How to stop bleed to power resistor

Thread Starter

xtal_01

Joined May 1, 2016
169
OK, so I posted earlier today about this circuit.

Found out R2 that was optional on the drawing (though I found it had been installed in all the boards) is used to trickle charge the battery.

So, new problem ... I see they yellow LED hat indicates the power is on, glows dimly even when the power fails and the beeper alarm is being fed from the 9 volt rechargeable battery (NiMH).

R2 is 1K ... to the battery goes through three 1K resistors when the power fails ... so about 3 mA

I like that the designer (think I mentioned in the first post ... he has long since retired and passed) is checking to see if the power is on by putting the LED off the second regulator.

Just not sure if there is a way of stopping the "glow" of the yellow power LED?

Thanks!

Circuit Board Schematic-1.jpg
 

crutschow

Joined Mar 14, 2008
34,050
if there is a way of stopping the "glow" of the yellow power LED?
Move the R3 connection to the output of REG 1.
It will then go out when the AC power is lost.
If that makes the LED too bright, increase the value of R3.
 
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Thread Starter

xtal_01

Joined May 1, 2016
169
Move the R3 connection to the output of REG 1.
It will then go out when the AC power is lost.
If that makes the LED too bright, increase the value of R3.
I thought about that but then the circuit would still function if the second regulator fails. It is good the way they have it as checks for power loss and also if the second regulator is working.
 

crutschow

Joined Mar 14, 2008
34,050
I thought about that but then the circuit would still function if the second regulator fails. It is good the way they have it as checks for power loss and also if the second regulator is working.
Okay, then how about putting a resistor across the LED with a value that shunts enough current so the LED doesn't light under fault conditions.
Something around 1kΩ should likely work.
You may have to then reduce the value of R3 some to retain the desired brightness when it's ON.

Or put two diodes in series going from REG 2 to the battery, and one diode in series with R2 to block the reverse current.
To minimize the forward drop of the two diodes in series, you could change them to Schottky types such as the BAT54.
The two Schottkys in series should have a forward drop similar to or slightly less than one junction diode.
 

Thread Starter

xtal_01

Joined May 1, 2016
169
Okay, then how about putting a resistor across the LED with a value that shunts enough current so the LED doesn't light under fault conditions.
Something around 1kΩ should likely work.
You may have to then reduce the value of R3 some to retain the desired brightness when it's ON.

Or put two diodes in series going from REG 2 to the battery, and one diode in series with R2 to block the reverse current.
To minimize the forward drop of the two diodes in series, you could change them to Schottky types such as the BAT54.
The two Schottkys in series should have a forward drop similar to or slightly less than one junction diode.
Awesome idea!

Either one seems like it will work.

I am hoping to order the parts this weekend ... I will prototype the circuit before I get the boards made. I can try both ideas and see it one seems better.

Thanks!
 

crutschow

Joined Mar 14, 2008
34,050
I forgot to mention that, using the three Schottky diodes means you likely will have to reduce the value of R2 to about half its present value to get the same trickle-current.
 

Thread Starter

xtal_01

Joined May 1, 2016
169
I forgot to mention that, using the three Schottky diodes means you likely will have to reduce the value of R2 to about half its present value to get the same trickle-current.
Got it! I am planning to show all my calculations and put notes all over the drawing as to why and how. This should help the next guy doing this job 10 years from now when I am gone.

Thanks so much for all the advice !!!!!!!!!!!!!

Mike
 

Dodgydave

Joined Jun 22, 2012
11,243
OK, so I posted earlier today about this circuit.

Found out R2 that was optional on the drawing (though I found it had been installed in all the boards) is used to trickle charge the battery.

So, new problem ... I see they yellow LED hat indicates the power is on, glows dimly even when the power fails and the beeper alarm is being fed from the 9 volt rechargeable battery (NiMH).

R2 is 1K ... to the battery goes through three 1K resistors when the power fails ... so about 3 mA

I like that the designer (think I mentioned in the first post ... he has long since retired and passed) is checking to see if the power is on by putting the LED off the second regulator.

Just not sure if there is a way of stopping the "glow" of the yellow power LED?

Thanks!

View attachment 307839
Okay so remove R2 and see if it the glow stops when power is removed, it should still work when the power is on.
 

Thread Starter

xtal_01

Joined May 1, 2016
169
Okay so remove R2 and see if it the glow stops when power is removed, it should still work when the power is on.
It will stop glowing but I want to leave R2 in the circuit to trickle charge the battery. I checked and every one in the field does have R2 installed even though it is listed as optional (honestly, don't even know where that note came from ... and notice the bottom says "see note 2" and there is no note 2).

Just want to clean this up as much as possible .. board say rev 3 on it ... and I think the original dates from the 80's

Thanks!
 

Thread Starter

xtal_01

Joined May 1, 2016
169
Have you tried putting a diode in series with R2 ?
Les.
I( have not tried that yet ... I was thinking about it but then you have the same voltage drop across both lets and I don't think it will trickle .. but maybe it still will. I will have to think about it ... or just try it.
 

MisterBill2

Joined Jan 23, 2018
17,827
OK, so I posted earlier today about this circuit.

Found out R2 that was optional on the drawing (though I found it had been installed in all the boards) is used to trickle charge the battery.

So, new problem ... I see they yellow LED hat indicates the power is on, glows dimly even when the power fails and the beeper alarm is being fed from the 9 volt rechargeable battery (NiMH).

R2 is 1K ... to the battery goes through three 1K resistors when the power fails ... so about 3 mA

I like that the designer (think I mentioned in the first post ... he has long since retired and passed) is checking to see if the power is on by putting the LED off the second regulator.

Just not sure if there is a way of stopping the "glow" of the yellow power LED?

Thanks!

View attachment 307839
This circuit schematic is very confusing and hard to follow . In addition it is cut off at the top.Thus we have no hint as to what the assembly is or what it is intended to do.
 

LesJones

Joined Jan 8, 2017
4,173
Looking at the circuit more closely, to achieve what you want I think you would need to add two diodes. One in series with R2 and one in series with D5. You would also need to adjust the output from the LM317 to compensate for the extra forward voltage drop caused by the two diodes. So you would need to adjust VR1 to give 9.6 volts on the output of the LM317.

Les.
 

MisterBill2

Joined Jan 23, 2018
17,827
An explanation of what the circuit is actually supposed to accomplish would be useful.
What I see from the circuit in post#1 is a complex implementation of a simple function.
 
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