Power/signal clamping with linear regulator?

[vlad777]

Since I know that the output of lin regulator is emmiter
of NPN transistor (and it can't conduct in reverse),
I want to ask what is the right way to clamp signal
(or inductive load), when using lin regulator?

**broken link removed**

Many thanks.

 

[ericgibbs]

hi vlad,
I would never consider using solution #2.

What are you signal and inductor details.?

 

[vlad777]

I am just wondering about the theory of solution #1.
Now I also wonder why not sol #2.

What if the signal is 12V and needs to be clamped to 5V, and it has lot of frequency,
(or is high most of the time) so it charges up the capacitor?

 

[ericgibbs]

hi Vlad,
In solution #1, If the 'signal' voltage was always higher than the +5V output from the 7805, the correct method would be to use a resistive voltage divider [ across the lower diode] so that the voltage to the IC would be +5V or less.

For inductive suppresion, the series resistor would not be present.

If you have a circuit where the output voltage of the 7805 can be 'forced' higher than +5V output, one method used to protect the 7805 is to connect a 1N4001 diode across the 7805. This will ensure that the 'reverse' voltage across the 7805 will not exceed approx 0.7V [diode anode to the 7805 output, cathode to input]

 

[KeepItSimpleStupid]

If this is for automotive use, there are chips that do a better job in this environment. See www.linear.com. For an inductive load, it's best to clamp close to the load. The automobile environment is particularly hostile. -200 to +50 is what I remember, but then I've read better analysis that even suggests 24 VDC continuous. Jump starting introduces some unusual faults.

Transients at the regulator input can usually be handles with TVS diodes and a reverse biased diode.

 

[vlad777]

I am not interested in anything too specific.
I am just wondering where is this excess charge suppose to go?
If the load is not sinking enough current, this charge will build up
the voltage on the main filtering capacitor.


Edit:

I think I get it: The current from the clamp would have to be bigger
then the load current, for the voltage on cap to increase.