I would look into a dedicated PS or find a laptop or similar SMPS and us a dc-dc step-down converter with good filtering .. If you want stay lineral supply the transformer will have to be change out and that will not be cheap and pretty big and heavy.. But you always could mod a microwave trans for needs and provide the current needed..

 

A redesign may be in order.
Will let you know how I get on.
I could have probably bought something cheaper by this stage, but I'm bull headed about this now.

What is your objective? Are you trying to learn how to design a power supply? Or do you just want a 12V@5A supply so you can get on with your real project?

 

12 V, 15 V, one 4007 in series, two 4004's - whatever. Above, I assumed a value of 1 V for the maximum allowable ripple across the filter capacitor. Here are the real numbers.

For any given transformer output voltage and current, and a regulator circuit's output voltage and current, you can calculate the exact amount of filter capacitance needed. It is messy and gets into intersections between parts of a distorted sine wave and parts of an exponential capacitor discharge curve into a constant power load. OK, make that *really* messy. A first-order approximation that is surprisingly accurate is to assume a square wave power source and a constant current load. Solving the integral equation for the voltage across a capacitor yields this little gem:

e x c = i x t

Rearranging, we get this: c = i x t / e --> The value of the holdup capacitor needed to maintain the ripple voltage to a specified value equals the output current times the period of the input power waveform divided by the ripple voltage. In your case:
2N3055 Vbe @5 A = 1.2 V The chart on the 3055 datasheet shows the base-emitter voltage as just over 1.15 V for 5 A collector current
Vpeak = 15 V Peak voltage across the bulk filter cap

12 V zener + one 1N4004 = 12.7 V effective zener voltage
12.7 - 1.2 = 11.5 V output voltage at full load - Too low

12 V zener plus two 1N4004's = 13.4 V effective zener voltage
13.4 - 1.2 = 12.2 V output voltage at full load - Let's use this option

15 V - 13.4 V = 1.6 V This is the regulator headroom. All input voltage variations, including ripple peak-to-peak voltage, must be less than this
i = 5 A
t = 0.00833 s Period of full-wave rectified AC (1 / 120 Hz)
e = 1.6 Vpp Again, any more ripple than this and the circuit will pull out of regulation and you'll see ripple on the output.

5 x .00833 / 1.6 = 0.0261 F = 26,100 uF This is not an outrageous number. One 27,000 uF/25V part will work and Digi-Key has them for under $5. 33,000 uF would be better, but the regulator still will be on the hairy edge of dropout because of the the low transformer voltage. If the input AC power fluctuates downward, ripple will appear on the output. A larger filter cap will help, but not as much as a different transformer. For example, a transformer with a 15 V secondary adds 4 V to the headroom. This lets you cut the filter capacitor by 50% while having more operating margin and a cleaner output. A simple zener-and-emitter-follower regulator is good enough for many applications, but everything has limitations on what it can do with what it is supplied, and you've run smack into your transformer's limits.

ak

 

What is your objective? Are you trying to learn how to design a power supply? Or do you just want a 12V@5A supply so you can get on with your real project?

Originally I wanted to build a small size power supply for a particular project. But the more I get into this the more I want to learn about how create these things properly. In some cases I guess it's horses for courses. It's good to learn new things yeah.

 

12 V, 15 V, one 4007 in series, two 4004's - whatever. Above, I assumed a value of 1 V for the maximum allowable ripple across the filter capacitor. Here are the real numbers.

For any given transformer output voltage and current, and a regulator circuit's output voltage and current, you can calculate the exact amount of filter capacitance needed. It is messy and gets into intersections between parts of a distorted sine wave and parts of an exponential capacitor discharge curve into a constant power load. OK, make that *really* messy. A first-order approximation that is surprisingly accurate is to assume a square wave power source and a constant current load. Solving the integral equation for the voltage across a capacitor yields this little gem:

e x c = i x t

Rearranging, we get this: c = i x t / e --> The value of the holdup capacitor needed to maintain the ripple voltage to a specified value equals the output current times the period of the input power waveform divided by the ripple voltage. In your case:
2N3055 Vbe @5 A = 1.2 V The chart on the 3055 datasheet shows the base-emitter voltage as just over 1.15 V for 5 A collector current
Vpeak = 15 V Peak voltage across the bulk filter cap

12 V zener + one 1N4004 = 12.7 V effective zener voltage
12.7 - 1.2 = 11.5 V output voltage at full load - Too low

12 V zener plus two 1N4004's = 13.4 V effective zener voltage
13.4 - 1.2 = 12.2 V output voltage at full load - Let's use this option

15 V - 13.4 V = 1.6 V This is the regulator headroom. All input voltage variations, including ripple peak-to-peak voltage, must be less than this
i = 5 A
t = 0.00833 s Period of full-wave rectified AC (1 / 120 Hz)
e = 1.6 Vpp Again, any more ripple than this and the circuit will pull out of regulation and you'll see ripple on the output.

5 x .00833 / 1.6 = 0.0261 F = 26,100 uF This is not an outrageous number. One 27,000 uF/25V part will work and Digi-Key has them for under $5. 33,000 uF would be better, but the regulator still will be on the hairy edge of dropout because of the the low transformer voltage. If the input AC power fluctuates downward, ripple will appear on the output. A larger filter cap will help, but not as much as a different transformer. For example, a transformer with a 15 V secondary adds 4 V to the headroom. This lets you cut the filter capacitor by 50% while having more operating margin and a cleaner output. A simple zener-and-emitter-follower regulator is good enough for many applications, but everything has limitations on what it can do with what it is supplied, and you've run smack into your transformer's limits.

ak

Right yes now I see. Thanks AK
This will help me to work it out for myself in the future.
It all comes back to the transformer in the end. If I beef this up I have a much better chance of getting a stable voltage out of it no matter which way I go.
Can I ask, how do you work out the figures required for a regulator such as the LM338?
Is there a similar method?
I followed a suggested link and found a simple circuit for one using a 40V transformer, but would really like to know how to work it out for myself.

Regards
New

 

Here's a nice chip for $1.33 @ www.mouser.com
It only needs 1/2 volt of headroom.
There are ways to do this by building your own low dropout regulator, but I shudder to think of the instructions it would require.

http://www.mouser.com/Semiconductor...0wd5eZ1z0wbdjZ1z0w79mZ1yx5k7vZ1z0shyuZ1z0z63x

Thanks for the link and info #12. I think 3A may be sailing a bit close to the wind for my purpose here but handy to know for possible future apps.

Regards
New

 

There are also 5 amp low drop out chips, but I made a mistake. You need to learn how to use the parametric search engines on the vendors sites.

So. Voltage out + voltage the regulator needs + lowest dip in voltage at the filter capacitor is your minimum required DC voltage.
The power transformer labeled voltage is multiplied by 1.414 to get the peak voltage. Subtract 1.4 volts from that for loss in the rectifier. That's your peak DC voltage.
Now you know how much voltage droop you can allow. Frequency of a full wave rectified supply is 2X the power line frequency.
C = I / (1.414) (voltage droop) (120 Hz)

If the answer is absurd, you need a higher voltage transformer.

 

I admit I skimmed through the post and read the initial one, why would you need a regulated supply for a DC motor with electronic control? This is rarely ever needed in the real world?
Max.,

 

I admit I skimmed through the post and read the initial one, why would you need a regulated supply for a DC motor with electronic control? This is rarely ever needed in the real world?
Max.,

In this case, you could just slap a big capacitor on the output of the bridge rectifier, tell the motor it's running off a 12 volt lead acid battery with a float charger, and the motor will never figure out you lied to it.

 

In this case, you could just slap a big capacitor on the output of the bridge rectifier, tell the motor it's running off a 12 volt lead acid battery with a float charger, and the motor will never figure out you lied to it.

TM motors and KB DC motor drives run off a bridge without a cap even!!
Max.

 

There are also 5 amp low drop out chips, but I made a mistake. You need to learn how to use the parametric search engines on the vendors sites.

So. Voltage out + voltage the regulator needs + lowest dip in voltage at the filter capacitor is your minimum required DC voltage.
The power transformer labeled voltage is multiplied by 1.414 to get the peak voltage. Subtract 1.4 volts from that for loss in the rectifier. That's your peak DC voltage.
Now you know how much voltage droop you can allow. Frequency of a full wave rectified supply is 2X the power line frequency.
C = I / (1.414) (voltage droop) (120 Hz)

If the answer is absurd, you need a higher voltage transformer.

Thanks #12
I can understand that perfectly.

Much appreciated

 

TM motors and KB DC motor drives run off a bridge without a cap even!!
Max.

There's 12V timers and relay circuits as well as solenoids. The motor drive speed control has low voltage cutout and reset circuits, including a chip. I'm worried that if the voltage is too high or unstable something will get killed.

 

Can I ask, how do you work out the figures required for a regulator such as the LM338?

What do you mean by "figures". A schematic? The resistor values needed to get 12V?

Is there a similar method?

???

I followed a suggested link and found a simple circuit for one using a 40V transformer, but would really like to know how to work it out for myself.

A 40VAC secondary would present a different set of problems. Using a linear regulator would cause 80% of the power to be dissipated in the regulator. You'd want to use a switching regulator in that case.

Computer power supplies would be a readily available, inexpensive solution for your 12V @ 5A requirement.

 

@newtopsupply
You have a couple of problems. Your transformer voltage is to low and it's current not high enough. But maybe not so bad. You won't be able to get the full 12 volts out of it and at 5 amps the transformer will get hot. But if you don't run it for to long at 5 amps maybe ok. So here is the plan.
Move the 4700 Ufd cap at the output of the regulator to the input side. That will raise the voltage a little bit. Then make the output cap a little bigger if you have one.
See if you can't find a 2N2222 transistor and add it as shown in the schematic to increase the base drive to the 3055.
Use 2 diodes in series with your zener. You won't get quite 12 volts at 5 amps, but maybe 11 or more.

 

Or you could add some feedback with a voltage follower and have some actual regulation. This is your original circuit with the beta multiplier suggested by ronv and an opamp to regulate the voltage. Since this is no longer a shunt regulator, you can use a 1/2W zener and the actual voltage isn't critical (6.8-12V will work). R3/R4 form a voltage divider that feeds 1/2 of the output voltage to the opamp. The voltage on the non-inverting input becomes 1/2 of the desired output voltage. Caps omitted for clarity...

You still need to address your small input-output differential because the opamp output needs to be able to get to the desired output voltage plus the 2 base emitter drops. If there isn't enough headroom for LM358 (can't get closer than a couple volts to +V), you'll need to substitute a rail-to-rail opamp.

 

dl324, a minor point. The original circuit never was a shunt regulator. His zener does the same thing yours does, provide a reference voltage for the output amplifier. His needed almost 200 mA at 12 V, but still...

Also, the only power source for the LM358 is the 15 V on the filter caps (minus ripple), and the 358 has to make almost 14 V out to have 12 V on the 3055 emitter. Don't think a 358 can swing that close to the rail.

ak

 

What do you mean by "figures". A schematic? The resistor values needed to get 12V?
???
A 40VAC secondary would present a different set of problems. Using a linear regulator would cause 80% of the power to be dissipated in the regulator. You'd want to use a switching regulator in that case.

Computer power supplies would be a readily available, inexpensive solution for your 12V @ 5A requirement.

You're probably right but like I said I got bull headed about this now. I'll learn how to do it if it kills me
THe 40V transformer setup was for a adjustable 5 - 30V output circuit

 

dl324, a minor point. The original circuit never was a shunt regulator. His zener does the same thing yours does, provide a reference voltage for the output amplifier. His needed almost 200 mA at 12 V, but still...

Also, the only power source for the LM358 is the 15 V on the filter caps (minus ripple), and the 358 has to make almost 14 V out to have 12 V on the 3055 emitter. Don't think a 358 can swing that close to the rail.

ak

Thanks everyone

Whats your thoughts ak, about capacitance here?
Yours was 27 - 33K uf. This one of course is only a third of that.
Is more better here?

 

Usually, but there are consequences. While your transformer can make 5 A steady state, a large filter cap can make 10 times that as a transient surge into a short circuit. This is the main reason not to have huge filter caps directly on the output. But even before the regulator, large filter caps equal less ripple but more surge current. Life is choice.

ak