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Foldback current limit circuit

by likephysics
Tags: circuit, current, foldback, limit
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likephysics
#1
Dec22-10, 12:28 PM
P: 615
I need a good explanation of foldback current limit ckt. I understand just current limit using BJT, but I can't find any foldback current limit ckts online. can someone either explain how it works or point me to a website.
I already looked at art of electronics ckt. I need another explanation.

Also, what happens if the load keeps decreasing(but not short ckt) in a regular current limit circuit?
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berkeman
#2
Dec22-10, 12:40 PM
Mentor
berkeman's Avatar
P: 40,639
Quote Quote by likephysics View Post
I need a good explanation of foldback current limit ckt. I understand just current limit using BJT, but I can't find any foldback current limit ckts online. can someone either explain how it works or point me to a website.
I already looked at art of electronics ckt. I need another explanation.

Also, what happens if the load keeps decreasing(but not short ckt) in a regular current limit circuit?
I find using Google Images for searches like this to be a good way to speed up the search for good circuits:

http://www.google.com/images?hl=en&s...=&oq=&gs_rfai=

The basic idea is that once current limit has been reached, the current demand has to be reduced a lot before the current limit behavior of the supply is released. This is necessary in many higher-power linear regulators.

With a regular current limit circuit, you can end up with too much power being dissipated in your source elements, causing overheating.
Averagesupernova
#3
Dec22-10, 03:02 PM
P: 2,497
I tend to think of foldback limiting using the output voltage of the power supply as the reference level of the limit. In other words, the current gets to a trip point, the supply then reduces the output voltage which lowers the reference point which further lowers the output, etc.

likephysics
#4
Dec24-10, 03:18 AM
P: 615
Foldback current limit circuit

How do you decide the bias resistor value for the series pass transistor.

Vsuppy - IbRb - Vbe =0?
Averagesupernova
#5
Dec24-10, 11:06 AM
P: 2,497
Post a schematic and let's start there. You will need to know the minimum beta of the transistor and how much current you want the supply to be able to source.
likephysics
#6
Dec24-10, 11:52 AM
P: 615
SChematic attached.

Series pass Transistor is 2N3055 - Min beta: 20
Ic = 1A (this is also the required current limit)

So Ib = Ic/20 = 50mA
Attached Thumbnails
BJT current limit.jpg  
Averagesupernova
#7
Dec24-10, 12:13 PM
P: 2,497
I don't see a voltage reference. Or is the intended output voltage to be around 20 to 22.5 volts?
likephysics
#8
Dec24-10, 12:32 PM
P: 615
Quote Quote by Averagesupernova View Post
I don't see a voltage reference. Or is the intended output voltage to be around 20 to 22.5 volts?
Yes, 20 to 22.5 will do.
Averagesupernova
#9
Dec24-10, 01:13 PM
P: 2,497
It takes 1/20th (worst case beta being 20) of the emitter current going into the base to make the transistor function the way you want it. So, that means .05 amp. The 100 ohm resistor in your schematic passing .05 amp will lose 5 volts right off the bat. So if your battery voltage starts at 22.5 volts (best case) the most you can expect on the base would be about 17.5 volts. The voltage on the emitter would be at least .7 volts less than that so lets just say about 16.5 volts. You would need to tweek the value of the 100 ohm resistor down to about 30 ohms to be able to get the voltage up close to 20 volts on the emitter. That is not a problem until the circuit goes into current limit mode. At that point you will have a full 22.5 volts across a 30 ohm resistor. Hmmmm, that's over 16 watts dissipated in that resistor and way more current than the little 2N2222 will handle. BTW, this circuit does NOT have foldback current limiting. It is just a pass transistor with no voltage reference with simple current limiting. How about if you tell us more about what you are doing? There could be an easier way of doing it. What type of project is this?
likephysics
#10
Dec24-10, 01:25 PM
P: 615
Dang! When the circuit goes into current limit mode, wouldn't the series pass transistor drop some voltage?
This whole thing started coz I tried to fix a broken Battery charger. I built a off line SMPS, but when I connect the battery, the power supply resets(battery trying to draw too much current). So now, I need to limit the current going into the battery.
What about choosing another transistor instead of the 3055, with a higher hfe?
Adjuster
#11
Dec24-10, 01:56 PM
P: 205
The series transistor does indeed drop voltage in current limit mode, but that's not the problem. The issues are that the transistor can't drop less than a few volts at somewhat lower currents, and that the dissipations in R4 and Q2 can be inconveniently big.

A Darlington type transistor for Q1 should work better, or you could add something like another 2N2222 (Q3, say) in Darlington configuration with a 2N3055 as Q1. You could then use rather more than 100 ohms for R4, and the dissipations would be reduced. Q1 would still require adequate heat sinking to handle at least 11W in case the output was shorted.
Averagesupernova
#12
Dec24-10, 03:12 PM
P: 2,497
Look around for a circuit that is similar to the one you've shown but is true foldback limiting. I can't recall exactly off the top of my head how it is configured. We can go from there.
likephysics
#13
Dec24-10, 11:33 PM
P: 615
I found this ckt which also has fold back. Basically the base current is a function of the sense resistor plus a voltage divider.
Not sure how to design though.

The equation I found is :
Ilim = Vbe[(Ra+Rb)/RbRs] +Vout (Ra/RbRs)

short ckt current = Vbe[(Ra+Rb)/RbRs]

Ra is upper part of pot R4 and Rb is lower part.

I should decide the short ckt current first and then plug in the numbers?
Attached Thumbnails
BJT foldback.jpg  
Adjuster
#14
Dec25-10, 08:48 AM
P: 205
Quote Quote by likephysics View Post
I found this ckt which also has fold back. Basically the base current is a function of the sense resistor plus a voltage divider.
Not sure how to design though.

The equation I found is :
Ilim = Vbe[(Ra+Rb)/RbRs] +Vout (Ra/RbRs)

short ckt current = Vbe[(Ra+Rb)/RbRs]

Ra is upper part of pot R4 and Rb is lower part.

I should decide the short ckt current first and then plug in the numbers?
Where did you get that circuit from? It does not look correct, are you sure that you have copied it correctly?

If R3 represents the output load, then with the pot R4 wound down to ground, Q2 base-emitter would be in parallel with the output, and so would get the full output voltage across it. This would exceed the 2N2222 reverse Vbe rating, which is only about 6V. Q2 base could then pass a large reverse current.

The situation could be more serious if the output were connected to a battery, as really big currents could then flow. Possibly Q2 would eventually go open-circuit e.g. due to a blown wire bond, but I would not wish for your safety to be depending on that.

No matter what circuit you are going to use, I would strongly suggest putting a fuse in series with your battery to avoid any possibility of dangerous currents passing. Batteries may deliver large currents if overloaded or short-circuited. A large current is a fire hazard in itself, and may result in overheating or bursting of the battery, possibly releasing corrosive and / or toxic materials.
likephysics
#15
Dec25-10, 12:00 PM
P: 615
Quote Quote by Adjuster View Post
Where did you get that circuit from? It does not look correct, are you sure that you have copied it correctly?
Here's the link - See page 128

http://users.ece.gatech.edu/rincon-m...s/ldo_book.pdf

Art of electronics has a similar ckt -
http://images.elektroda.net/4_1200125223.jpg
Adjuster
#16
Dec25-10, 01:01 PM
P: 205
Quote Quote by likephysics View Post
Here's the link - See page 128

http://users.ece.gatech.edu/rincon-m...s/ldo_book.pdf

Art of electronics has a similar ckt -
http://images.elektroda.net/4_1200125223.jpg
Yes, but those circuits used fixed resistors for the potential divider where you have the pot R4. It's OK to use a pot for adjustment, but you must never let the resistance Rb to ground become too small. That could easily happen if the pot were inadvertently turned too far - then there could be trouble!

I would recommend adding a fixed resistor in series with the ground end of the pot R4. If this resistor were made perhaps four times the value of the R4 pot it should be safe, and you should find it easier to set up the ratios you would need.
likephysics
#17
Dec25-10, 02:22 PM
P: 615
The pot divider was only for illustrative purpose and maybe a bit of tweaking.
I don't understand how the Ra/Rb divider helps and I also don't know how to find Ra, Rb.
Should I start by fixing the short ckt current? Maybe 1.5A.
Averagesupernova
#18
Dec25-10, 04:23 PM
P: 2,497
The last posted circuit is still simple limiting and not foldback.


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