prevents the output voltage to go extremely high.

R3, R4 and VR1 act as a variable voltae divider.

The wiper goes to Q2 base. The voltage - in regulated state - at this node is considered to be V_z + V_BE = 5.1V + 0.6V = 5.7V (with tolerance).
This voltage is fix, the regulator always regulates the output voltage in a way that this node voltage is 5.7V.

Now if you connect this node directly to the output voltage, then the output voltage would be 5.7V.
if you have an 1:1 voltage divider you get 5.7V across the bottom resistor and 5.7V across the top resistor .. resulting in a total output voltage of 5.7V + 5.7V = 11.4V

if you have a 1:2 voltage divider you get 5.7V across the bottom resistor and 2x 5.7V = 11.4V across the top resistor .. resulting in a total output voltage of 5.7V + 11.4V = 17.1V

and so on... the rest is just math.

(for sure in detail one needs to consider BJT behaviour, base current, thermal drift ... and so on)

Klaus

Now it is clear. If R2 is for safety and the resistors R3, R4 and VR1 act as a variable voltage divider then it is much easier to understand how we can set the Vout using these resistors.

Regarding BJT behavior, thermal drift etc. I was also thinking to have OpAmp.

I understand that Q2 transistor causes lazy turning ON and OFF the transistor Q1.

I have seen OpAmp in place of Q2 in some linear regulator where it is often called Error Amplifier. I guess with OpAmp in place of Q2 there will be hard ON/OFF the output of the OpAmp connected to the base of Q1.

In case of having OpAmp in place of Q2. The non-inverting input of the OpAmp is connected to the voltage reference which is ZD1 (5.1V) and the inverting input of the OpAmp is connected to the negative feedback, in other words connected at the mid-point of the voltage divider from Vout. The voltage divider mid-point which is connected at the inverting input of the OpAmp should be set at 5.1 V for a given set voltage of Vout.

As the Vout drops below the set point. The voltage divider mid-point will be below 5.1 V which is inverting input of OpAmp, causing the OpAmp output high because the non-inverting input is connected to the reference 5.1 V. This will turn on the Q1, causing the Vout to increase.

As the Vout increases beyond the set point, the mid-point of the voltage divider will be higher then 5.1 V connected at inverting input of the OpAmp. This will make the OpAmp output low causing the Q1 to turn off.

Which option is better Q2 transistor or OpAmp ?

I guess with OpAmp in place of Q2 there will be hard ON/OFF the output of the OpAmp connected to the base of Q1.
This is not how it should work. You always want to design it to keep it in linear mode. --> no ON/OFF!

The non-inverting input of the OpAmp is connected to the voltage reference which is ZD1 (5.1V)
I would probaly use a true - ready to buy - voltage reference. For precision (low drift) reasons.

The voltage divider mid-point which is connected at the inverting input of the OpAmp should be set at 5.1 V for a given set voltage of Vout.
should be --> will be
for a given Vout --> for any Vout

One should design in a local feedback fro the OPAMP for stability reasons.

Which option is better Q2 transistor or OpAmp ?
"better" is meaningless as long as you don´t state in which regard.

Regarding
* precision: yes.
* cost: no
* stability: it depends

Klaus

The last thing I would like to add in this post is where do we connect the sense wire ?

Consider the experimental setup is located at some meters away and there is voltage drop across the power cables. We will have the sense wires from the experimental setup load.

The question is where we will be connecting/attaching the sense wires in the circuit which is already attached above.

The last thing I would like to add in this post is where do we connect the sense wire ?

Consider the experimental setup is located at some meters away and there is voltage drop across the power cables. We will have the sense wires from the experimental setup load.

The question is where we will be connecting/attaching the sense wires in the circuit which is already attached above.
I guess you ignored the message that this design is a good one to forget (in spite of how many hobbyist websites might have it) and just use a better one now ..., after you provide us all your application load specs which includes long cables. Types of cables can matter so if more than say a PC DC power cable , you might want to have the regulator attached to load.

Why use a regulator with an output impedance in the range of 1 ohm (= Rb / hFE best case) when we have decades of experience on ones with tens of millohm output impedance or less.

By the way, output impedance/ load ratio of the regulator is the same as of load regulation error (%)

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Regarding the sense wires.

I think I got it. The resistors R3 and R4 can be physically present located at the linear power supply position but the ends of R3 and R4 which are connected to +Vout and GND respectively. I need to run them longer using two wires and attach them to the point of load to sense the voltage there which can be couple of meters away. This will compensate the voltage drop in the current carrying wire.

Regarding the sense wires.

I think I got it. The resistors R3 and R4 can be physically present located at the linear power supply position but the ends of R3 and R4 which are connected to +Vout and GND respectively. I need to run them longer using two wires and attach them to the point of load to sense the voltage there which can be couple of meters away. This will compensate the voltage drop in the current carrying wire.
The drop in the regulator due to load error gain and output impedance is likely far in exceed of your imaginary power loss unless you are using undersized wire.

I am not trying to be super critical, but you have not been exactly as expected for an EE to specify mandatory specs like load currents, acceptable ripple etc.

> how do we calculate the Vout with the values of resistors R2, R3, R4, Potentiometer, and reference voltage 5.1 V
Use Kirchoff's Laws to find node voltage and change load resistance to determine effects.
Does this require more training?

The drop in the regulator due to load error gain and output impedance is likely far in exceed of your imaginary power loss unless you are using undersized wire.

I am not trying to be super critical, but you have not been exactly as expected for an EE to specify mandatory specs like load currents, acceptable ripple etc.

> how do we calculate the Vout with the values of resistors R2, R3, R4, Potentiometer, and reference voltage 5.1 V
Use Kirchoff's Laws to find node voltage and change load resistance to determine effects.
Does this require more training?
The purpose of this post to understand the internal circuit of the linear power supplies. What are the main components and their functions. How do we set the output voltage, which resistors are involved in setting the output voltage. In case the linear power supply is couple of meters away from the point of load then where we attach the sense wires in the power supply circuit.