5
\$\begingroup\$

I am trying to design a current limiter or my LM317 based power supply using this schema.

The power supply must be able to provide a maximum of 1.5A@30V. I made some calculations and I would like to understand if I am on the right way. Goal is to find suitable values for the resistors and the right BJTs.

1) Q1: Assuming the power supply is working at its limit, on Q1 we will have a Ic=1.5A and Vce=Vcc-Vq2be=30-0.7=29.3V. A suitable BJT able to support this values could be the D44H8.

2) Value of R1: The D44H8 has a worst case current gain of 10. So, on Q1, if Ic=1.5A the Ib=1.5/10=0.15A. Being near the limit of 1.5A, we can assume Vq2be=VRsen=0.7V, thus: VR1 = 30-0.7-0.7=28.6. This lead to R1 = 28.6/0.15=190 Ohm

3) Value of Rsen: On Q1: Ic=1.5, Ib= 0.15, Ie=1.65 Voltage drop across Rsen is circa 0.7. Then: Rsen=0.7/1.65=0.42 Ohm

4) Q2: Q2 must be able to support at least 30-0.7-0.7=28.6V on its collector, but much smaller current. I choose a BJT in the BC337 family, which will guarantee 45Vce and a max collector current of 800mA

Keeping R1 constant and varying Rsen I can set different maximum current values on the load.

Am I more or less right?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
4
  • 2
    \$\begingroup\$ Did you calculate the power dissipation in your transistors? If RL is 0 Ohms, Q1 will dissipate 40 Watts or so. I didn't read all your text. But this circuit is an OK way to limit current through RL. Note that if you plan to use it as a current limiter for a supply, you might want to re-arrange it so it operates on the high side (using PNP instead of NPN). If the 30V supply is floating, then it doesn't really matter. But if the 30V supply is earth grounded (or could be), then it could be dangerous to put the limiter on the low side because the user might earth ground the collector of Q1. \$\endgroup\$
    – user57037
    Commented Oct 13, 2015 at 22:46
  • 2
    \$\begingroup\$ This will not give you a precision current limit, and the actual limit will vary quite a bit with the temperature of Q2. \$\endgroup\$
    – user57037
    Commented Oct 13, 2015 at 22:47
  • \$\begingroup\$ @mkeith I simulated the circuit with EveryCircuit, and I actually noticed that for small changes of R1, very big changes on Ic occur. So this was actually my next question, I somehow had the feeling this system is not so precise. Thank you for confirm this. \$\endgroup\$
    – Enrico
    Commented Oct 14, 2015 at 7:12
  • \$\begingroup\$ Definitive calculation was made here planetanalog.com/… \$\endgroup\$
    – kimstik
    Commented Dec 29, 2021 at 15:51

1 Answer 1

Reset to default
4
\$\begingroup\$

You are more or less right, but this is not a good solution. Your voltage at the collector of q2 will never get much above 1.4v. And your 190 ohm R1 will dissipate 30x0.15 = 4.5W

Because you have feedback you can use a fet to pass the current and make R1 much bigger. You will probably have to use heat sinks on Rsense (1W+), and your pass transistor q1 will have to take the full whack of power. That is near enough 45W. Also, your limit will vary by at least 300mA across the temperature range, maybe 1.3A minimum to 1.6A maximum.

There are many more efficient and more accurate solutions. What other design criteria do you have? Eg Cost, Accuracy, board space?

Followup to comments: Below, Enrico asked to see some alternatives. One possibility is to use Lmp8646 which is a current sense amplifier from TI. You would need Rsense of about 0.1R. Set the gain accordingly then feed the output into a comparator. Use the comparator to turn off the supply (does your regulator have an enable input?). For this circuit you would need a fixed supply less to power the ic, comparator, and reference. The resistor would dissipate just under 0.25W. I would suggest putting a lot of hysteresis on the comparator and also a capacitor on the output to stop the thing switching on and off very fast during fault condition. You could instead latch the supply off until a reset button is pressed.

This IC is also suitable for more sophisticated current control, such as variable current limiting and folding back the supply voltage by connecting the output into the regulator feedback. It is just the first one I came across - other brands / circuits are available.

\$\endgroup\$
8
  • 2
    \$\begingroup\$ You might need to be careful with power mosfets in this essentially linear setup. Their power ratings are defined for switching applications, and iirc hot spots can appear in the silicon in high power resistive regime. \$\endgroup\$
    – Nicolas D
    Commented Oct 14, 2015 at 6:14
  • \$\begingroup\$ Thanks, I didn't know that. Are you saying that a power mosfet operating in linear mode can be damaged even though it stays within its power ratings and continuous drain currents? I had taken for granted that current limit was a fault condition (ie temporary) but if this is a bench supply for development them maybe not. \$\endgroup\$
    – Loganf
    Commented Oct 14, 2015 at 6:41
  • \$\begingroup\$ It would be very interesting for me to know which are some possible alternative solutions. I am currently learning, building my workbench supply for other projects, so I don't actually have big limits. I could decide to implement every solution. In the original project current limiting was done with simple resistors. \$\endgroup\$
    – Enrico
    Commented Oct 14, 2015 at 7:10
  • 2
    \$\begingroup\$ Mosfets don't use realistic power ratings. The max power is typically based on maintaining a case temperature of 25C. But of course, when dissipating 40W, the case will not be at 25C. You have to go through the thermal calculations using thermal resistances from the data sheet. Check to see whether the junction will exceed its maximum allowable temperature. \$\endgroup\$
    – user57037
    Commented Oct 14, 2015 at 7:40
  • \$\begingroup\$ Okay thanks I am familiar with that process. I thought maybe you were referring to their junction structure or something. For example, I vaguely recall that a high dvdt can damage fets because the current is concentrated in a small area despite all other factors being within the recommended operating conditions. Anyway... @enrico do you want to run in constant current mode indefinitely? If so the best approach is probably to adjust your regulator output. Any power dumping solution is going to cause your supply to get hot \$\endgroup\$
    – Loganf
    Commented Oct 14, 2015 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.