Very Confusing Circuit. - Page 1
Forum section: electronics / projects

author: kvresto
author_rank: Regular Contributor
timestamp: February 17, 2015, 08:26:28 am
content: 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) = 6.3V1) So I’m wondering if my calculations are correct? And why the difference at VF3?2) Why is VF2 = 5.19V3) 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 5Vrail? These are DC conditions always applied to the cct.Thankskvresto.

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: February 17, 2015, 08:35:18 am
content: 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.

author: SteveyG
author_rank: Unknown
timestamp: February 17, 2015, 09:01:36 am
content: Quote from: JohnnyBerg on February 17, 2015, 08:35:18 amAnd VF2 is approximately one diode drop lower than the voltage at VF3.

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: February 17, 2015, 09:06:01 am
content: Quote from: SteveyG on February 17, 2015, 09:01:36 amIf there was current flowing through D1, one could even say that VF4 = VF2

author: kvresto
author_rank: Regular Contributor
timestamp: February 17, 2015, 10:02:55 am
content: 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 = VF3Kv

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: February 17, 2015, 10:13:33 am
content: 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 simplyA 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.

author: kvresto
author_rank: Regular Contributor
timestamp: February 18, 2015, 11:26:25 am
content: 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?cheerskv

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: February 18, 2015, 11:34:28 am
content: Quote from: kvresto on February 18, 2015, 11:26:25 amNot 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

author: Howardlong
author_rank: Super Contributor
timestamp: February 18, 2015, 11:40:48 am
content: Meanwhile, I hope that R1 is sufficiently well rated, 1/2W at least.

author: kvresto
author_rank: Regular Contributor
timestamp: February 18, 2015, 12:13:31 pm
content: 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.cheerskv