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As a starting point I will go with 200 ohms 0.5W and a 1% tolerance resistor. As the maximum base voltage is 5 Volts for the MJ11015 transistor, 1 Volt to turn it on seems very low. And 5mA current for the l7812 also seems very low.

As I don't understand transistor physics very well.... At one end of the extreme I've been told 25 ohms 0.5W will drop 2.5 Volts across R7, and a current of 100mA. And at the other higher end of the spectrum I'm told 200 ohms 0.5W will drop 1 Volt across R7 and a current of 5mA.

The less work for the regulator l7812 sounds good, if 1 Volt is sufficient to turn the transistor on. But as there where doubts that 25 ohms 0.5W would turn on the MJ11015 transistor, I'm at a loss in understanding how lower input values (Voltage and current) would drive it.
 
Well there's got to be a starting point somewhere, and a do remember 5mA on the input of the l7812 with the TIP2955 transistors. I think I get confused on the input current and output current of the l7812 I'm sure that's where I'm going wrong. Anyway I see at 200 ohms 0.5W nothings going to go pop bang, so from there I can only come down in resistance if its needed, hopefully not. Its a taste it and see thing I think. As said before, and in case anyone else is contemplating this circuit, I will post some practical results after building it. Thanks again for your help, that's for everyone !
 
I know this thread seems a bit dragged on... But I will do the practical stuff in the next few days.

I've looked at the quiescent current for the lm7812 and the 78S12cv the 2 Amp regulator, and there quiescent current is 5mA@12 V and [email protected] V so the current through a 200R resistor is unlikely to turn on the transistor forward bias at such low levels ? Is the current borderline or have I got it wrong there ? I will do the practical in a day or so, although it would be good to get the value of R7 right first time.
 
I've looked at the quiescent current for the lm7812 and the 78S12cv the 2 Amp regulator, and there quiescent current is 5mA@12 V and [email protected] V so the current through a 200R resistor is unlikely to turn on the transistor forward bias at such low levels ?

True, but don't care. The current boosting external transistors do not have to be active in the circuit all the time, just when *you* want to take some of the thermal and current load off of the 7812. When that happens is a design tradeoff, not some hard and fast design rule. That is why I recommended setting R7 so the external transistors came on at a higher current level. I think pushing the circuit such that the boost transistors come on at very low currents is a mistake.

ak
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
yeah, but it will only have 2.5 volts across it at pretty huge currents.

As an example if you have 2 Darlington transistors with 0.1 ohm emitter resistors then they will each be dropping about 1.2 volts, equating it to 12A through each of the transistors.
 
As I don't understand transistor physics very well.... At one end of the extreme I've been told 25 ohms 0.5W will drop 2.5 Volts across R7, and a current of 100mA. And at the other higher end of the spectrum I'm told 200 ohms 0.5W will drop 1 Volt across R7 and a current of 5mA.

True and true. It's resistor physics:

E = I x R
1 V = 0.005 x 200
2.5 V = 0.1 x 25

Again, it is your call.

*Decide* the 7812 current level at which you want the boost transistors to start to conduct. Note that this is not an instant thing; the transistors will assume more and more current as the voltage across R7 increases because the conduction "knee" of the base-emitter junction is not very sharp and it is softened even more by the ballast resistors.

*Determine* the voltage that starts conduction from the boost transistor datasheet.

*Calculate* R7.

*Calculate* the current through R7 at full circuit load, and hence the peak power dissipation in R7.

ak
 
Thanks for coming back guys ! I see 2.5 Volts @ 100 mA, and 1 Volt @ 5mA. Which will work ! I'm not sure yet, I understand ohms law, and some basic formulas. But when I sit there trying to diggest transistor formula and physics its just not happening, not from text anyway

I can't see where Steve got 12 Amps from ? I think both senarios are the input current the regulator dissipates... Think that is ! What are the pros and cons of 2.5 Volts @ 0.100mA over 1 Volt @ 0.005mA ? One means the regulator turns the transistors on later, whilst the regulator does the early work. The other means the transistors are turned on early, and the regulator works far less hard.

Well I know heat equals loses, so its has to be the regulator doing less of the work, and saving it performance for better voltage control, whilst running cool. Sorry if I sound like I don't understand transistor physics very well yet, but I'm trying, believe me I am. I've been back over all 6 pages of this thread a few times now, more than a few lol.

I've tried a base transistor calculator, that seemed a flop. And still reading, re reading over and over the formulas to determine values of base transistor resistors, ebv , ecv, etc. Its not getting easier. Voltages across, currents through every junction. And still I've not got a grip of it...

Just to mention, there's a guy on another forum, he's names Matt, cutting the story short, he can only draw schematics in his own point to point style so he understands it. Nothing wrong with that in my book, the guy has made some amazing stuff I've followed.

This feels similar in understanding transistor formula and associated physics, if you get what I mean. From your posts above, from my limited knowledge, I would think overall performance would be better with the regulator working lite duty. Simply as less heat is created, and less degration of the component regulator.

As to whether 200 ohms @ 1 Volt 5mA will forward bias the transistor, I'm not sure ! I do know if it doesn't, when I load the circuit with 4 Amps or so, the l7812 will throttle back the voltage to protect itself.
Hope some of my jargon makes sense.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
I get 12A because there is an emitter resistor and a base-emitter junction (or multiples thereof) in parallel with R7.

If there is 2.5V across R7, then there is 2.5V across the emitter resistor and the base-emitter junction.

Since with a Darlington the base-emitter voltage will be in the order of 1.3 volts, this leaves 1.2 volts across the emitter resistor. If the emitter resistor is 0.1 ohms, then this means the current through it is 12 amps.

Because the has majority of this current is going through the collector we can assume the collector current is also 12A.
 
I get 12A because there is an emitter resistor and a base-emitter junction (or multiples thereof) in parallel with R7.

If there is 2.5V across R7, then there is 2.5V across the emitter resistor and the base-emitter junction.

Since with a Darlington the base-emitter voltage will be in the order of 1.3 volts, this leaves 1.2 volts across the emitter resistor. If the emitter resistor is 0.1 ohms, then this means the current through it is 12 amps..

In post #26 I spelled out my reasoning. Based on the datasheet for this and other similar power darlingtons, my estimate of the combined Vbe at 5 A Ic is 2.0 V. 5 A per boost transistor was an early design goal. So 2.0 V across Vbe and 0.5 V across the ballast resistor.

ak
 
1. Do not eliminate R7. R7 sets the load current level at which the external transistors start to assist the 7812.

2. From the point of view of this circuit, the11015 and 11033 transistors are identical. Both have enormous gain at 5 A collector current, and approximately the same Vbe. Unfortunately, both are equally poorly documented, so your first pass at this circuit should work but the point at which the external transistor start to conduct might be off.

3. Please re-read post #26. The value of R7 depends on the value of the ballast resistors, the number of external transistors in parallel, the transistor Vbe (which is not known exactly), and your choice of the maximum current through the 7812.

ak
Why would the point at which the transistors conduct be off AK ? This bit I don't understand. If these values are known then why can't the transistor turn on be worked out.

And thanks Steve I can see where the 12 Amps is now. And AK I have been over post #26 quit a few times now. Its hard to get my head round the huge differential between 25 ohms and 200 ohms and how the circuit is affected by both in terms of power values (Voltage and current) and the doubts I've seen in other posts, as to whether its possible to bias these transistors at all.

Don't forget there have been two separate posts from two different members saying you don't need R7 at all. Although I've worked out that's a bad idea now. But early on in the thread, for a while I did think "really" no R7. For a novice its very confusing, then some one said the regulators queicent would bias the transistors ! "Really" I thought queicent current was the regulators no load resting state. So you see that's how it gets confusing.

Then Steve says increase the resistance to 200 ohms, although I know Steve knows he's stuff, again another member said 1.4 volts in his senarios would be unlikely to bias the transistors, let alone 1 Volt. Although I can clearly see the maximum base voltage for the MJ11015 transistor is 5 Volts, so I think it would, think that is at this stage. Its only the controversy sourounding this Darlington configuration that seems to make the choice of R7 difficult.

So this is why I've asked lots of questions, failed to grasp completely the maths in its entirety so far. Hopefully this will change. I can take voltage and current measurements in circuits easily, yet I can't pin down a value for R7 on paper for the MJ11015 transistor. Hope some of that makes sense from a novices perspective. Thank you for helping me.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
There are two interpretations to "you don't need R7".

One is that you replace it with a short circuit. In that case there is nothing to turn the transistors on.

The other interpretation is to replace it with an open circuit. In this case the regulator's quiescent current will turn on the transistors to some extent.
 
Surely short circuit is no R7 in series with the l7812 input, if that constitutes a short, although I can't see how its a short, so if its voltage straight in that's a closed circuit, and open circuit is a break in continuity so no voltage flows across the R7 path. Only one can, or should say might turn the transistors on, as the other is open circuit. Open circuits don't conduct anything, do they.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Are you agreeing with me?

A short circuit and a closed circuit mean the same thing in this case. It simply means CURRENT can flow unimpeded (and thus there will be no voltage drop) across those points.
 
I'm still to determine why AK might think the turn on for the transistors "might be off".... Although I am going to start with a 200R 0.5W R7 resistor. From that measure current and voltage as load goes from 0 - 4 Amps, hopefully the transistors will start up early and carry the current. I'm just waiting on a TO3 heatsink to turn up, as previous builds using the TIP2955 transistors I've used CPU heatsinks, and TO3 packages are just not compatable with that type of heatsink.
 
Are you agreeing with me?

A short circuit and a closed circuit mean the same thing in this case. It simply means CURRENT can flow unimpeded (and thus there will be no voltage drop) across those points.
No Steve absolutely not, my understanding of a short circuit is a fault situation ! I see you mean it as no R7 but rather a clear path through to the transistors base and regulator input.

PS Wasn't sure if you meant arguing or agreeing, but short meaning no R7 so if agreeing, yes I am. And definatly not arguing, appolagies if I read your post wrong.
 
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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
A short circuit is a fault if a short circuit is not what is required.

If you are intending to replace a resistor with a piece of wire it is clearly intentional. :)

Basically, removing a resistor can mean either removing the resistance (making it a short) or removing the path of conductance (making it open). Neither are desirable in this circuit.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
I'm still to determine why AK might think the turn on for the transistors "might be off"....

He is saying that the rough calculations we've done as to when the transistors will turn on may not be exactly as calculated. ("not exactly as calculated" == "might be off").

I think we're having a language barrier problem. I would still recommend R7 being 100 ohms, but 200 ohms should be OK. AK believes that the value should be lower so as to give the best transient response, but I'm not totally convinced it will make a huge difference. My point all along was simply to suggest that you change from one type of transistor to the other and change nothing else.
 
He is saying that the rough calculations we've done as to when the transistors will turn on may not be exactly as calculated. ("not exactly as calculated" == "might be off").

I think we're having a language barrier problem. I would still recommend R7 being 100 ohms, but 200 ohms should be OK. AK believes that the value should be lower so as to give the best transient response, but I'm not totally convinced it will make a huge difference. My point all along was simply to suggest that you change from one type of transistor to the other and change nothing else.
Yes I got what you mean Steve with short circuit terminology now, my misunderstanding. And thanks for explaining what AK was meaning back in his previous posts. I think from what's been said 200 ohms 0.5W is a good starting point, well good for less losses in heat and wasted energy, if it fails to turn on the transistors I have 180R 0.5W and 100R 0.5W resistors. But would have to get some 25R 0.5W resistors if needed, but looking at the currents above with a value that low, its no looking like the best option. Thank you again for helping me.
 
Why would the point at which the transistors conduct be off AK ? This bit I don't understand. If these values are known then why can't the transistor turn on be worked out.

Because the point is NOT a point. As the voltage across the base-emitter junctions increases, the transistors turn on over a range of values anywhere from 0.3 V for a single small transistor to 1.5 V for a giant power darlington. AND, that value increases as the collector current increases. The standard approximation that Vbe = 0.6 V (0r 0.7 or 0.65 or whatever) for a single transistor is wrong 99% of the time. That is because it is an approximation describing the average behavior of thousands of part numbers, not any one part. Your circuit will use one part, it is a darlington, not a single small-signal part, and a power device to boot. These details matter way more than any rule of thumb. Many people spout these "rules" as if they are actual rules. They are not, and knowing that separates the professionals from everyone else.

Its hard to get my head round the huge differential between 25 ohms and 200 ohms and how the circuit is affected by both in terms of power values (Voltage and current) and the doubts I've seen in other posts, as to whether its possible to bias these transistors at all.

Of course it's possible. Steve and I present two different possibilities. Neither is right or wrong, something you clearly are looking for. Give it up; there is no one right answer. If the 7812 is in a place where it can't touch a heatsink, then adjust its current lower. If you want it to control the output more directly by increasing the ratio of its static (quiescent) current to its output current, adjust it higher. I keep pointing out that these are decisions to be made by the designer, you, based on how you want the circuit to perform.

Don't forget there have been two separate posts from two different members saying you don't need R7 at all. Although I've worked out that's a bad idea now. But early on in the thread, for a while I did think "really" no R7.

Eliminating R7 is just plain stupid. The 78xx parts have been made by dozens of manufacturers around the world for decades, and every datasheet that has this circuit has R7 in there for the same reason, so I'm glad to see you stick with that body of knowledge.

then some one said the regulators queicent would bias the transistors !

Technically, it can. But all possibilities are not equally smart. And, inside each power darlington transistor there already are resistors across the base-emitter junctions. If the 7812 were actually a 78L12, with a max output current of only 100 mA and very poor thermal properties, then we would be discussing using those built-in resistors as part of the biasing circuit, another uncontrolled variable that wiould affect the overall circuit performance. But fortunately we don't have to.

"Really" I thought queicent current was the regulators no load resting state.

Quiescent current *is* the regulator's resting state, the static or no-load current it needs to function. Internally the regulator has a voltage reference, an error amplifier, a power amplifier, a current sense circuit and a temperature sense circuit. All of that takes a small amount of current to function even when the output is disconnected from any load, and that current flows from the input pin to the ground pin. It also varies from part to part, and with temperature, and with the load current, which is why I don't think it should have a major role in determining when the boost transistors start to kick in. Working from memory, the 78xx series static current is around 8-10 mA. Choosing R7 so the boosters start up at 100 mA means that their startup is determined 90% by the load current and 10% by the static current. Again, a design choice based on my opinion, based on my experiences.

I can take voltage and current measurements in circuits easily, yet I can't pin down a value for R7 on paper for the MJ11015 transistor.

Then stop trying. There is no one right value. There wouldn't be even if the boost transistors weren't darlingtons (which does make things less precise). You have a range of acceptable values. Pick one, take real voltage and current measurements, let us know the results, and we can discuss any adjustments. It is T.H.E O.N.L.Y W.A.Y to determine how the circuit performs in the real world.

You don't know me, or Steve, or Colin, etc., which makes this risky. Rather than just spout unsupported assumptions, Steve and I have presented two positions that are the boundaries for R7, complete with the supporting math. Now, as it has been from the beginning, it's up to you.

ak
 
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