Very Confusing Circuit.

[kvresto]

Hi Everyone.
I came across this circuit today which has me asking some questions, and I hope someone can shed some light on the subject. Its been a long day and maybe I’m confused.

From the cct pictured in the image, 24V is applied to R1, and 5V to the cathode of D2. Also R1 and R2 form a voltage divider and by my reckoning the point at VF3 should be:          12V – 5.0 – 0.7V(Vf of D2) = 6.3V 

1)   So I’m wondering if my calculations are correct? And why the difference at VF3?
2)   Why is VF2 = 5.19V
3)     Also why would anyone design such a circuit, and as this is actually implemented on a functioning board, when is it ever acceptable to pump about 6V into the 5V
rail?  These are DC conditions always applied to the cct.

Thanks
kvresto.

[JohnnyBerg]

It is a basic clamping circuit. As soon as the voltage across R2 rises above 5,7 V the diode starts conducting. This prevents the volage to rise any further, because of the charasteristic of the diode.

[SteveyG]

It is a basic clamping circuit. As soon as the voltage across R2 rises above 5,7 V the diode starts conducting. This prevents the volage to rise any further, because of the charasteristic of the diode.

And VF2 is approximately one diode drop lower than the voltage at VF3.

[JohnnyBerg]

It is a basic clamping circuit. As soon as the voltage across R2 rises above 5,7 V the diode starts conducting. This prevents the volage to rise any further, because of the charasteristic of the diode.

And VF2 is approximately one diode drop lower than the voltage at VF3.

If there was current flowing through D1, one could even say that VF4 = VF2 

[kvresto]

Thanks everyone.

What about current flowing into the 5V rail through the conducting diode(D2), doesn’t that upset the 5V regulation in any way, ie: perhaps overheat the regulator?
BTW the resistors R1 & R2 are 10k not 1k, the current delivered by the 24V supply is:
(24V – 5.65V(VF5) ) / 10k = 1.84mA?, Sooo I think 1.84mA – (VF3/R2) = current traveling into the 5V rail ignoring the piddlyamps into the buffer?

what does that sound like?

Yes I know VF5 = VF3

Kv

[JohnnyBerg]

If there are enough consumers on the 5V rail, that is not a problem. The current that "tries" to lift the 5V rail is consumed first. To say it simply 

A well know "trick" is to put a zener diode with a slightly higher voltage then 5V (5V6) over the rail, so when the rail gets lifted, it cannot be lifted over the zener diode.

[kvresto]

Thanks for the info JohnnyBerg.

I have a problem with this cct, I cant believe that the designer would accept the risk of the 5V rail lifting is ok.
This is a "Digital" board so perhaps its not so critical, but if there was sensitive analog cct's being powered of the 5V rail, I think this arrangement would not be acceptable. From memory the input to the buffer is the output of a 24V hall proximity sensor(24V=off, 0V=on/active),.... so weird?

cheers
kv
 

[JohnnyBerg]

I have a problem with this cct, I cant believe that the designer would accept the risk of the 5V rail lifting is ok.
This is a "Digital" board so perhaps its not so critical, but if there was sensitive analog cct's being powered of the 5V rail, I think this arrangement would not be acceptable. From memory the input to the buffer is the output of a 24V hall proximity sensor(24V=off, 0V=on/active),.... so weird?

Not at all. Current through R1 can be a maximum (estimate) of 24-5 / 1K = 19 mA. Current through R2 will be 5 / 1K = 5 mA.
So, current left to lift the rail = 19 - 5 = 14 mA. That kind of current is likely to be consumed by the rest of the electronics. A power on led will do the trick. Or the designer was realy smart, and put a zener on the 5V rail 

[Howardlong]

Meanwhile, I hope that R1 is sufficiently well rated, 1/2W at least.

[kvresto]

I see what you mean, you've cleared up the cct operation for me, must go off and look for a zener next time I pick this board up.

cheers

kv