7809 Current boost + SC protection

Hi,
attached is a schematic I found on the 7809 datasheet. When I am simulating it with different loads and short circuit protection kicks in,
the current stays at maximum while the voltage drops - something which follow Ohm's Law.

I read on a book how this configuration works: when current exceeds a certain value and 0.7V develop across Rsc, Qlimit switches on forcing all the current through the voltage regulator. The thermal shutdown of 7809 kicks.

Is this the effect of thermal shutdown of the regulator that is causing the current to stay at max while dropping the voltage?

Is this the effect of thermal shutdown of the regulator that is causing the current to stay at max while dropping the voltage?

Click to expand...

You should firstly think about current limiting action of the voltage regulator. Thermal shutdown possibly comes in addition, depending on the operation parameters, e.g. input voltage. As a point to consider, you should check if Q1 is operated safely under current limiting conditions.

No. Thermal shutdown kicks in only when the regulator temperature exceeds a safe limit as internally set by the manufacturer. Another thing is that Q1 does not completely shut off. If it did, there would no longer be any voltage drop across Rsc to turn Q2 on. An equilibrium will be maintained to limit current to a value just enough to start to turn Q2 on.

The regulator also has a current limiting feature which is not the same as the thermal shutdown action. Thermal shutdown, when activated, shuts the regulator down completely until the IC has cooled down. The current limiter reduces the output voltage to a level that will force the current to stay below a certain set limit.

i did not understood the feed back through PNP 6132 can some please explain me why to use this.

---------- Post added at 01:20 ---------- Previous post was at 01:12 ----------

Q1 will remain off during operation PNP is active low. then why to use it in feed back

It only starts to conduct when the voltage across R1 reaches it's Vbe, from then on it diverts current around the regulator straight to the output. When the current causes less than Vbe to be dropped across R1, the regulator alone carries all the current.

Brian.

I understood to operation of the transistors. After Q1 is left to conduct the maximum current while Q2 is conducting the rest of the current. The rest of the current has to pass through the regulator.. until current limit of the regulator kicks in. In this manner, we can say that we can never set a current limit with only Q1 and Q2 but we have to consider the current limit of the regulator finally. Does this make sense?

No. It works like this:

The regulator does all the work up to a certain current. That current is decided by the resistor in series with it's input (R1). As current flows through R1, a voltage is dropped across it, when it reaches the turn-on voltage of the B-E junction of Q1, the transistor starts to conduct and the remainder of load current flows through it.

Q2 is there only to provide short circuit protection. The current through Q1 causes a voltage drop across Rsc, when this reaches about 0.8V Q2 starts to conduct and it clamps the B-E voltage available to Q1 to prevent excess current flowing.


The advantage of wiring a regulator like this is that you can make use of the regulators inernal shut-down circuit. When it reaches a certain temperature the cut-out operates and limits or in some cases stops any output current flowing. Because the current from the regulator is limited, so is the input current to the bypass transistor so it also stops conducting. If they are both mounted on the same heat sink the regulator thermal shutdown will also sense the transistors temperature.

Brian.

Okay, so what will change if the three devices are mounted on a very large heatsink that they will never exceed say 50 degrees Celsius no matter if it has a big load or short circuited.

Nothing will change except the over-temperature threshold will never be reached. It will regulate the voltage as before and the short circuit current will be the same as before.

I's probably best NOT to mount Q2 on the same heatsink because normally it runs cold unless you overload the supply. To be honest it wouldn't make a lot of difference if they shared a heatsink though.

Brian.

The link above is a guide for biasing the described regulator circuit with the addition of a normal 0.7Vf diode to share the current at a calculated ratio at all times. The guide states that this configuration works best by ensuring that the transistor is working at the active region.

Is there any way I can prove it as the biasing of the transistor is in an odd manner. I tried doing the calculations of a based biased transistor with the regulator replaced by a diode of forward voltage 3.26V but I got to nowhere. I took all the voltages of the based biased transistor with respect to the 9V reg output.