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current limiting

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ultrasoundguy2

ultrasoundguy2

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I am using a diode to bias a BJT in order to regulate voltage to a load. See attached image (top)
This top circuit in the attached figure works well and this concept of diode regulation cant be changed.

The problem is that if the load fails as an open circuit, all the current is forced through the BJT
The BJT can get very hot, even with a heatsink as there is 7 Volt drop over it.
To be honest, Im not sure why there has to be a drop here but there is.

Id like to come up with a method to bypass the entire BJT if the load open circuits.
An example concept is in the attached image, bottom, but this wont work.
The idea is that the induced voltage drop over a 1V shunt resistor can switch on a JFET.
Of course, it wont work as once the BJT is bypassed, there is no current on this resistor.
It would only work if the FET latched.
Whatever circuit elements are intruded need to be regulated internally, mean, there is only 6.8volts until it bypassed.

thanks
 
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The reason why you get around 7 V drop across the transistor is that base current doesn't flow until you reach zener voltage plus base-emitter voltage.

I am not so sure that this is an optimal solution to what you want to do. On the other hand, I am not so sure what you want to do.

But if it helps to understand why you have the voltage drop - there you are.

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
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If you don't care about wasting power and eventually lowering the reliability of the BJT, as well as lowering the reliability of the power supply, then the original circuit does the job, more or less. The concept of the original circuit is to bleed current from the power supply until the power supply voltage just barely turns on the BJT, but that could still be dumping a lot of current, and hence, power. This is why the BJT gets hot, and why it has to get hot. because it's a very brute force circuit and a bit of a kludge. Rather than patching a kludge, why not just get a switching power supply that doesn't have these problems, or even just a regular 3-terminal analog regulator?

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com:
 
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To be honest, Im not sure why there has to be a drop here but there is.

It is doing exactly what it is supposed to do. There are better options beside a shunt regulator. Even with a shunt regulator, some of the power is dissipated in a resistor in series with the transistor so you don't just heat it till it fails. Explain what your power source is and exactly what the load is. You need to step back and rethink this.
 
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You would have no voltage applied to the load if you didn't have 7V across the transistor.

To word it another way, the load must require 7V hence that regulator is set to maintain 7V applied to the load. The regulator doesn't care if the load is connected or not so it just maintains that 7V regardless.

What you are asking would require the bypass circuit to latch which would then require turning off the power source to reset. Then, you'll have the issue that the voltage isn't regulated if the load is just reconnected without first turning off power.

Honestly, this smells like a school assignment - "Improve this voltage regulator circuit"
 
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I appreciate everyones efforts, no one was able to come up with a solution though.
Like I said previously, the power source needs to be a shunt regulator, I know there are other sources.

I really don't care about the reason why the BJT get hot, I knew the basics, it side tracked the topic.

Sometimes there just are no solutions for issues.
Sometimes I do get great unexpected advice from the forums and I keep trying.

 
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OK, but some problems simply have silly answers.

You need a circuit that senses current in the load line, and switches in a dummy load. When the real load starts to draw current, the dummy load is shut off. So, you've basically tripled the complexity because your instructor gave you a silly constraint.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com:
 
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usg: "no one was able to come up with a solution though"

Now, that's a gross underestimation of our skills. And probably an overstimate of yours.

There are dozens of ways to current limit. But you haven't defined what you really are trying to do. Please, specs first. Without that Rube Goldberg attachment. Then we will flood the site with simple and reliable solutions.

I think that you should start with an honest "I'm sorry". Then, we will see...

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
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sorry I offended anyone.
I guess I was alittle offended by someones 'homework assignment' comment.
my comment should havfe been better stated.

what I post is intentionally vague, it would make sense if you had more details.
we will have to leave this as something that cant be answered.

ive gotten a lot of great help from irstuff and you over the last 6 or maybe even 10 years or so under many different usernames
 
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"To be honest, Im not sure why there has to be a drop here but there is." - Exactly. Why not put a resistor in series with the collector to help dissipate power? I.E. move the voltage drop.

You never stated what your variation is on VCC. if it too high then series resistance between VCC and the top circuit would also help.

Z
 
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It's a crowbar, though; adding a resistor lessens its efficacy. A shunt regulator operates through the brute force of playing the compliance of the transistor against the compliance of the power supply.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com:
 
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"To be honest, Im not sure why there has to be a drop here but there is."

It sure didn't sound like you understood the basics. You also don't seem to understand that shorting the transistor just means the heat is dissipated in the upstream series resistance of the source instead of the transistor. You have to dissipate the heat caused by the voltage drop somewhere. This is why, generally speaking, a shunt regulator is a really crappy way to regulate voltage.

Take a look at a SCR equivalent circuit. Good luck.
 
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You will be chasing your tail here a bit. As already suggested, the shunt regulator is doing just what it is supposed to do. But now you want it to not do its job if it's doing it's job too well.

Perhaps instead of some sort of latch circuit that tries to disable it on some tricky condition, you could just limit its range? Assumably you're happy for the regulator to take some current while it's doing its job, so why not just limit that current?

Stick a current limiter like this thing in series with the BJT. Set Rsense so 0.6/Rsense = Ilimit. Keep in mind the extra 0.6V drop across the limiter circuit. Then if the regulator hits the current limit, Vcc will just rise according to the supply, but it sounds like you're okay with that.
 
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Hi Ultrasoundguy2...

I am not sure if you're still reading this post but here goes:

It looks to me as if there is something you can't tell us (confidential perhaps). On that basis and the fact that you have a desire to keep a circuit in similar vein to your posting - would it be sensible to wire a 12V bulb (not LED) in series with the transistor collector?

That way the transistor will not be burdened with the bulk of the power. Most of the time the bulb should not even glow but in the event of open circuit at the load, the power will be directed at the bulb. Part of the bulb selection might need to be on filament strength. You don't have to stick with a bulb - you may be able to just use resistive wire (cheaper/easier to dissipate power in some applications).

Does that help?
 
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