Does this PS look right?

[obscurity]

I am going to try my hand at building a power supply. This is really just a first try and is a little smaller than I would need to build for a 3 axis router. Its main purpose is to prove to myself that I can build one. I am using off the shelf radio shack components for this one because they are easier to get

My questions are:

Does this look right?
What am I missing?
Are my calcs right?
How do you size the Bleed resistor?

Thanks for any help
John W.

[vladdy]

Just off the top of my head, maybe 10k 10w for bleed... depends on how long you want it 'live' before 99% bleed down... ? about 3 minutes for above [pretty much a WAG, sorry..]

What some others do is drop in a 'circuit live' LED and let that bleed down slowly, looks nice too..

fancier would be throw in a SCR and on maybe ? 50% normal bleed off, then crowbar it down to virtually zero..

[dgolka]

It looks good. The calc for the Capacitors is C = (80000 * I) / V. I used a 1K 10W for my bleed resistor and it takes about 30 seconds to bleed down so the LED goes out. I used 4 4700uf caps because thats all I could get so it is well above what I needed. Here are a pic of the board and schematic showing the inclution of a 7805 to get 5V available as well.

[H500]

Make sure the caps are rated 50v or more.

[gar]

051130-1651 EST USA

dgolka:

You should never operate at maximum ratings.

Note the National Semiconductor www.national.com maximum voltage rating is 35 v. The thermal resistance of the TO-220 to ambient is 50 deg C/watt. Maximum junction temperature is 150 deg C. You really do not want to operate at maximum junction temperature. Thus I think 1 watt is reasonable for no heat sink. At 50 deg C ambient ( (9/5 x 50) + 32 ) or 122 deg F and 50 deg C rise the junction would be at 100 deg C.

I question your supplying the 7805 from 36 v. This is above the maximum rating.

Assume a maximum dissipation in the 7805 of 1 w as described above. This can be increased with heat sinking.

If you were to work with the 31 v drop ( 36 - 5 ), then at 1 watt maximum current would = P/V = 1/31 = 0.032 A = 32 ma. But the 36 v input is too high.

You would be much better off with a source voltage of 9 v at maximum load current at nominal line voltage of 120 v. This 9 v would drop to about 7.1 v at 95 v input. Under the 95 v low line voltage condition the drop across the 7805 is 7.1 - 5 = 2.1 v and this is just above the maximum dropout voltage at 1 amp. If you really had to worry about a 95 v low line voltage, then a somewhat high nominal voltage at 120 input might be used. At 135 v input the 9 v rises to about 10.1 volts. Here the difference voltage across the 7805 is 5.1 v and at a 1 watt criteria the maximum load current is = 1/5.1 = 0.20 amps = 200 ma.

If you do not need much load current, then to avoid another power supply you could use a series zener in series with the 7805 input to lower its input voltage. Zener current rating and power dissipation have to be adequate. Or a small zener supplying the base of a power emitter follower. It is easy to get transistors with higher voltage ratings.

It is very useful to learn how to use data sheets and keep in mind that manufactures use specmanship in their general description of a part. Bascially in the real world you can not operate at maximum values and get good reliability.

.

[pminmo]

Gar is correct on the 35V, and unloaded the transformer may put you above 35V. At rated load you will be slightly under with the diode drops. My recommendation is to drop the 7805 if all it is going to do is run the LED. Run the LED at 2ma, a red led at 2ma is almost indistinguisable from one running 10ma. Series resistor then is R=E/I or 35V- 2v(Led drop)/ .002 or around 16.5K 1/4 watt. Closest standard value 15K.

[dgolka]

Thanks gar I do have a clip on heat sink on this and it seems to be ok thus far but I would sooner be safe than sorry any suggestions for the zener. I am a millwright machinest still trying to get my head around the electronics.

[gar]

051201-0801 EST USA

dgolka:

How much current load do you want to put on the 7805?

.

[smarbaga]

36 volts is nearing the maximum input for a 78 series regulator.
with approx 20,000 uf., the filter will be good for 10 amps.
just figure 2000 uf per amp and you will be ok with a linear supply
*** with your previous calculation , subtract the diode drop after converting the ac to dc____ ac X 1.414 - 0.7
but if the circuit voltage was that critical, one would use a precision voltage regulator.

[pminmo]

Originally Posted by smarbaga *** with your previous calculation , subtract the diode drop after converting the ac to dc____ ac X 1.414 - 0.7

Actually since its a bridge, it's two diode drops DC = AC *1.414-1.4. And at 4 or 5A it will be closer to 2V drop.

[dgolka]

gar originally I was looking to powrer a breakout board that consists of a single 74LS541N that would be suppling step direction and enable signal to 3 L297's so I dont know what the requirement for this would be but I believed it would be well under the 1 amp max of the 7805. After reading all the info that has been poted hear I think I need to go back to the drawing board and look at a one of the high voltage chips like Phil has posted on some of the boards on his site @ http://pminmo.com

[smarbaga]

your power supply wiill give you clean dc and a good regulated 5 volts for the ls541.
you can power about 5 or 6 of those chips with the 7805.
just don't think you can suck 3 or 4 amps out of a 2 amp transformer without the voltage dropping.
use an lm338 if you want to regulate and vary the motor voltage.
this way you can use any transformer up to about 25 volts ac