Current behaviour through current mirror with CC driver and an open string

Thread Starter

SP180

Joined May 10, 2020
3
Hi,

I am starting an LED project for my aquarium where I will have 6 channels using MW LDD-1500H 1500mA CC Drivers powering 12no "3W" (3.4-4Vf 750mA) LED's per channel in a series/parallel configuration of 4 strings of 3 LED's. I am looking to put c. 375mA through each string (1500/4) but to avoid thermal runaway have been researching current mirrors.

I have found a technical note by ST which details a suitable balancing circuit 'LED balancing circuit 3' described on page 4 with a diagram and equations on pages 5 and 6. Link to note: https://www.st.com/resource/en/technical_note/cd00145612-led-balancing-circuits-stmicroelectronics.pdf

LED balancing circuit.jpg

I am struggling to grasp exactly how the circuit is working alongside the CC and my maths isn't strong enough to go through all of the equations listed.

In the description it states:
If a LED (for example D1) in one string is open, the current through the string (D1 and D2) is zero. The base to emitter junction of Q1 behaves like a diode and a small amount of current passes through R1. The reference voltage Vr drops, the other base to emitter junction voltages drop, and less current flow through the other strings (see Equation 10).

Specifically I am struggling to understand how much current would be running through the three remaining strings with D1 open and where the remaining current / power would be dissipated given the CC driver will always kick out 1500mA.

Any help or pointers would be much appreciated.

Many thanks,
Steve
 
Last edited:

ci139

Joined Jul 11, 2016
1,463
depending on the specific type of the D7 , D8 and the value of the Rr it may regulate the current "quite well"
it works basically the same as the current mirror . . . basically . . . if you disconnect D1/D2 it will put a bit less current through the Q2 & Q3

the D7 & D8 have relatively little voltage change when the current through them changes ⇒ if the voltages on the R1 ... R3 are not changing -- the base currents keep the same thus the Ic -s are near constant . . . if the Q1 no more has Ic it's BE diode has lower drop than Q2 Q3 that further limits their Ic

  • if the 1500mA is the CC limit it won't be pushing that to circuit
  • otherwise (inductive charge pump) your Vin would fast grow causing the Q2 and Q3 to break down ((the breakdown pulse destroying the LED-s)) (or less likely the D7 & D8 to fail ... and then the Q2 Q3 to break down )
a poor simulation http://tinyurl.com/ybyapq4j giving an apx. idea of the opt.2
 
Last edited:

Thread Starter

SP180

Joined May 10, 2020
3
depending on the specific type of the D7 , D8 and the value of the Rr it may regulate the current "quite well"
it works basically the same as the current mirror . . . basically . . . if you disconnect D1/D2 it will put a bit less current through the Q2 & Q3

the D7 & D8 have relatively little voltage change when the current through them changes ⇒ if the voltages on the R1 ... R3 are not changing -- the base currents keep the same thus the Ic -s are near constant . . . if the Q1 no more has Ic it's BE diode has lower drop than Q2 Q3 that further limits their Ic

  • if the 1500mA is the CC limit it won't be pushing that to circuit
  • otherwise (inductive charge pump) your Vin would fast grow causing the Q2 and Q3 to break down ((the breakdown pulse destroying the LED-s)) (or less likely the D7 & D8 to fail ... and then the Q2 Q3 to break down )
a poor simulation http://tinyurl.com/ybyapq4j giving an apx. idea of the opt.2
Thanks for the response and simulation!

The technical note stated what you described further down, which I missed despite reading over it multiple times.
The power supply source, which has a constant current source, tries to supply the
required current, and will raise the output voltage up to the power supply output voltage limit.


On the below circuit I work out (if my maths are correct) to having a PD worst case of 15.3V over each transistors if one string is open. Given the Ic mA will drop as demonstrated by your simulation will the PD (Vc*Ic) not be sufficiently within the range of the BJT?

The PTOT Tc of the BD139 is 12.5W @ 25°C or 5W @ 100°C.

LED balancing circuit 2.jpg

Thanks.
 

Thread Starter

SP180

Joined May 10, 2020
3
One other thought, what if I put an LM2596 Adjustable DC/DC Step Down between the power supply and drivers setting the Vout of the buck converter to Vft? So if a string opened the CC drivers current would be limited by the mirror and the the CC drivers voltage would be limited by the buck so the voltage dissipation happens over the buck.

Of course this would be more inefficient as the buck would have to dissipate the extra voltage 24/7 whereas the BJT would only have to dissipate the extra voltage if a string was open until the LED was replaced.

https://hobbycomponents.com/power/215-lm2596-dc-dc-3-35v-adjustable-step-down-power-supply-module

Thanks,
Steve
 

ci139

Joined Jul 11, 2016
1,463
a MOSFET v. http://tinyurl.com/ycfb262g in real the mosfets have to trimmed individually with pot resistor dividers (because they vill have different V.gate I.drain dependencies)
basically the circuit is adaptable for BJT-s -- with the difference that you need voltage reference and take from there CC feed for bases (each transistor needs to be trimmed individually again . . . + you can add 0.33 to 1 ohm emitter resistors for T compensation) . . . like http://tinyurl.com/ycojwg5k
 
Last edited:

ronsimpson

Joined Oct 7, 2019
680
I found a LED constant current power supply that will drive all your LEDs in series. 300mA not 375 but it makes things simple. (12 LEDs x 4V=48V)
LED driver
Input is 110 or 220 vac
Out is 300mA and the different models have different ranges of voltage. DC24-42V or DC36-63V
1589818800840.png
 
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