Help: Regulated Power Supply

[hardcore misery]

Quote:
Originally Posted by hardcore misery
we were not required to use an IC to our project...

so is ti possible to step down a 220VAC into 15VDC? then by using a voltage multiplier, 15VDC can be 60VDC? that is my design... and it should be regulated...


In theory, but it's not generally a practical solution - forget about voltage multipliers, they are a waste of time!.





sir...can you suggest an alternative to voltage multiplier? how can i produce a 60V DC output from a 220VAC with a load current of 1A and by using discrete components? hoping for help here... sorry for my bad english... godbless to all

 

[Nigel Goodwin]

You use the correct voltage transformer, then (if it needs to be regulated) you add a discrete component regulator on the output - the transformer will need to be a different voltage for regulated or not!.

For the 15V rail either use a seperate winding or a seperate transformer, it's really too big a drop for a linear regulator.

 

[hardcore misery]

You use the correct voltage transformer, then (if it needs to be regulated) you add a discrete component regulator on the output - the transformer will need to be a different voltage for regulated or not!.

For the 15V rail either use a seperate winding or a seperate transformer, it's really too big a drop for a linear regulator.



sir, what do you mean by correct voltage transformer? so it is impossible to step down a 220VAC into a 60VDC or 15DC? the DC output should be regulated...

 

[Nigel Goodwin]

sir, what do you mean by correct voltage transformer? so it is impossible to step down a 220VAC into a 60VDC or 15DC? the DC output should be regulated...

A mains transformer outputs AC, for an unregulated DC you would require a transformer a little over 40V AC to give roughly 60V DC after rectification and smoothing. For a regulated supply you need more than 60V to feed the regulator - so you would need a higher voltage transformer. The actual voltage depends on the design of the regulator, and on the specifications it needs to meet.

 

[hardcore misery]

so i can't use a 220VAC input? but that is the required input voltage for our designed PSU...i have no option but to choose 60V (or 15V but to multiply it or 12V but to make it 42V to be unique with my other classmates) as an output DC voltage...

 

[Nigel Goodwin]

Yes, of course you can (and should), 220V is the PRIMARY of the transformer, it's the secondary that gives the output voltage.

If your knowledge is so limited I suggest you go and talk to your teacher, either you haven't been paying attention in class, or you've been ill and missed all the lessons!.

 

[Hero999]

I've already shown you where to see a schematic for a discrete voltage.

But here's the Googles I've used to get the results.

https://www.google.com/search?num=1...=en&q=voltage+regulator+schematic&btnG=Search
https://www.google.com/search?clien...tor+discrete&sourceid=opera&ie=utf-8&oe=utf-8

 

[hardcore misery]

**broken link removed**

please ignore the captions on the image...

sir.. is the 220VAC a rms? or a peak to peak voltage?

all i did was to assume that as a rms voltage and step down it to 60Vrms
which i converted the 60Vrms to Vpeak( 60V * 1.414 ) then
this voltage runs though 2 diodes ( Vpeak - 0.7 - 0.7 )
then the resulting voltage is the Vsupply which is rectified...

after that.. i'm going to solve for R1 which is (Vsupply - Vz all over Iload + Iz)
zener diode is based on the expected DC output..so i've chosen 56V/1W zener voltage...

how can i assume a Iz and a Vz? our load current is from 750mA - 1A
so if i computed for R1... and the Vdrop across R1 ( vsupply - Vz)

am i going to use voltage divider rule for the expected DC output?
(Vout = (R1/ (R1 + RL))*Vdrop)?

i have the complete details... i'm asking for your confirmation against my ideas given on this design...

godbless to all of you...

anyway, i'm sorry for my bad english...

 

[audioguru]

Your 1k resistor has 27.5mA in it without a load. So if the load draws 27.5mA then the zener diode doesn't regulate and if the load draws more than 27.5mA then the output voltage drops.

A zener diode is a lousy voltage regulator because it wastes power when the supply doen't have a load.
Change the 1k resistor to 27 ohms. Then the current in it and in the zener diode is 1.02A without a load. The power in the 56V zener diode is a whopping 27.9W! But you can load it with 1A and it will still have 56V regulated.

A zener diode is a shunt regulator. Use a transistor as a series regulator (instead of the 1k resistor) then it won't waste much power when there isn't a load. A zener diode can be from its base to ground and the transistor is an emitter-follower. The zener diode and the base of the transistor will need a small amount of current from the 83V unregulated to function that is supplied by a suitable resistor.

 

[Hero999]

This is getting frustrating. . .

You need to read up on this before you ask any more questions.

https://www.electro-tech-online.com/custompdfs/2006/10/EE332LE5rev3.pdf
http://en.wikipedia.org/wiki/Linear_regulator
**broken link removed**
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/emitfol.html
**broken link removed**
http://ludens.cl/Electron/Ps20/Ps20.html
http://www.williamson-labs.com/480_xtor.htm

http://en.wikipedia.org/wiki/Bandgap_voltage_reference

 

[muki55]

Here's what you want a transistorized regulated power supply.
I'll post soon the parts list if this circuit helps.

note: voltage display (error) reverse the switch.
voltage display corresponding in voltage divider.

 

[Nigel Goodwin]

It's rather crude, and only slightly regulated - using a couple of foward biased silicon diodes as a reference doesn't make for good regulation.

 

[redserpentone]

electronic techs are a dying breed, killed by managers that demanded techs to help unqualified and unschooled workers to learn easy rules of thumb, then you get fired because the managers believe that they now have a competent worker that knows how to find the answers at a fraction of the pay ... it hasn't changed I see ... I do security work now ...
redserpent1

 

[Roff]

Here's what you want a transistorized regulated power supply.
I'll post soon the parts list if this circuit helps.

note: voltage display (error) reverse the switch.
voltage display corresponding in voltage divider.

The base-emitter junction of Q2 will break down for output voltages above about 7.5V (assuming 6V breakdown voltage). This will degrade the beta of Q2 over time, and will also increase the current through your reference diodes (D2, D3) and cause the output voltage to be somewhat higher as a result.

The temperature coefficient of your reference is about -6mv per degree C (D2 & D3 plus Q4 Vbe), which is not a problem if the ambient temperature is relatively constant (and you don't have the diodes or Q4 mounted close to Q1). If they do drift, the tempco of the output is multiplied by the feedback ratio. For example, on the 12V setting, the output tempco will be about -40mv per deg. C.

 

[muki55]

Transistor Q1 is the series-pass transistor;Q3 is a darlington configuration. This improves the response of Q1 with respect to the error signal. Transistor Q4 acts as a comparator or a differential voltage amplifier. D2 and D3 are silicon diodes in forward biased connection. Q4 and R4 supply enough current to the diodes and maintain an almost contant voltage drop of about 1.4 volts, the reference voltage. Capacitor C3 provides pre-regulation to the comparator circuit. The base of Q4 is connected to the voltage-dividing network of R6 and R7 and the combination of R7 with any or all of resistor R8to R12 Any change of the output voltage appearing at the top of the voltage divider is sensed at the base of Q4 and afterwards amplified. This results to the error or control signal at the base of Q3 and the series-pass transistor Q1 responds immediately by changing its conduction to maintain the desired output voltage.
An increase in output voltage will be feedback to the base of Q4 which, in turn conduct more current. This reduces the drive to the load, effectively reducing the output voltage. The reverse will happen when the output voltage decreases. Thus, the output voltage is properly regulated

 

[Roff]

I agree that it's well-regulated. It just has a crappy tempco, and a short-circuit protection transistor which is operating with the b-e junction in avalanche.

 

[muki55]

A short circuit protection is a current-limiting with foldback type. Foldback is desirable since the short circuit power dissipation of the series-pass transistor Q1 can be limited, preventing damage to it. The emitter of Q2 is connected to the output which effectively reverse biases transistor Q2. If the negative and positive output terminals are accidentally short circuited, the emitter of Q2 will be grounded. Current flowing through R2,R4< and R5 will force saturation, thereby limiting the current to almost zero. On the other hand, when the short circuit is removed, the output voltage will be back again since Q2 becomes reverse biased again.

 

[Roff]

A short circuit protection is a current-limiting with foldback type. Foldback is desirable since the short circuit power dissipation of the series-pass transistor Q1 can be limited, preventing damage to it. The emitter of Q2 is connected to the output which effectively reverse biases transistor Q2. If the negative and positive output terminals are accidentally short circuited, the emitter of Q2 will be grounded. Current flowing through R2,R4< and R5 will force saturation, thereby limiting the current to almost zero. On the other hand, when the short circuit is removed, the output voltage will be back again since Q2 becomes reverse biased again.

The problem is, when the Q2 emitter is at, say, +12V, and the base is connected to the reference diodes through R5, the base-emitter will potentially have about 10.6V reverse bias across it. Look up the breakdown voltage spec on Q2. I doubt that it is more than 6V. This means that the reverse current flows through the junction, which will degrade beta over time. Depending on the value of R5, the degradation may occur slowly. Still, it is not a good design from an engineering standpoint.

 

[audioguru]

The first thing I was told at my first job (at Philips) was to never allow an emitter-base junction to have avalanche breakdown.

When did Philips buy Lumileds, the makers of Luxeon very high power LEDs?

 

[eblc1388]

Will the circuit work satisfactorily by adding the following changes:

1. extra diode(in4001) in the base connection of short circuit protection transistor Q2.

2. add one more diode in series to existing diodes(reference) or replace the three diodes using a 2.5V reference like LM336

3. recalculate the output voltage divider

4. forget the above and just use a three terminal regulator like LM317 or its higher current brothers