# Foldback current limit circuit

1. Dec 22, 2010

### likephysics

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?

2. Dec 22, 2010

### Staff: Mentor

I find using Google Images for searches like this to be a good way to speed up the search for good circuits:

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.

3. Dec 22, 2010

### Averagesupernova

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.

4. Dec 24, 2010

### likephysics

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

Vsuppy - IbRb - Vbe =0?

5. Dec 24, 2010

### Averagesupernova

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.

6. Dec 24, 2010

### likephysics

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 Files:

• ###### BJT current limit.jpg
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7. Dec 24, 2010

### Averagesupernova

I don't see a voltage reference. Or is the intended output voltage to be around 20 to 22.5 volts?

8. Dec 24, 2010

### likephysics

Yes, 20 to 22.5 will do.

9. Dec 24, 2010

### Averagesupernova

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?

10. Dec 24, 2010

### likephysics

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?

11. Dec 24, 2010

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.

Last edited: Dec 24, 2010
12. Dec 24, 2010

### Averagesupernova

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.

13. Dec 24, 2010

### likephysics

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 Files:

• ###### BJT foldback.jpg
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14. Dec 25, 2010

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.

15. Dec 25, 2010

### likephysics

16. Dec 25, 2010

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.

17. Dec 25, 2010

### likephysics

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.

18. Dec 25, 2010

### Averagesupernova

The last posted circuit is still simple limiting and not foldback.

19. Dec 25, 2010

You should aim for a smaller short-circuit current (Isc) than 1A, since the maximum current (Ilim) will be larger.

One way to begin the problem would be to decide that you want Ilim to be a certain number N times Isc.

You gave two equations: Ilim = Vbe[(Ra+Rb)/RbRs] +Vout (Ra/RbRs), Isc = Vbe[(Ra+Rb)/RbRs]

Putting Ilim = NIsc, we get Vbe[(Ra+Rb)/RbRs] +Vout (Ra/RbRs) = NVbe[(Ra+Rb)/RbRs]

Vout (Ra/RbRs) = (N-1)Vbe[(Ra+Rb)/RbRs], so Vout/(N-1)Vbe = [(Ra+Rb)/RbRs]/(Ra/RbRs) = (Ra+Rb)/Ra, or Ra = (Ra+Rb)(N-1)Vbe/Vout

Example, N=3, Vout=20V, Vbe=0.6V, using a 1000ohm pot with a 4700 ohm fixed resistor to ground for safety. This would give (Ra+Rb) = 5700ohms

Ra = 5700*(3-1)*0.6/20 = 342 ohms. So Rb = 5700 - 342 = 5358 ohms. (The 1k pot would be set about to about one -third down from the top in your diagram)

It remains to find Rs, in order to get the currents right. Isc = Vbe[(Ra+Rb)/RbRs], so Rs = Vbe[(Ra+Rb)/IscRb]

Isc = Ilim/N, For the example N = 3, Ilim = 1A, Isc = 0.333A, so Rs = 0.6*[(5700)/0.333*5358] = 1.92ohms.

Does this give 1A Ilim as expected? Ilim = Vbe[(Ra+Rb)/RbRs] +Vout (Ra/RbRs), = 0.6*(5700/(5358*1.92) + 20*(342/5358*1.92) = 0.332A + 0.665A = 0.997A

For a practical set-up you could choose Rs to the nearest preferred value, say 1.8ohms or 2ohms. Note it will dissipate 2W for a 1A limit!

Then adjust the R4 pot to set Ilim. Isc won't then be exactly 1/3 of Ilim, but that's probably not very important.

20. Dec 25, 2010