Step 5Create a Voltage Divider Circuit
From my experimenting I've discovered that many inverters expect 12V+ input to drive the inverter and and around 5V+ to "enable" and set the "dimming" level. Check this generic spec sheet for some guidance: http://www.lcdinverter.co.uk/MH-1405A04-spec.htm. From my testing, the circuits are quite robust and can accept between 4.5 and 7 volts on the enable and dim pins and work properly. Above 7 volts a suspicious whining tends to be emitted.
A 12V DC power source can be easily adapted to power the inverter by using a voltage divider circuit. In a voltage divider circuit, use two resistors to drop the voltage as needed. In this circuit R1 drops the voltage 7V then R2 an additional 5V. Notice that 7V+5V=12V. The schematic below shows the circuit I used to create the voltages I needed to drive my inverter.
In this circuit, C is Ground, A is 12V+, B is 5V+. For an inverter that has enable and and dim pins, connect those to 5V+ to turn the screen on.
Prototype your circuit using a breadboard. Measure the voltage between C and A; it should be 12V+. Measure the Voltage between C and B and it should be 5V+. If you get values within 10-20%, you should be OK.
If you need assistance in choosing resistors for your particular voltage source, check out the Circuit Design Tutor.
Some tips to keep in mind:
- If the sum of the ratio of the resistors equals the input voltage, your design will be much simpler. For example Vin=12V, ratio of resistors is 50:70 or 5:7 - 5+7=12.
- Remember you can simply add resistors together in series to create a single resistor (read up on Ohms Law for help here.
- If your input voltage is 18V, R1 should be 130K Ohm, R2 should be 50K Ohm.
A: 12V+
B: 5V+
C: 12V-
R1: 70K Ohm Resistor
R2: 50K Ohm Resistor
K1: SPST switch
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