Hi
So we have a curcuit as shown below. It's a part of a power supply (LM317) and amp meter (LM3915) - signnal at pin 5 is proportional to current.
There is an overload protection - T4 and T5 and I can't understand how it works.
I know that when current is too high (pin 10 is active) T4 is turned on, T5 goes into saturation and LM317's voltage drops to 1.25V (because we bypass P1).

VCC for 3914 and LEDs is 8V if I remember correctly, and R value in pin 7 is 2.2k so it give us (1.25/2k2)*10 = 5.6mA of current for each LED.
I wanted to use different LED current in higher VCC so I wanted to calculate resistors for that T4-T5 circuit but I just can't understand how it works - to be specific I don't know how to calculate resistors values.

Here is what I figured out:
1. Can we assume that because of pin 7 setting the current going into pin 10 is 5.6mA?
If so than Ur12=5.6m*180 ~ 1V
assuming Ueb=0.6V
Ur17 = 0.4
Ie = 0.4/22k = 18.2uA (???)
That would give very small Ib, it simply doesn't look right

2. On the other hand if T5 is saturated, assuming Ucesat ~ 0 than we have in collector circuit R1 on which we have 1.25V ref volt from LM317 so we have
Ic=1.25/510 = 2.5mA
assuming Ic/Ib = 10
we would need Ib ~ 250uA

I think that I don't quite understand how 3914's outputs operate Could someone please explain how both BJTs are biased in this circuit?
Also what is the of R8=100 if 3914 can set it's outputs currents?

 

A na polskim formum nie dało sie oto zapytać??

 

Hi
So we have a curcuit as shown below. It's a part of a power supply (LM317) and amp meter (LM3915) - signnal at pin 5 is proportional to current.
There is an overload protection - T4 and T5 and I can't understand how it works.
I know that when current is too high (pin 10 is active) T4 is turned on, T5 goes into saturation and LM317's voltage drops to 1.25V (because we bypass P1).

VCC for 3914 and LEDs is 8V if I remember correctly, and R value in pin 7 is 2.2k so it give us (1.25/2k2)*10 = 5.6mA of current for each LED.
I wanted to use different LED current in higher VCC so I wanted to calculate resistors for that T4-T5 circuit but I just can't understand how it works - to be specific I don't know how to calculate resistors values.

Here is what I figured out:
1. Can we assume that because of pin 7 setting the current going into pin 10 is 5.6mA?
If so than Ur12=5.6m*180 ~ 1V
assuming Ueb=0.6V
Ur17 = 0.4
Ie = 0.4/22k = 18.2uA (???)
That would give very small Ib, it simply doesn't look right

2. On the other hand if T5 is saturated, assuming Ucesat ~ 0 than we have in collector circuit R1 on which we have 1.25V ref volt from LM317 so we have
Ic=1.25/510 = 2.5mA
assuming Ic/Ib = 10
we would need Ib ~ 250uA

I think that I don't quite understand how 3914's outputs operate Could someone please explain how both BJTs are biased in this circuit?
Also what is the of R8=100 if 3914 can set it's outputs currents?

Your analysis looks correct. You are also correct in thinking that Ic(T4) is way too low.
I thought R8 might be for sharing power dissipation with the LM3914, but in dot mode, it only drops about 0.5V, so... ?
Why does your schematic show a 3915? Which are you using?
See the attachment for another question.

 

A na polskim formum nie dało sie oto zapytać??

Ponieważ jest to English language forum, prosimy zadać pytanie w języku angielskim, Dziękuje !

(As this is an English language forum, kindly ask question in English!)

 

HI Ron.
Here you have a full circuit diagram
http://elportal.pl/pdf/k04/14_16.pdf
The original circuit comes from Polish electronics magazine. This circuit was original design long time ago (1997) and work as a simplifies laboratory power supply.

 

well the diagram I posted was sent to me, I think it's slightly diffrent from what Jony posted, but it's pretty clear that, what I got was based on it, the T4-T5 part looks similar though. I'm not home right now so I can't look at that modified version (part of which I posted), but I'm quite sure it doesn't have T3, among other things.

 

I don't see how T5 can ever turn on. R13 will have about the same voltage across it as R17 (≈0.4V).