Voltage regulator short circuit protection (7 - Regulators)
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- เผยแพร่เมื่อ 17 ม.ค. 2024
- How to design overcurrent protection in a voltage regulator using a transistor and a resistor. This is a common way to implement short circuit protection. A transistor automatically switches off when the voltage across the base-emitter junction gets to high.
Aaron Danner is a professor in the Department of Electrical and Computer Engineering at the National University of Singapore.
danner.group
Video filmed and edited by Cheryl Lim.
@randomcheryl
Very well done, been using this circuit since around 1971, never fails to give a bit of safety.
Fantastic explanations. Thanks
Thank you for another great video.
Thank you, i can think of many applications.
It would be great if you gave us specific transistors and opamps to make the circuit at home and try it!
Thank you for these lessons!
In the case of a short circuit on the output, the action of R3 and Q2 does limit the current through Q1, but now the collector to emitter voltage across Q1 is much higher than it is in the normal operating case, leading to additional power dissipation in Q1. It is important to size Q1 and its heat sink to handle this power dissipation.
But the current is Zero, so no power is dissipated.
@@W1ZY Be sure to realize that Q2 only turns on in this example when the voltage across R3 is 0.6 volts. To have the voltage across R3 be 0.6 volts, the current through R3 must be 0.5 Ampere, and that current (plus the Q2 base current) comes from Q1. With a short circuit on the output, and assuming the unregulated supply voltage in this case is at least 15 volts, the power dissipation in Q1 is roughly (15 - 0.6) * 0.5 = 7.2 Watts, (or more if the unregulated supply voltage is higher).
In summary, the action of Q2 places an upper limit on the current in Q1, but Q2 doesn't cause the current in Q1 to "fold back".
Very clear explanation paced very well with no time wasting. Like it!
This is a good presentation of a current limiting circuit. It applies well for circuits that have their current increased as they heat up, like Power LEDs. Without a current limiting circuit they would burn out.
However, it is not an efficient solution for a short-circuit situation as you propose. During a short-circuit the current keeps flowing (even if limited) and all the unregulated power supply voltage gets dumped on the pass transistor Q1 to a maximum dissipation of 12W for a maximum voltage drop of 24V.
Also, during a normal operation with a current limited to 500mA, the Q2 collector current would drive the operational amplifier current to a much higher value than the 3mA it was designed to carry so, if the Q1 transistor does not burn, the operational amplifier certainly will.
A better solution would be an Electronic Circuit Breaker that shuts off the load. You might consider the design presented in this TH-cam video “Electronic Circuit Breaker Part 3” from Palomas Estate (link th-cam.com/video/hUShvpRlItE/w-d-xo.html).
I have simulated the circuit with a 24V power supply, 12V output, and 500mA current limit. Without the current limiting circuit (Q2 and R4) and adding the circuit breaker, the maximum power dissipation is 7W when the current reaches 500mA and the circuit shuts off. Considering that at 500mA the power dissipation on the pass transistor is 6W (24V in - 12V out, times 500mA), the circuit breaker solution only dissipates 1W more when the current reaches 500mA, in a short-circuit situation.
Thanks.
Also the old uA741 nowaday has a short circuit protection
@@stefano.a if you read the uA741 manufacture’s datasheet (I read from three different ones) you will see that the short circuit protection they use is exactly the same technique used in this video.
If the output goes to ground (like in an output short circuit) a 25ohm resistor starts to turn ON a transistor that will lower the base current of the pass transistor as the collector current rises.
This happens when the current reaches 25mA (25ohmX25mA=625mV), which is the typical current specified for the “short-circuit output current (Ios)” in the datasheet.
When the output current reaches 40mA (as per datasheet) there will be 1 volt (25ohmX40mA=1000mV) across the resistor causing all the remaining voltage from the power supply to be on the pass transistor (like it will happen in this video).
For the maximum power supply voltage recommended (15V), this will make the uA741 dissipate 600mW (40mAX15V=600mW). However, the total power dissipation of the uA741, without heat sink, is 500mW.
Conclusion: The power supply cannot be higher than 12.6V (a fully charged car battery) for the short circuit current not to fry the uA741 (40mAX12.6V=504mW). That is why they mention “unlimited” for the duration of the output short circuit (as long as you don't use 15V power supply).
Thanks.
But there is no current through Q1, and therefore no power dissipated by it. The only potential problem would be current from the IC, which has current limiting protection internally.
Great video. Thanks👍
There is a great deal of information here, but your delivery is very quick and therefore hard to follow. But thank you for posting
if you thermally connect both transistors, would this give you some over temperature protection, too? (the idea is that Vbe(on) drops with temperature)
Good old LM723.
Thanks for that comment. The circuit configuration in LM723 is almost the same as in the video. That`s actually a good practice to look at the datasheet of it to understand the subject better.
ok you will save q1 for the live of an opamp.
opamp do not like too much current in long term.
my english is verry bad but.... you have to Design the circuit so that never ever the opamp reach füll current;).
A lesson i get from repairing faulty power supplies.
I had experience as you say. The pass transistor was fine, but the OP amp failed.
almost all op amps have short circuit protection