Power supply for my TTL clock

Thread Starter

jgessling

Joined Jul 31, 2009
82
I've gotten my TTL clock all working on my solderless breadboard and I'm thrilled about that. All built from flea market or scrounged parts. It draws a little more that 500ma according to my bench supply at 5v. I got a 9v dc 1000ma wall wart to use as a power supply (It actually reads 12v without load) hooked to a 7805 regulator with caps to make the 5v. Everything works fine, but the 7805 is pretty hot. I have a heatsink on it so I guess it will run but I am concerned about the heat especially when I put it into an enclosure. If I got a lower voltage wall wart, say 7v, would the 7805 run cooler? Any other suggestions. Thanks.
 

Wendy

Joined Mar 24, 2008
23,408
The 7805 may not be able to handle voltage that low. Just a suggestion, but look for an old cell phone power supply for a car. Most cell phones use 5V, and the current is about right too. Since they are switchers they will pull about half the current on the 12V side that they give on the 5V side.
 

Thread Starter

jgessling

Joined Jul 31, 2009
82
Are you saying to run the 12v from the wall wart into the car cellphone adapter and then into my clock? Sounds interesting, those cell phone things are cheap and available. Should I protect my 5v TTL with a 5.1v zener or something in that case?
 

Wendy

Joined Mar 24, 2008
23,408
About the protection, maybe. Be aware than zeners can vary quite a bit (±5%), so it is possible you could get a zener that conducts at 4.85 Volts, and have it screw everything up. It may not be worth it.

But as for the rest of it, yeah. You will likely get two different current measurements, input and output.
 

mik3

Joined Feb 4, 2008
4,843
If you use a 7V wallwart the 7805 will dissipate less heat and thus will run cooler. However, you are on the limit of its drop out voltage which is 2V (it requires at least 2V input voltage above the output voltage to opetate normally). If the wallwart stays always above 7V it will be fine otherwise you need to buy a low dropout voltage regulator. Another solution is to use your 12V wallwart with a bigger heat sink and a vented enclosure.
 

AllVol

Joined Nov 22, 2005
55
500 mA sure seem like a lot of current for a TTL clock. That may be once reason your 7805 is running hot.

I do a lot of breadboard projects, using little psu's I've made up with 7805's or 317's. I drilled holes in little scraps of aluminum for heat sinks, but they really don't need them. They never get more than just slightly warm.
 

Thread Starter

jgessling

Joined Jul 31, 2009
82
I wonder if that is a lot (500ma). Here's the chip count

4 - 7447 seven segment drivers
1 - 74390 dual decade counter
3 - 7490 decade counter
1 - 7432 quad OR
1- 7414 schmit trigger NOR gates
1 - 7492 counter
1 - 7408 quad AND
1 - 7442 decimal encoder (for the bargraph)

4 seven segment LED displays
1 10 bar led bargraph thingy
1 plain LED
crystal osc from battery powered clock

Taking the 74390 for example it has 42ma supply current and 7490 at 29ma (typical) then 13 chips * (say) 30ma is 490 ma. Throw in the display which I don't really have data on and I guess we're in the ballpark. I just kept building it and watching the current gauge on my bench supply.

Some of these choices my seem a bit odd, but I wanted to use what I had and what I had was gathered pretty randomly.
 

Thread Starter

jgessling

Joined Jul 31, 2009
82
I do have multiple 7805's and similar available. Is there a way I can use more than one to power my clock? Like one do half the chips with one regulator and another regulator handle the other half? Or would the lack of a common point just cause trouble?
 

BillB3857

Joined Feb 28, 2009
2,570
Does your clock strobe the display one digit at a time? If not, that will eat a lot of power. Your thought of using multiple 7805 regulators has been done before. One for the logic and the other to drive the displays.
 

Thread Starter

jgessling

Joined Jul 31, 2009
82
I think you are asking if the digits are multiplexed. They are not, whatever segments that are on are on until the next tick when they might change. I am reserving that technique until later, this is my first real electronics project and I wanted to keep the logic simple. Good suggestion though, maybe my next project needs to come along sooner.
 

Thread Starter

jgessling

Joined Jul 31, 2009
82
From doing some more research I realize that the 7805 has got to produce heat when it reduces voltage. One site said that the difference between the input and output voltage times the output current is the number of watts to be dissipated. (is that right?) So the simplest thing to do is to lower the input voltage, keeping in mind the headroom required by the regulator. Of course in my junk box was another wall wart, this one labelled 9v 600ma. Measured it's unloaded voltage and is was about 10v, the other one was 12v. So I plugged in the new one and the loaded input voltage went down from 10.5v to 8.5v. Bingo, the 7805 is running a lot cooler, comfortably warm instead of hot and the clock is ticking away.

First wall wart - 10.5 -5 * 0.5a = 2.75 watts
second one - 8.5 - 5 *0.5 = 1.75 watts

I guess it makes a difference, glad to learn something from all this.
 
Last edited:

AllVol

Joined Nov 22, 2005
55
Sharing the load with two or more regulators sounds doable. Also, multiplexing or strobing the displays could save a lot of current.

Good luck.
 

Wendy

Joined Mar 24, 2008
23,408
I've been working on a 12 to 5 V switching regulator for a while, it isn't even close to prime time. It was a cell phone adapter that got me to thinking.
 
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