He can use switching voltage regulator to get 80-90% efficiency but its more complex then linear reg.

 

Using the above formula you can get what heatsink size to use.You would need to use at least 10W heatsink on the regulator since it can output 1A.In case bypass transistor is used it would require heatsink too.

Transistor heatsink:
5Vx3A=15W
10Vx3A=30W
15Vx3A=45W

Regulator with bypass transistor:

How do you get that you need at least 10W heatsink since it can output 1A?

First off, heatsinks are NOT rated in watts, they are rated in temperature rise per watt, °C/W.

Second, the heat that the regulator has to sink is only indirectly related to the current it can deliver. The LM7805 has a typical dropout voltage of 2 V, so with a supply voltage of 7 V and delivering 1 A the regulator needs to dissipate 2 W. But the supply voltage can be as high as 35 V and, under those conditions, an output current of just 100 mA would require the regulator to dissipate 3 W.

For a supply voltage of 12 V and an output current of 1 A, the regulator must get rid of 7 W of heat, so using a heat sink that can get rid of 10 W is not unreasonable (and I see where you probably got your 10 W number from), but you don't go and buy a 10 W heat sink because that's not how they are sized.

The LM7805 from Fairchild has a thermal resistance junction to case of 5 °C/W and a thermal resistance to air of 65 °C/W. With a maximum junction temperature of 125 °C, if the air is room temperature then without a heat sink you could, at the limit, dissipate about 1.5 W of heat without a heat sink. If we want to operate at the limit then our case temperature would be 125 °C - (7 W)*(5 °C/W) = 90 °C and at 25 °C ambient air we would need a heat sink with a thermal resistance of (65°C)/(7W) = 9.2 °C/W. But it's seldom a good idea to run at the limits. So let's use a 10 W design target, which means that the junction will be 50 °C hotter than the case. If we want to hold the junction to, say, 100 °C then we need to hold the case to 50 °C. That means that we need to get a heat sink that can dissipate 10 W while having about 25 °C across it, or 2.5 °C/W, when properly mounted on a TO-220 case. If we want to hold the junction to 85 °C (a common design target) then we need to keep the case below 35 °C which means that we will need a thermal resistance in our heat sink of 10 °C / 10W or 1 °C/W.

Now, 2.5 °C/W is a very aggressive heat sink (for a TO-220 case) and have fun finding a 1 °C/W heat sink for a TO-220 case. A quick check of DigiKey shows that their most aggressive TO-220 heat sinks for natural air is 2.6 °C/W, which should work given the margins that we built into the design. If you want to keep the junction down in that 85 °C range (at 10 W) then you would need to go to forced air cooling.

 

There is a better solution that a 5V regulator.

Please enlighten/elaborate?

 

How do you get that you need at least 10W heatsink since it can output 1A?

First off, heatsinks are rated in watts, they are rated in temperature rise per watt, °C/W.

Second, the heat that the regulator has to sink is only indirectly related to the current it can deliver. The LM7805 has a typical dropout voltage of 2 V, so with a supply voltage of 7 V and delivering 1 A the regulator needs to dissipate 2 W. But the supply voltage can be as high as 35 V and, under those conditions, an output current of just 100 mA would require the regulator to dissipate 3 W.

For a supply voltage of 12 V and an output current of 1 A, the regulator must get rid of 7 W of heat, so using a heat sink that can get rid of 10 W is not unreasonable (and I see where you probably got your 10 W number from), but you don't go and buy a 10 W heat sink because that's not how they are sized.

The LM7805 from Fairchild has a thermal resistance junction to case of 5 °C/W and a thermal resistance to air of 65 °C/W. With a maximum junction temperature of 125 °C, if the air is room temperature then without a heat sink you could, at the limit, dissipate about 1.5 W of heat without a heat sink. If we want to operate at the limit then our case temperature would be 125 °C - (7 W)*(5 °C/W) = 90 °C and at 25 °C ambient air we would need a heat sink with a thermal resistance of (65°C)/(7W) = 9.2 °C/W. But it's seldom a good idea to run at the limits. So let's use a 10 W design target, which means that the junction will be 50 °C hotter than the case. If we want to hold the junction to, say, 100 °C then we need to hold the case to 50 °C. That means that we need to get a heat sink that can dissipate 10 W while having about 25 °C across it, or 2.5 °C/W, when properly mounted on a TO-220 case. If we want to hold the junction to 85 °C (a common design target) then we need to keep the case below 35 °C which means that we will need a thermal resistance in our heat sink of 10 °C / 10W or 1 °C/W.

Now, 2.5 °C/W is a very aggressive heat sink (for a TO-220 case) and have fun finding a 1 °C/W heat sink for a TO-220 case. A quick check of DigiKey shows that their most aggressive TO-220 heat sinks for natural air is 2.6 °C/W, which should work given the margins that we built into the design. If you want to keep the junction down in that 85 °C range (at 10 W) then you would need to go to forced air cooling.

now I'm not trying to make my problem yours, I just ask for advice, these are the heatsinks I can buy at my local store, my far away store might have a bigger range, but I don't really have transport so my further away store is actually not a possibility (mantech.co.za) Here's the heatsinks I can buy: http://www.fort777.co.za/index.php?main_page=index&cPath=715_1066

 

studiot said: ↑
There is a better solution that a 5V regulator.Click to expand..

.Please enlighten/elaborate?

I also said rather more in that post #40.

I am very concerned that your responses indicate you don't properly understand the basics.

Now that is not a problem to me but I don't want you to go wasting money, time and effort on false ideas.

Other respondents here have rushed on offering more and more sophisicated solutions without asking the most basic question.

What are the specifications to the 12V supply you wish to divide?
That is where I would start.
Then we could talk about the supply you wish to create, but we need to get some facts about current and power straight first.

Others have guessed wrongly at my alternative solution.
In fact I would probably recommend a variable voltage divider, but not the sort you had in mind.

 

I also said rather more in that post #40.

I am very concerned that your responses indicate you don't properly understand the basics.

Now that is not a problem to me but I don't want you to go wasting money, time and effort on false ideas.

Other respondents here have rushed on offering more and more sophisicated solutions without asking the most basic question.

What are the specifications to the 12V supply you wish to divide?
That is where I would start.
Then we could talk about the supply you wish to create, but we need to get some facts about current and power straight first.

Others have guessed wrongly at my alternative solution.
In fact I would probably recommend a variable voltage divider, but not the sort you had in mind.

It's a 12V 2.5A Switching power supply, I want to do the digital course at www.play-hookey.com

I'm very interested in your alternative solution, I wan't to learn as much as I can.

 

OK, here is a quote from their advice on power supplies:

The requirements for a power supply depend primarily on what you intend to do with it. A power supply for digital IC circuitry must have an output voltage as required by the ICs of the logic family you're using. The most common modern requirement is +5 volts dc, as required by TTL devices and quite suitable for CMOS devices. However, RTL ICs will want +3.6 volts, while ECL ICs are designed to work with a -5.2 volt supply (-5 volts will work, and so will +5 volts if you remain consistent).

On the other hand, a power supply for analog circuits such as radio receivers and audio amplifiers will depend on the design requirements of the circuitry. A pocket transistor radio is specifically designed to operate at 9 volts, so it can be powered by a single 9-volt battery. However, any circuit employing operational amplifiers (op amps) will require both a positive and negative power supply, in the range of ±12 to ±15 volts. Such voltages are also quite satisfactory for experimental analog circuits, and so are practical values to have for a breadboarding system.

Which is pretty much what I was going to tell you next.

I note that they have introduced twin supplies (eg +12_0 _-12) other voltages such as 9, 3.6, -5.2 and so on.

They also say that CMOS logic chips will work on TTL logic voltages (5volts) but that implies they will work on others as well, which is what I meant by saying they will work directly on 12 volts.

I further note their text is not specifically based around TTL, but generic so could equally well be conducted at 12V with CMOS chips (which are cheaper anyway).
Nearly all the TTL chips have a CMOS equivalent.

I see they have a '555' section, I'm sure you will want to experiment with. The 555 works on a wide ranges of supply voltages.

As to the voltage divider one way is to use a pair of emitter followers controlled by a potentiometer that can be set to all the different votlages you would require. The emitter followers would be power transistors or even power darlingtons that could provide all our needs.
I used to build these for model railway enthusiasts who said, could I convert my single speed 12V Hornby supply to make the train go forwards, backwards and if possible vary the speed?

Another alternative, I built a long time a go when I was learning like yourself, was to provide several fixed supply values from a twin 12V supply because you sometimes need more than one for circuits. This used several standard fixed regulators as have been suggested here and was built into an old British Airways digital clock case.

 

I also said rather more in that post #40.

I am very concerned that your responses indicate you don't properly understand the basics.

I've done a bit of work on the basics and admittedly are only on the basics level, is there a guide that I might follow that'll solidify my knowledge about the basics?
R=V/I
P=VI; P=I^2R; P=V^2/R
Parallel Resistance
Rtotal = R1xR2/(R1+R2)
Rtotal = 1/(1/R1 + 1/R2 + 1/R3 ..)
Vout = Vin*R2/(R1+R2)
Series, just add the values.
T=R*C
Parallell Capacitance= C1+C2
Series Capacitance= 1/Ct=1/C1+1/C2+1/C3...
Voltage divider
V1= Vs*R1/Rt
current divider
I1=It*R2/(R1+R2)

Thats the theory I know, applying it, well I haven't gotten that far. I have no money for school, barely money for components, Through the WEB I will learn...

 

That theory is fine, but there's more to it than just formulae, it's also what they tell you.

For instance Ohm's law tells you that for a given circuit or even a single component or length of wire you cannot specify both the voltage and the current.

That circuit will have a particular resistance so you can specify the voltage and the circuit resistance will force a particlar current to suit

or

You can specify a particular current and the circuit resistance will drop a voltage to suit.

Many beginners misunderstand this and ask questions like I have a 12V, 2.5 amp supply. Will it be OK for a 1 amp bulb?

Did you understand what I said about the wall socket?

 

That theory is fine, but there's more to it than just formulae, it's also what they tell you.

For instance Ohm's law tells you that for a given circuit or even a single component or length of wire you cannot specify both the voltage and the current.

That circuit will have a particular resistance so you can specify the voltage and the circuit resistance will force a particlar current to suit

or

You can specify a particular current and the circuit resistance will drop a voltage to suit.

Many beginners misunderstand this and ask questions like I have a 12V, 2.5 amp supply. Will it be OK for a 1 amp bulb?

Did you understand what I said about the wall socket?

What you drive from the wall socket determines the load, given the wall socket's rating is not lower than the load you want to "burn"
That's what I meant with my pipe analogy, The wall sockets pipe can deliver (arguments sake) 10A of water, but the load is the lamp, and it uses 5A water, fits through the pipe, load is load of lamp, if the load being "burned" is higher than the wall sockets available pipe size, the water flowing through the pipe is at its maximum 10A and can't drive a 20A load, the pipe is not thick enough, the PSU can't deliver enough current?

 

What you drive from the wall socket determines the load, given the wall socket's rating is not lower than the load you want to "burn"
That's what I meant with my pipe analogy, The wall sockets pipe can deliver (arguments sake) 10A of water, but the load is the lamp, and it uses 5A water, fits through the pipe, load is load of lamp, if the load being "burned" is higher than the wall sockets available pipe size, the water flowing through the pipe is at its maximum 10A and can't drive a 20A load, the pipe is not thick enough, the PSU can't deliver enough current?

The curent (water) is determined by the device that loads, or makes water flow through the pipe, not the pipe size (10A)

 

What about the lamp determines that it draws 5 Amps?

 

The operation of the lamp (and its specification makes it possible for you to know if you have a thick enough pipe), not the wall socket, that only makes it possible to drive the lamp, if the wall socket can deliver 5A, 4A -- pipes too thin, lamp won't work; 6A, pipe can handle 5A of flow, lamp works

 

I don't know what your mains voltage is but mine is usually 250V

So Ohm's law says that the lamp has a resistance of 250/5 = 50 ohms.

As far as the mains is concerned, if I connected two identical lamps to the mains then they would want to draw 10 amps.

Would the connection be in series or parallel?

If I connect a third lamp they would attempt to draw 15 amps and blow a fuse in a 10 amp supply circuit.

If each lamp had its own 5A fuse and the supply circuit had a 10 amp one, which fuse or fuses would blow?

What I am trying to do is connect those formulae you stated with real equipment and circuits so that you can instantly recognise whether they are series or parallel and tell what will happen.

 

I don't know what your mains voltage is but mine is usually 250V

So Ohm's law says that the lamp has a resistance of 250/5 = 50 ohms.

As far as the mains is concerned, if I connected two identical lamps to the mains then they would want to draw 10 amps.

Would the connection be in series or parallel?

If I connect a third lamp they would attempt to draw 15 amps and blow a fuse in a 10 amp supply circuit.

If each lamp had its own 5A fuse and the supply circuit had a 10 amp one, which fuse or fuses would blow?

What I am trying to do is connect those formulae you stated with real equipment and circuits so that you can instantly recognise whether they are series or parallel and tell what will happen.

I know its parallel, R=V/I=250/10=25 Rt=50*50/2*50=25 I=250/25=10. And it should be the 10amp one, current passes through that first. I'm not sure what the logical order of calculations should have been, maybe I was hacking. But I immediately knew they were parallel, just the theory lacks (as you suspected). Maybe give me step by step with reason why that step in that order... *humble*

 

But I immediately knew they were parallel

That's good and what we are aiming for.

Yes its the 10A fuse that blows, but not because its first in line, but because its the only point where the current exceeds (or tries to) the rating.
The rating would not be exceeded in any one of the parallel branches containing a lamp.

What would happen if the two lamps were connected in series?

Do you have any sort of meter, if so what is it?

 

5A flows through the first bulb, and then the (same) 5A flows through the second lamp. But ohms law will have a voltage drop across lamp 1 and lamp 2 making both lower voltage and less bright. But I'm not exactly sure of this, I think the voltage over the first is higher than the voltage across the second? (making the first lamp brighter than the second) I think V=R/I if I and R is constant would mean second lamp shines less brightly, because the voltage dropped over the first lamp first. If I'm wrong: please explain why both will shine equally less brightly.

 

Each lamp is the same so each lamp has a resistance of 50 ohms.

You are correct that since they are in series the same current flows through each one in turn, but is this current 5A?

Say the current is Xamps

So the voltage across each one is , by Ohm's law, 50X.

We know that they are in series and the voltage is 250 so the voltages add.

So (50X + 50X) = 250

100X = 250

X = 250/100 = 2.5 Amps

So the voltage across each lamp, by Ohm's law, = 2.5 * 50 = 125 volts

So the lamps form a potential divider.

At least as important as the calculation is the drawing.
How are you with circuit diagrams?

Can you draw these simple situations from my descriptions/ (there is no need to show me)?

Also I asked about a meter. This is to make progress on your power supplies as well as other things. Are these power supplies old mains adapters from computer or other equipment?

The meter will be good to test these supplies and any additions you make.

If you have no meter a simple logic probe will be needed to test logic circuits.

 

Each lamp is the same so each lamp has a resistance of 50 ohms.

You are correct that since they are in series the same current flows through each one in turn, but is this current 5A?

Say the current is Xamps

So the voltage across each one is , by Ohm's law, 50X.

We know that they are in series and the voltage is 250 so the voltages add.

So (50X + 50X) = 250

100X = 250

X = 250/100 = 2.5 Amps

So the voltage across each lamp, by Ohm's law, = 2.5 * 50 = 125 volts

So the lamps form a potential divider.

At least as important as the calculation is the drawing.
How are you with circuit diagrams?

Can you draw these simple situations from my descriptions/ (there is no need to show me)?

Also I asked about a meter. This is to make progress on your power supplies as well as other things. Are these power supplies old mains adapters from computer or other equipment?

The meter will be good to test these supplies and any additions you make.

If you have no meter a simple logic probe will be needed to test logic circuits.

OK, I can use TinyCAD to draw a simple single-shot 555 monostable multivibrator, got the multivibrator from the internet, but added the "single shot" capacitor myself, and I also added the driver (2N3904 NPN) for the buzzer myself. That was my first and only project (so far), and other than the RC time constant over pin 6 and 7, I don't really know why a specific pin does what it does. And I've got VeeCAD full version, so I've auto routed the TinyCAD circuit diagram, and now I need to solder. Got myself a hand drill and a decent (Magnum 1220 - I think it's South African brand) soldering iron, regulated, variable temperature but its not a soldering station. THAT was TOO expensive, although I know it's waay better.

When I was a kid 9yrs old (1994), I built a couple of circuits (kits for kids), and I still have my multimeter I got back then, measures Volts, Volts AC, Amperes, Resistance and hFE and Diode but I don't know how the Diode bit works.

I got a Raspberry Pi for my birthday, so immediately it wasn't enough, too easy to do some things on the Pi than on "real" electronics, so I said, well age 9 I tought myself programming upto adept level year or two later, and super adept by now. Now the same with electronics, it's too expensive NOT to know electronics, If I can integrate my programming knowledge with electronics, well the possibilities are just that much higher

 

The meter is plenty good enough to start with.

And you have some basic familiarity with circuit diagrams. TinyCAD huh? they don't need to be pretty but hey, why not.

So you are well on your way and we will soon have you up to speed.

What about the 12V power supplies, is my guess correct, are they old wall warts? (That's not a problem)

And do you have a source of some old power transistors or would you have to buy some?
Scrap equipment is a good source of many useful parts.

Here is the basic idea of a suitable voltage divider add on that will swing the output voltage Vout from nearly +V to nearly -V.

If we can stack two of your 12V supplies in series you can obtain any voltage between about +11 and -11 from it, which should be plenty.
It will require a few extra bits to complete.

So if your power supplies are as I suspect can you measure the resistance between the output jack and the prongs on the mains input lead (whilst they are disconnected).

This resistance should be too high for your meter to measure.