How to change voltage in a circuit

Thread Starter

whathaveidonenow

Joined Feb 9, 2020
9
I am replacing an LCD screen with an OLED screen and the supply voltage is supposed to be 5 volts. 4.8 min to 5.3 max. Is it possible to change the voltage for pin 14 from 6 to 5 volts? There is a stable 9 volts feeding this circuit. Is it possible to make the changes in the red highlighted area? And could you provide an explanation of what is happening in that section of the diagram? Thanks in advance for any help LCD2.PNG
 

sagor

Joined Mar 10, 2019
757
A couple of ideas. The current DA1 provides about 1.4V to 2V voltage drop at around 10mA. So, pin 2 of C26 would have about 7V when there is a 10mA load on it. Not quite the 6V as the schematic, but that may be an "approximate" value.
To get to 5V, you would need to know the current draw at pin 14. Measure it (current load of OLED) first with a bench 5V supply. Then, you could add more diodes in series with DA1 to provide further 0.7V drops per diode until you get close to 5V. You have to measure the voltage with the same OLED load as when pin 14 was tested.
The other option is to install a small 5V regulator like a 78L05 replacing DA1. Put a bypass capacitor on the input pin of the 78L05. The output C25 and C26 should be good enough for output regulation. The current limit of a 78L05 is 100mA, make sure the OLED does not take that much power. If it does, you will have to use a regular LM7805 regulator that can handle about 1A with a heatsink, less without a heatsink.
EDIT: At 10mA load, R5 will also produce a voltage drop of 2.2V. So, voltage feeding DA1 will vary depending on the current load. Too high a current, and R5 may have to be changed to a lower value, to produce a smaller voltage drop.
In summary, at 10mA, R5 gives 2.2V drop, DA1 gives at least 1.4V drop, total drop of 3.6V from 9V = 5.4V at pin 14. It all varies with the load current!!
 
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Thread Starter

whathaveidonenow

Joined Feb 9, 2020
9
A couple of ideas. The current DA1 provides about 1.4V to 2V voltage drop at around 10mA. So, pin 2 of C26 would have about 7V when there is a 10mA load on it. Not quite the 6V as the schematic, but that may be an "approximate" value.
To get to 5V, you would need to know the current draw at pin 14. Measure it (current load of OLED) first with a bench 5V supply. Then, you could add more diodes in series with DA1 to provide further 0.7V drops per diode until you get close to 5V. You have to measure the voltage with the same OLED load as when pin 14 was tested.
The other option is to install a small 5V regulator like a 78L05 replacing DA1. Put a bypass capacitor on the input pin of the 78L05. The output C25 and C26 should be good enough for output regulation. The current limit of a 78L05 is 100mA, make sure the OLED does not take that much power. If it does, you will have to use a regular LM7805 regulator that can handle about 1A with a heatsink, less without a heatsink.
EDIT: At 10mA load, R5 will also produce a voltage drop of 2.2V. So, voltage feeding DA1 will vary depending on the current load. Too high a current, and R5 may have to be changed to a lower value, to produce a smaller voltage drop.
In summary, at 10mA, R5 gives 2.2V drop, DA1 gives at least 1.4V drop, total drop of 3.6V from 9V = 5.4V at pin 14. It all varies with the load current!!
Thank You for the quick response. I will need to think about this for a while. The spec sheet of the OLED states "supply current" at 5 volts, 35 mA typical with 30mA min to 42 mA max. Would you need to know the load of the old LCD screen?
 

sagor

Joined Mar 10, 2019
757
No, you really need to know the load of the new display. Assume 42ma max. That would give over 9V drop over R5, so that will not work.
At 35ma, a 56 ohm resistor for R5 will match what existed before. almost 2V drop plus 1.4V drop from diodes, giving 5.6V final. Still too much.
Lets back up a bit. At 30ma, you want 5.2V (below 5.3V max). Count 1.4V for diodes, makes it 6.7V after R5. 9v-6.7V leaves 2.3V drop required by resistor at 30ma. That gives R5 of 76.66 ohms, so a 75ihm will match that closely.
Now, at 43ma maximum, the 75 ohm resistor will give a voltage drop of 3.225V. Add the 1.4V of the diodes, any you get 4.625V total drop from 9V, giving a final voltage of 4.375V. That may be too low now...
Thus, using a voltage dropping resistor and diode combination as shown will not work for the range of 30ma to 43ma loads.
Next solution would be to add another silicon diode (0.7V drop) and rely less on the resistor. So, assume total diode drop of 2.1V (3 diodes in series). Re-do the math....
30ma to get 5.2V - 2.1V from diodes means 5.2+2.1 = 7.3V From 9V, you need 1.7V drop from resistor. That makes it a 56 ohm resistor. Then, at 43ma, you get 2.4V drop plus 2.1V drop 3 diodes = 4.5V drop. From 9V source, you get 4.5V left for display. Again, not enough voltage....
So, add one more diode (4 in total). That makes a 2.8V drop (on average). At 5.2V plus 2.8V, you get 8V after R5. That means R5 needs to drop 1V at 30ma. Math says it becomes a 33 ohm resistor. Now, at 43ma, it gives 1.42 volt drop. Add the 2.8V diode drop, you get 4.22V drop at 43ma. That leaves 9-4.22 = 4.78V. This might work!
You need to test the diode voltage drops at the 30ma and 43ma load levels. It looks like a 33 ohm resistor for R5 and 4 diodes in series may give you an operating voltage range that will work.

After all that, if too complicated, go with the 78L05 voltage regulator. I would suggest changing R5 with the 78L05 to a 33 ohm resistor as well.

Good luck.
 

dcbingaman

Joined Jun 30, 2021
720
Replace DA1 with a LM7805 Voltage regulator (input to 'pin 1', output to 'pin 2'). Replace R5 with a 0 ohm resistor. Connect the 7805 middle terminal (ground) to circuit ground. The voltage is getting down to 6V in the original circuit by relying on so much current going through DA1 and R5. This is a very bad circuit design as any changes in load current will change the voltage. With the 7805 you can have all the changes in load current you want with no change in voltage (far superior) IMHO.
 
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