It could if there was a fault on the board.ı can measure the input supply but ı think your guess is not correct. because the regulator has 2.2 A for short circuit limit. the lcd couldnt be draw that much current, ı think.
It could if there was a fault on the board.ı can measure the input supply but ı think your guess is not correct. because the regulator has 2.2 A for short circuit limit. the lcd couldnt be draw that much current, ı think.
okey you are right. it couldIt could if there was a fault on the board.
Thank you for finally sharing your circuit. The problem is that you have a floating ground. There is no DC path from your loads back to the power source.ı share my circuit right below. apologize for being late...
Your supply is all messed up. As you try to power something with the 5 V output the return current has no where to go. That messes up the balance between your capacitors and will pull your supplies in all kinds of strange directions.okey you are right. it could
but ı have a question. if regulator drops short circuit level, could ı be able to still measure 3.2 volts at the output of the regulator when lcd board is connected. ı think regulator gives zero volt in that situation.
just curious.
in our main regulator board, we have transformer that drops 220v ac to 17 volts ac. then, bridge rectifier converts that 17 v ac to dc. then the regulator use it.Do you have a centre-tapped transformer, with the centre tap connected to ground?
Do you need a -5 V output? If you do, then you really should use a center-tapped transformer. If not, then you need to use one side of the transformer as your ground.in our main regulator board, we have transformer that drops 220v ac to 17 volts ac. then, bridge rectifier converts that 17 v ac to dc. then the regulator use it.
yes.Do you have a centre-tapped transformer, with the centre tap connected to ground?
we used center tapped transformer with center tap connected to gnd. we have two +15V output and common gnd.Thank you for finally sharing your circuit. The problem is that you have a floating ground. There is no DC path from your loads back to the power source.
If you want to have bipolar supply outputs, then you need to either use a center-tapped transformer, or you need to use one side of the transformer output as your ground and use the other to drive a rectifier to pump the two sides on alternative halves of the cycle.
Then why doesn't your schematic show that? It's really hard to help when you show us a schematic for a circuit that you are not using.yes.
we have center tapped transformer with center tap connected to gnd. we want to get +5 and -5 volts output.Do you need a -5 V output? If you do, then you really should use a center-tapped transformer. If not, then you need to use one side of the transformer as your ground.
Actually ı know that and fixed it with connecting a resistor series in the input of the regulator. So, we have 17V at the input of the regulator .Your original schematic is missing the center-tapped transformer.
The rectified transformer produces almost plus and minus 20VDC which overheats the voltage regulators that need no more than
7VDC.
What value of resistor?Actually ı know that and fixed it with connecting a resistor series in the input of the regulator. So, we have 17V at the input of the regulator .
I KNOW THAT BUT FOR COMPLETE THE TESTS, I FIXED LIKE THAT TEMPOARARY. I WILL REVISED WHOLE CIRCUIT. I NEED TO SPECIFY ALL CORRECTIONS RIGHT NOW.The extra voltage causes lots of heating in the resistors and regulators. It also costs extra for the wasted electricity power.
1k ohmsWhat value of resistor?
How did you calculate 1k?1k ohms
you might be right but it is not my problem. my problem is about voltage drops when connecting the lcd board. if the current is below the quiescent value, then my regulator would not give 5 volts when nothing connected. but it gives 5 volts. so, ı think the problem is not about the current value.How did you calculate 1k?
I think that might be the problem. I note that it is conspicuously absent from your circuit diagram. You have limited your output current to about 4mA (the resistor limits it to 10mA and the 7805 has 6mA quiescent current)