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Topic: voltage spike/indication circuit (Read 1 time) previous topic - next topic

TomGeorge

#15
Feb 18, 2018, 10:00 pm Last Edit: Feb 18, 2018, 10:03 pm by TomGeorge
Hi,
The 10k to ground may be a problem, as it will prevent the transistor from being biased OFF when the fuse is in tact.



Tom... :)
PS, Just had my morning java.. :o
Everything runs on smoke, let the smoke out, it stops running....

DudeGuy

i'll give that a try. my thoughts were that the 10k to fuse would supply positive to the PNP and when the fuse blew then the 10k to ground would supply the negative for the PNP.  I used the two resistors off the base of the PNP to supply a difference of potential pending the state of the fuse.

JohnRob


It might help if you drew the schematic with and without the fuse.

In TomGeorges schematic there is not path for the base current.

Look at it this way:

The fuse is gone.
The input voltage is now below the 16V of the zener so the zener is not conducting.

There is no path for the base current.

Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

TomGeorge

#18
Feb 19, 2018, 04:47 am Last Edit: Feb 19, 2018, 04:48 am by TomGeorge
It might help if you drew the schematic with and without the fuse.

In TomGeorges schematic there is not path for the base current.

Look at it this way:

The fuse is gone.
The input voltage is now below the 16V of the zener so the zener is not conducting.

There is no path for the base current.

Good point...

If Vout is connected to a load, then Vout pin will go to near gnd, base current will flow through emitter/base.

BUT Vout will have to have a load.

Putting a 1K from Vout to Gnd would ensure base current even without a load.

Tom... :)
Everything runs on smoke, let the smoke out, it stops running....

JohnRob

@TomGeorge,


I don't see the current path you are describing.   If the base is essentially open circuit, no current is going to flow.

Go back to the sketch (crude) in post #10,  Remove the fuse and Zener, you will see a current path from emitter -> base -> R2 -> Ground

Putting the fuse and zener back in.   The fuse essentially shorts out the base emitter junction, not allowing any base current to flow out of the transistor.

John
Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

TomGeorge

#20
Feb 19, 2018, 09:49 pm Last Edit: Feb 19, 2018, 09:50 pm by TomGeorge
@TomGeorge,


I don't see the current path you are describing.   If the base is essentially open circuit, no current is going to flow.

Go back to the sketch (crude) in post #10,  Remove the fuse and Zener, you will see a current path from emitter -> base -> R2 -> Ground

Putting the fuse and zener back in.   The fuse essentially shorts out the base emitter junction, not allowing any base current to flow out of the transistor.

John

The fuse keeps the transistor OFF, so the LED dos not light, which is what the OP wants.
As I said in the update post, the load or addition of a resisitor across the zener will make the transistor turn ON when the fuse blows.

Tom.. :)
Everything runs on smoke, let the smoke out, it stops running....

JohnRob

@TomGeorge,

Now I see what you were saying.  I was reading it as add a resistor across the load (C to Gnd).


I think the OP should drop the 10k to 5k (roughly) else the PNP will be on the edge of full conduction.

In switching applications I always start with a force beta of ~10.

John

Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

TomGeorge

I think the OP should drop the 10k to 5k (roughly) else the PNP will be on the edge of full conduction.

In switching applications I always start with a force beta of ~10.

John


Yes, good idea.
I tell you by the time the OP sees this we will have  it designed, prototyped and in production. :) :)


Tom.... :)
Everything runs on smoke, let the smoke out, it stops running....

DudeGuy

Hahahaha yeah. I just tried it as per the latest revision. My resistor values aren't what you guys said yet. I'm using a 2.2k for Vout.  The circuit works as it should.

Now the only thing left is feed it a little over 16vdc and go for smoke

Thanks guys!!!

JohnRob

Quote
Hahahaha yeah
You will find the circuit will function with a very wide range of values.


I'm glad you have it working (with help from the forum).

Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

DudeGuy

yes, now can I ask, what is the science behind moving the resistor (R2) from one side of the zener to the other? 

TomGeorge

Hi,
yes, now can I ask, what is the science behind moving the resistor (R2) from one side of the zener to the other?  
With the fuse in-tack, and R2 from the base to gnd, you have a path for base current to gnd, this why your LED was ON all the time.
With R2 moved, you now have no path for base current to gnd while the fuse is good, this because the base and emitter are connected together.
When the fuse blows, Vout is now at or near gnd, so provides a path for base current, R2 now ensures that there is a path in case the load goes open circuit.
Tom... :)
Everything runs on smoke, let the smoke out, it stops running....

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