[SOLVED] Linear Power Supplies

Status
Not open for further replies.
E

engr_joni_ee

Advanced Member level 3
Joined
Nov 3, 2018
Messages
831
Helped
2
Reputation
4
Reaction score
7
Trophy points
18
Visit site
Activity points
6,955
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)
1.3 Voltage Readout
1.4 Current Readout
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 ?

Thanks in advance.

 
Last edited:
Sort by date Sort by votes
engr_joni_ee said:
I am sorry I do not have any specific design goal.
Click to expand...
Most power supplies don´t have sensing capabilities.
Many power supplies are also meters away from the load.

So if you ask for special "accurate voltage regulation methods" ... (voltage sensing) .. it is quite straight forward (for me) to ask why you think you need this special method.

Usual reasons are:
* your load is very sensitive to supply voltage change
* load that changes the drawn current in a wide range
* there is something special we don´t know yet.

We simply don´t understand why you accept a drift of the output voltage by ambient temperature...
while you are worried about output voltage drift caused by voltage drop on the wires.

You ignore the recommendation to adjust wire gauge to get low voltage drop ... but you are willing to install extra wires for sensing.
It does not make sense to us.

****
If we professions do this job with numbers and calculations ... we don´t do it just out of mood.
If we can expect an output voltage drift by temperature of 100mV max ... and we expect an output voltage drift by 30mV max caused by the load current... then we focus on the highest expectable error reason--> the thermal drift.
This are just example numbers. We don´t have realsitic numbers, but you should have. You are the designer.
You should have numbers for your use case and you should have numbers for the maximum error you allow.

***
An example: If I want to bulid a house ... and I ask for an earthquake proof foundation .. but the house is not meant to be built in an earthquake region .... then someone may ask for the reason behind it. And if I don´t have a valid reason ... it´s quite expectable that someone recommends me to forget about the extra solid foundation. It´s just a waste of anything.

***

Klaus
 
Upvote 0 Downvote
Klaus ST filled in all the gaps perfectly above about that circuit..

As i tried to asses Your electronics skill level so i can give You information that might help You on Your level, i came to conclusion (and D.A.(Tony)Stewart above seemed to note the same..) that You would benefit greatly if You would go out in the google and really get the Kirchoff's law under Your belt and also connect it with Ohms law

..meaning..do like few of those 'resistor puzzles' that are sometimes in exams succesfully..You would have instantly recognized the voltage divider had You practiced such puzzle..(there are resistors connected seemingly randomly together and you have to figure out how to calculate some missing value..etc..).

That must not be regarded as some newbie stuff..but stuff that absolutely everything revolves around.. and sooner it is Your other nature, the sooner You'll start getting a real grip on the circuits that You look and need to understand..! Learn these and i see nothing standing in Your way in a lifelong process with electronics..
 
Upvote 0 Downvote
engr_joni_ee said:
I am sorry I do not have any specific design goal. The purpose was to understand the basic liner power supply circuit and how it function.
Click to expand...
Then the purpose of your experiment should be to define the measurable node voltages in simulation, compare with math and logic with your understanding of Kirchoff's Laws and transistors then list test results so you understand how to design by specs here reverse engineering in lab tests and/or simulations.

Try harder.

Critical specs you would be expected to know in a day-to-day design job and fix are;
Adjustment range or sensitivity.
No/Full-load noise. (Vpp < 20MHz with 10:1 probe or AC coupled into 50 Ohms.) Then understand why.
Load regulation error % at various loads or worst case.
Power Loss vs load (Ohm's Law) and temperature rise vs theoretical thermal resistance in free air with case temperature.
Max current. wire guage and mOhms per meter thus voltage drop. This is critical for laptop chargers that have fine flexible cables and sense wires at the plug on a 1m cable. But not likely on a backplane or 2 m AWG20 with < 1 A.

Have you specified the wire ohms/m and current yet?
Would you expect it to affect anything on this regulator with remote sensing or cause more issues with sensitivity to nearby noise (parasitic) like a soldering iron relay clicking on/off transient?

With practice you will be able read plots and schematics as fast as cartoons. ')

Then ask a better question.
 
Last edited:
Upvote 0 Downvote
A linear power supply works by using a pass transistor (Q1) and a feedback loop to regulate the output voltage. The transformer (T1) steps down the AC voltage, and the bridge rectifier (D1-D4) converts it to DC. A smoothing capacitor (C1) reduces ripple. The output voltage is controlled by the voltage reference (ZD1) and a voltage divider formed by resistors (R2, R3, VR1). The feedback transistor (Q2) adjusts Q1 to maintain a constant output voltage.

The output voltage VoutV_{out}Vout is calculated as:

Vout=Vref×(1+R2+VR1R3)V_{out} = V_{ref} \times \left( 1 + \frac{R2 + VR1}{R3} \right)Vout=Vref×(1+R3R2+VR1)
Where VrefV_{ref}Vref is the reference voltage (e.g., 5.1V from ZD1).
 

Upvote 0 Downvote
Anovalight said:
The output voltage VoutV_{out}Vout is calculated as:

Vout=Vref×(1+R2+VR1R3)V_{out} = V_{ref} \times \left( 1 + \frac{R2 + VR1}{R3} \right)Vout=Vref×(1+R3R2+VR1)
Where VrefV_{ref}Vref is the reference voltage (e.g., 5.1V from ZD1).
Click to expand...
Kindly explain how you find it. Rest is clear what you said before.
 

Upvote 0 Downvote
Linear regulators are a combination of series pass emitter followers to lower Zout with shunt reference voltages {Vz + Vbe} in this bad example instead of a bandgap Vref.
I was able to reduce the ripple with extra ordinary efforts to nearly 16 mV at 1A pulse load indicating a successful low Zout of 16mV/1A = 16 mohms which would be expected for a 1A regulator with an excellent load regulation error of 16mV/9V=0.178% (1% typ. Is good)
But this required brute force bulk cap to reduce the raw ripple to 866 mV@ 1A with an error reduction due to loop gain of 866mV/16mV=54x which is very low even considering I bypassed the R divider DC to null Vac drop with a 10uF feedback cap, dropped the feedback base resistor and still very low loop gain compared to an Op Amp. normally included in a linear regulator.


If you think you can improve the cost/performance ratio of a commercial Linear Regulator then you might get a designer job in analog circuits, but you have a long haul of learning how IC design is mastered to keep current with IC designs.

Conclusions
The critical design parameters are PSRR for ripple reduction , load regulation from source / load impedance ratio where source output impedance is reduced from bulk emitter follower Zout by negative FB gain (beta) and thermal losses from optimal source Vrms/ Vdc ratio. Considering this example uses 60 year old technology , modern LDO’s use MOSFETs with low dropout high error FB gain and critical choice of smaller e-caps with stability issues the the ESR is too low suppressing too much ripple feedback and too high ESR that results in normal load regulation errrors.
--- Updated ---



Editing this schematic text on a smart phone with a fat finger is a real PITA
 
Last edited:

Upvote 0 Downvote
Since you put in so much effort into your question, I thought to do the same in another answer.

I think you understand how the negative feedback function cuts the shunts bias, this is because almost all Linear Voltage regulators are unipolar like Class A amplifiers, active in one direction of current and passive depending on R in the opposite direction. Since MOSFETS are voltage controlled resistance the ratio of R to RdsOn enables a higher gain for gain. Thus I was able to drive a 100% square wave into a 54 W load @ 9V before the minimum unregulated voltage started to create > 50 mV drop due to the threshold Vg-Vt difference that affects RdsOn.

To all who care to learn

If you can read scope plots with stats, name 3 significant results from this simulation
e.g. P.stored/P.load or reactive/passive power ratio from a 1 H step-down transformer.

This is not an optimal design for an LDO, just an enhancement of the initial question with low RdsOn FETs.

 
Upvote 0 Downvote
Linear power supplies are a type of power supply that use a linear regulator to convert AC power to DC power. They are known for their low noise and high efficiency, making them ideal for applications that require precise voltage regulation.
 

Upvote 0 Downvote
Upvote 0 Downvote
Let's see how this works.

To have a good starting point, redraw the circuit and break VR1 into two parts, namely, VR1a and VR1b. Connect VR1a parallel to R2 and VR1b in series with R4.

After doing that, you will find that VR4 + VVR1b = VZD1 + VbeQ2 = 5.1V + 0.7V = 5.8V. That means that VR2 (equals VVR1a) + VR3 = 9V - 5.8V = 3.2V.

I hope you can proceed from there.
 
Upvote 0 Downvote
Mehvish99 said:
Linear power supplies are a type of power supply that use a linear regulator to convert AC power to DC power. They are known for their low noise and high efficiency, making them ideal for applications that require precise voltage regulation.
Click to expand...
Yikes!!
I have never seen so much misinformation in a single post.
 

Upvote 0 Downvote
Let's consider this post close. There has been some good reply and comments in this post from those who have knowledge and experience in linear power supplies.

I myself apologies for not asking all the questions in the first posting and sometime not providing enough information in the beginning.

Thanks everyone for your comments.
 

Upvote 0 Downvote
engr_joni_ee said:
Let's consider this post close.
Click to expand...
you mean to "close this thread"? --> then simply press the [Mark as solved] button.

You are the OP of this thread .. so why should any other person decide when to close this thread?
Basically this is why the button exists...

Klaus
 

Upvote 0 Downvote
A typical ACDC 3A 9V flyback SMPS will be 88 % efficiency +/-4 and the most expensive part is the $3 low loss 4A Inductor,
L = 470 µH
DCR = 125 mΩ
IDC = 4 A
To improve on efficiency for a linear regulator the DC voltage ratio must be > 80%.
While no load bulk DC rises 40% (~sqrt(2)-1) from Vac rms, for a full bridge while line input tolerance is 10%.
So a linear regulator can only be more efficient if the input Vdc is fixed and a true LDO like from 10V to 9V.

An example from design tool TI's WebBench

( All SMPS or LC schematics must include this DCR (L) ,ESR (C) otherwise performance may not be the same)




Simple buck SMPS may have lower efficiency > 80 while boost regulators are 90~95% but complexity is a tradeoff for more efficiency.

Here is a simpler buck SMPS




If efficiency is most important, due to excess heat and temp rise in components, examine all the variables and minimum load requirements for ALL REGULATORS.

No load can cause stability issues with a pulsed load.


Low load pulses may cause less damping to overshoot.
--- Updated ---

Mehvish99 said:
Linear power supplies... They are known for their low noise and high efficiency,
Click to expand...

ONLY TRUE for DCDC LDO's where Vout/Vin > 92% for full load or at no load (sleep) otherwise this is FALSE due to high V drop from all variables ( %Vin, %Iout ) >> 10% loss and 50% loss means samer power of load is dumped in regulator.


Time for you to accept answers and close.
 
Upvote 0 Downvote
Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…