Building a stable 5v regulated power supply

[AlainB]

Basically the regulated design is comming from the datasheet. I just add a 2200uF (C1) to reduce the ripple.

I would like to build the 5 volts supply as a module that could be add to a suitable transformer as needed.

Before making the PCB I would like to know what you think of that design and how could it be improved for maximum stability.

Thanks

Alain

 

[Reloadron]

Looks like you want an unregulated 12 volts and regulated 5 volts. Why use two full wave bridges? You would only need a single bridge. Depending n what you want from the regulated 5 volts a heat sink would be in order.

Ron

 

[JimB]

As drawn, the two bridge rectifiers effectively put a short across the transformer secondary. (On edit, wrong! a bit of brain fade there, sorry).

Two bridge rectifiers are not necessary, see my attached modification of your schematic.

A better way to do it would be to have a transformer with two secondary windings and have completely separate supplies with the 0v lines only coming together where necessary on the equipment being powered.
where there is a common 0v line, high current pulses in the feed to the motors can put voltage spikes in the 5v sipply to the logic.

JimB

 

[AlainB]

The reason for the 2 bridges is that I would like the 2 voltages as separate as possible. It is often recommended to use 2 differents power supply when working with steppers motors. I was thinking that using 2 bridges would be enough. Also, as I said, I want the 5 volts supply to be a module.

I have hard time figuring out where is the short.

 

[JimB]

I have hard time figuring out where is the short.

When I first looked at the circuit, I thought I saw a short circuit route through the bridges via the 0v line.
On further inspection, it looks like I was wrong.

The short usually happens when someone tries to use two bridge rectifiers to make a bipolar power supply, but that is not the case here.

JimB

 

[MrAl]

Hi,

Alain:
If you want two power supplies that you can hook up any way you want, you need isolated supplies which requires two separate transformers or a transformer with two separate secondary windings.

If you intend to connect the two supplies to a circuit with a common ground, then you dont need isolated supplies but can use the same transformer output (and rectifier bridge) for both supplies. Simply use the regulator for one and no regulator for the other, or a lower voltage regulator for one and a higher voltage regulator for the other. I believe JimB had shown this diagram.

 

[crutschow]

An interesting option for two DC voltages, where one is 1/2 the other, is to use a center-tapped 12V transformer. Then a bridge connected to the 12V winding will generate a 12V rectified output and the center tap will have a 6V rectified output. This works because the bridge acts as a two-diode full-wave rectifier to the center tap.

 

[bountyhunter]

The only bad thing is the current pulses caused by driving a motor will put noise on the transformer secondary which will couple into the other 5V supply.

 

[AlainB]

I will use a DC adaptor (wall wart) to provide the lower voltage. The R1 will take off part of the heat from the 7805 but also, I hope, with C1 will form a filter as stated in the K179 Driver Documentation:

https://www.kitsrus.com/pdf/k179.pdf

It is said that "R2/C2 (R1/C1 in my schematic) form a low-pass filter to filter fast-rise switching transients from the motor"

As it is, the 5 volts supply is working well (as far as I can tell) but I have no idea if the C1 is doing something good at all.

Any ways to make it better?

 

[bountyhunter]

Looks OK, I would add more capacitance to the input and output of the 7805 to keep it clean and stable.

 

[Mr RB]

You don't need to use a separate wall-wart, as you are connecting the grounds together anyway. Just run it all from the one transformer.

 

[bountyhunter]

I like the idea of using a separate wart because I think he will get less noise and hash on the 5V line caused by the motors current steps.

 

[Mr RB]

That's a point, but he has already decoupled the 5v regulator somewhat with a 9 ohm resistor and large cap.

Ideally each stepper driver will have its own large cap on board. It's common and generally doesn't cause problems having the 5v regulator driven from the same transformer as the motor power providing some decoupling is used.

 

[AlainB]

Looks OK, I would add more capacitance to the input and output of the 7805 to keep it clean and stable.

I changed the 2200uF capacitor value in the schematic for a 1000uF because it was a mistake that I made in the drawing. In fact, I was using a 220uF. I also add a 100uF on the output side.

How is it looking now?

 

[davenn]

The reason for the 2 bridges is that I would like the 2 voltages as separate as possible. It is often recommended to use 2 differents power supply when working with steppers motors. I was thinking that using 2 bridges would be enough. Also, as I said, I want the 5 volts supply to be a module.
I have hard time figuring out where is the short.

connecting the 2 grounds together is negating all you stated. so if you are going to do that then you mite as well use JimB's circuit!!

Dave

 

[AlainB]

Thanks! But still, some points are not answered.

I am very much interested in opinions on the capacitors and filter and their respectives values to get an optimal regulated supply.

 

[ronv]

Optimizing

Interesting. The 6 amp load from the steppers is less of a problem than the transient response to the load change from the low current supply. I suspose it depends on how perfect you want to make it. It's pretty good as it is. See model. You could improve it slightly by making the 100Ufd. cap on the output of the regulator larger.

 

[mvs sarma]

i even suggest additional isolation to Jimb's mod , to have a 3 amp diode in series to rectified 12V (perhaps 18V now) and then add a 10uF at the output , in addition to disc cap shown. then i suppose the load could be better guarded from spikes.

 

[Mr RB]

It would be good design practice to use a large cap (2200uF?) on the input at each stepper driver, then decouple them to the main filter cap with 0.1 ohm power resistors, then the main filter cap value can be reduced quite a lot.

Stepper drivers are current regulated, so they have extremely good supply ripple rejection. There's no need for a massive main filter cap especially after the capacitance is distributed more toward each driver.

 

[AlainB]

Here is my completed 5 volts module. I put a bridge on it so that it can accomodate any reasonable input; a transformer or a wall wart, AC or DC.

I use a rheostat (0-40 Ohms, 15 watts) to find out, without calculation, what will need to be the value of the ballast resistor (R1) to give the voltage that I would like to have at the input of the 7805, in my case around 8,5 volts.

As it is in the last picture, a 10 volts DC wall wart adaptor is feeding the module. The load is 0.250A. The voltage prior to the bridge is 11.58 volts and 10.15 volts after. To get 8.5 volts at the input of the 7805, the rheostat value is at 6.2 Ohms. Ohms law would give 6,6 Ohms but it is close enough. The optimal value would be the one given by the rheostat anyway.

In this particular case, I would think that a resistor is not really essential, the heat sink should dissipate the heat. A jumper wire would then be used instead of the resistor.

(On edit) : With no resistor, the heat sink is at around 63C at a room temp. of 25C. Personnaly, I would put a resistor. 63C degree is not comfortable to the touch.