the best power regulator

[John Gutmann]

what is the best power regulator for 5 volts to use with TTL? I know there are some good ones in the TO-220 package but idk what ones. any ideas? links?

[Matt Krass]

Regulating down from what? You might want to check out Maxim ICs (www.maxim-ic.com) level shifters, nice stuff and they'll usually give you free samples.

[Mike Betts]

Originally Posted by sparksandtabs

what is the best power regulator for 5 volts to use with TTL? I know there are some good ones in the TO-220 package but idk what ones. any ideas? links?

Without more information as to exactly what your specifications are, the 7805 is the industry standard for TTL applications.

The 7805 is made by many manufacturers. Here is National's implementation.

Mike

[Hutch]

Ugh, sorry about the Puzzler post... Definitely left logged in and I didn't notice.

Anyhow, I'd say the best solution would be the regulator but if you don't have one, another (possibly cheaper) solution could be made with appropriately rated zener diodes.

EDIT: Found some decent diagrams and an explanation of how to do things the zener diode way, as well as how to do it with regulators (just keep hitting next slide):

http://www.mines.edu/Academic/course...t11/sld005.htm

[John Gutmann]

down from 12 volts

[John Gutmann]

so if i used a 12 volt in put and get +5volts as an output then what do i use as ground?

[BrianBSL]

Originally Posted by sparksandtabs

so if i used a 12 volt in put and get +5volts as an output then what do i use as ground?

On a 7805, and most regulators, the grounds are tied togeather. on a TO-220 package, the metal tab is usually internally connected to the ground as well (so make sure you don't short anything out on it).

[Hutch]

Yep voltage is all relative, so it's output 5 volts relative to the wire the chip's ground is attached to.

[John Gutmann]

so i use the output ground as the input ground, and i have seen schematics of 7805's that hav non-polarized capacitors between the leads and i was wondering if those are neccessary, if they are what size should I use? and if not should i use somthing else

[Hutch]

They aren't completely necessary, but they will prevent current spikes. When logic gates change states, they draw a lot of current (relatively), and the capacitors across the leads will keep the the current drawn from spiking as much.

[ConKbot of Doom]

Yes, the caps are necessary. The stabilize the regulator and keep it from oscillating. Larger caps however, will help out with current spikes. Larger caps in addition to the small ones, not instead of.

[Hutch]

Hmmm I didn't recall them being necessary. But I at least am 99% sure that you can have the large ones instead of small ones since, after all, parallel capacitors are like one bigger capacitor (hooray for narrow, worthless AP physics knowledge ).

[John Gutmann]

so for the ground i use, do i use the same ground as what i use for the transister or the one on the part on the back?

what size caps? should it be greater than a certain size or will the supply just be more stable if i use bigger ones? what is the max size i should use? any recommendations on sizes?

and are there any other transistors that can be used the same way, i think maybe the 7804, the 7803 and 7806, i am not sure so someone please correct me

also should i use a small heat sink with it? it is required or should i used one just for insurance?


thx
-john-

[Joe Ross]

The exact size of the capacitors is absolutely necessary for ensuring stability (unless you really know what you are doing). Unfortunately, the real world doesn't always work like AP Physics. The datasheet for your regulator will tell you what size capacitor and any other requirements like type (tantalum, ceramic, etc), locations, etc *.

The datasheet for the 7805/LM340 is available here: http://www.national.com/ds.cgi/LM/LM340.pdf. If the part you get is not manufactured by national, you should get the datasheet from that manufacturer. There may be a few subtle differences (not likely for the 7805, but you'll probably run across a difference in some other chip before too long).

As far as heatsinks, it depends on how much current you plan to draw and the package of the regulator. Looking at the national datasheet, at 25 degrees C, the TO-220 can dissipate about 2 watts, although for safety, lets use 1 watt. The equation for power dissipation can be approximated by Pdis~=(Vin-Vout)*Iin. Since Vin is 12v and vout is 5v, you can solve for Iin of 1/7 of an amp. Since that's Iin, your output current (Iout) could actually be 350ma. I have a feeling that's more then enough for your application.

All of the grounds should be hooked together, the ground from your source, the ground from your regulator (middle pin of the TO-220 as well as the case), and the ground of whatever the load or loads are (IE your TTL chips and anything else).

Note that I've never designed a regulator circuit (other then a bridge rectifier), so you'll be a step ahead of me. I have worked with a bunch, though.

* I've seen a person with an EE degree and several years industry experience that designed a regulator circuit that didn't quite work because he didn't follow the requirements listed in the datasheet.

[Manoel]

Originally Posted by Joe Ross

As far as heatsinks, it depends on how much current you plan to draw and the package of the regulator. Looking at the national datasheet, at 25 degrees C, the TO-220 can dissipate about 2 watts, although for safety, lets use 1 watt. The equation for power dissipation can be approximated by Pdis~=(Vin-Vout)*Iin. Since Vin is 12v and vout is 5v, you can solve for Iin of 1/7 of an amp. Since that's Iin, your output current (Iout) could actually be 350ma. I have a feeling that's more then enough for your application.

Joe,

Voltage regulators are not transformers, unfortunately. The regulation is obtained by means of power dissipation, as you pointed out. I agree I_{in} (is this LaTeX? ) is 1/7 of an amp, but Iout must be essentially the same, minus some small leakage to ground. To make 1/7 of an amp at the input become 350 mA at the output you considered that the input power is equal to the output power, as in an ideal transformer (Vin*Iin = Vout*Iout), but that contradicts your first statement about power dissipation and thus, Iout is not 350 mA.
1/7 of an amp is approximately 142 mA, which is still a good number for sparksandtabs's use (digital logic circuits).