reducing DC voltage in a circuit?

Thread Starter

jzeds1491

Joined Feb 5, 2020
16
Hi everyone, I'm new here and I hope that we have a great time here in this forum!
I have a mp3player that has blown the led backlight of LCD, so I want to fix this.
I measured the input voltage for leds and it was 24volts for 3 LEDs! (new leds get very hot under load and I know they might burn too) so I want to reduce voltage to 9 volts if it's possible.
I was thinking about shorting this inductor :
(it converts 3.7v to 8-9volts )

also the inverter transistor (D256A) next to this inductor gets very Hot without LEDs connected.
 

Beau Schwabe

Joined Nov 7, 2019
72
Shorting the inductor is likely to cause more issues ... you're better of removing it rather than shorting it, but personally I wouldn't do either. You say you measured 8V or 9 V on the other side of the inductor and that you would like to reduce the 24V to 9V so it seems as though there might already be a solution there to try. Albeit a band-aid patchwork approach, but one that will have a minimal destruction impact to your mp3player ... Dare I ask where the 24V is coming from?


Alternatively you can try a proper value resistor in series with the 24V output and the LED's

Rled = ( Vsupply - Vled ) / Iled

Rled = ( 24V - 9V ) / 10mA
Rled = 15V / 10mA
Rled = 1.5k resistor
 

Thread Starter

jzeds1491

Joined Feb 5, 2020
16
Shorting the inductor is likely to cause more issues ... you're better of removing it rather than shorting it, but personally I wouldn't do either. You say you measured 8V or 9 V on the other side of the inductor and that you would like to reduce the 24V to 9V so it seems as though there might already be a solution there to try. Albeit a band-aid patchwork approach, but one that will have a minimal destruction impact to your mp3player ... Dare I ask where the 24V is coming from?


Alternatively you can try a proper value resistor in series with the 24V output and the LED's

Rled = ( Vsupply - Vled ) / Iled

Rled = ( 24V - 9V ) / 10mA
Rled = 15V / 10mA
Rled = 1.5k resistor
Thanks for the reply!
Doesn't a resistor generate much heat too? my problem is the heat that is focused on either LEDs or D256A transistor. (which burns LEDs or drains battery)
I measured the inductor without LEDs and I read different numbers each time (variable from 40 to120volt!) that is connected to a schottky diode (output is 24 v at this diode)
the input voltage of inductor is 3.5v
take a look at this sketch
 

DickCappels

Joined Aug 21, 2008
6,659
... Therefore, don't mess with the inductor. When the correct LEDs are installed they will limit the voltage to the amount needed to get the required current through the LEDs.

1580907156174.png

It is a very simple circuit. The current through the LEDs is sensed as the IR drop across R2. Most likely one of the LEDs opened. If you are getting an output voltage that is higher than the input voltage, then probably everything is working except one of the LEDs.
 

Thread Starter

jzeds1491

Joined Feb 5, 2020
16
hi jz,
This is the datasheet for that IC, it is a SMPS switcher.
Resistors R1 and R2 set the Vout.
Calculate the values for a lower Vout.
E
View attachment 198340
I have been searching and asking every electronics forum for past two days, How did you recognize this chip?! this is exactly what I needed thank you so much, God bless you wherever you are.
I will use the datasheet and the formula to fix my issue and I'll post the results.
-only one last question since the other user(@DickCappels) said the chip will adjust voltage by itself, do I still need to replace resistor?

... Therefore, don't mess with the inductor. When the correct LEDs are installed they will limit the voltage to the amount needed to get the required current through the LEDs.

View attachment 198341

It is a very simple circuit. The current through the LEDs is sensed as the IR drop across R2. Most likely one of the LEDs opened. If you are getting an output voltage that is higher than the input voltage, then probably everything is working except one of the LEDs.
Thanks so much, I used normal sized smd LEDs before proceeding to replacement (and testing)of those backlight LEDs(because I only have 3 of these) , and are you suggesting that I don't need to worry when connecting these flat SMD LEDs? I don't need extra steps like changing resistor ? all adjusted by ic itself?
 

DickCappels

Joined Aug 21, 2008
6,659
If the LEDs you use for replacement are roughly the same as the ones you are replacing you should be fine. LEDs are best when driven at a constant current, not a constant voltage.
 

Thread Starter

jzeds1491

Joined Feb 5, 2020
16
Just put a 1.5K resistor 1/4W in series with the 3 LEDs, this will give you approx 10mA led current.
Hi, I added the 1.5k resistor and everything is good and cool at this part But that DC converter ic gets super hot and something whistles in the circuit. (maybe inductor)

If the LEDs you use for replacement are roughly the same as the ones you are replacing you should be fine. LEDs are best when driven at a constant current, not a constant voltage.
Update:
I connected the new screen leds, at startup it draw 0.4 amps after a couple of seconds when menu appears, it doubled the current through to 0.8A!
LEDs turned blue and I shut the power off. Fortunately LEDs survived.
Now I'm going to change R2 resistor
 

Thread Starter

jzeds1491

Joined Feb 5, 2020
16
hi jz,
By using a web search.
E
www.ic37.com › LM2703MF-ADJ-NOPB_datashee...
LM2703MFX-ADJ (TI) PDF技术资料下载LM2703MFX-ADJ ...

Translate this page
LM2703MFX-ADJ PDF技术资料下载LM2703MFX-ADJ 供应信息LM2703 www.ti.com SNVS172F – FEBRUARY 2002 – REVISED ... LM2703MFX-ADJ D256A.
You visited this page on 04/02/20.
Sorry to bother you again, I replaced the resistor with higher value (up to 100x) but no significant difference happened here, neither current or voltage ...
Now I am thinking maybe it's the schottky diode cause it's handling the current through. Datasheet says forward current is 350mA but I'm reading at least 800mA each time, How can I test it?

 

Dzoro

Joined Feb 1, 2019
161
Hi everyone, I'm new here and I hope that we have a great time here in this forum!
I have a mp3player that has blown the led backlight of LCD, so I want to fix this.
I measured the input voltage for leds and it was 24volts for 3 LEDs! (new leds get very hot under load and I know they might burn too) so I want to reduce voltage to 9 volts if it's possible.
I was thinking about shorting this inductor :
(it converts 3.7v to 8-9volts )

also the inverter transistor (D256A) next to this inductor gets very Hot without LEDs connected.
Just use an linear voltage regulator lm7809 with a big heataink beacose you will drop 15 volts accros it so it will get very hot without heatsink you just need 2 capacitors on the input and output for stabilization its not efficient but ita cheap unnamed.gifhire is a small circuit dont mind the bottom one just focus on the top one
 

Thread Starter

jzeds1491

Joined Feb 5, 2020
16
Just use an linear voltage regulator lm7809 with a big heataink beacose you will drop 15 volts accros it so it will get very hot without heatsink you just need 2 capacitors on the input and output for stabilization its not efficient but ita cheap View attachment 198480hire is a small circuit dont mind the bottom one just focus on the top one
thanks but I'm sure I can't fit lm7809 or 200mah battery cannot handle this amount of energy consumption.
-now the goal is to reduce the current from 0.8A to at most 100mA(max33mA per Led)
 

djsfantasi

Joined Apr 11, 2010
6,975
thanks but I'm sure I can't fit lm7809 or 200mah battery cannot handle this amount of energy consumption.
-now the goal is to reduce the current from 0.8A to at most 100mA(max33mA per Led)
Hi! New to this thread, but your last statement struck me. LEDs are current driven devices. You can’t as a practical matter reduce the current to them without reducing their performance. Having said that, an LED will operate as low as 5-10mA albeit as a dimmer light. It is possible to calculate a series resistor at the lower current.

The resistor value,R, is calculated as:
R=(Vs-n*Vf)/Id​
Where Vs is supply voltage, Vf is the forward voltage of the LEDs, n is the number of LEDs in series and Id is the desired current in amps.
 

Thread Starter

jzeds1491

Joined Feb 5, 2020
16
Hi! New to this thread, but your last statement struck me. LEDs are current driven devices. You can’t as a practical matter reduce the current to them without reducing their performance. Having said that, an LED will operate as low as 5-10mA albeit as a dimmer light. It is possible to calculate a series resistor at the lower current.

The resistor value,R, is calculated as:
R=(Vs-n*Vf)/Id​
Where Vs is supply voltage, Vf is the forward voltage of the LEDs, n is the number of LEDs in series and Id is the desired current in amps.
Thanks, but the thing is if I use resistor the ic gets hot, it's like heat goes from LEDs to IC
 

DickCappels

Joined Aug 21, 2008
6,659
I would check the voltage on pin 3 and see whether or not it is above 1.237 volts. If pin 3 is above 1.237 volts the regulator is broken.
 

Dzoro

Joined Feb 1, 2019
161
thanks but I'm sure I can't fit lm7809 or 200mah battery cannot handle this amount of energy consumption.
-now the goal is to reduce the current from 0.8A to at most 100mA(max33mA per Led)
You can make that with some tranzistors making them in constant current because leds work better with vonstant current than constant voltage

Just search on google about tranzistor constant current you will find a lot
 
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