When you have sized batteries is 1.5V or 3V, but need a voltage 9VDC, DC converter circuit to help you for sure!, This is 1.5v to 9v step up dc converter circuit using TL496
Some problems in the electronics job, can modify or bring devices come to apply available not difficult If we know the principles and limitations of those devices. As the project is being offered at this time, will be bringing sized batteries 1.5 volt connected through an electronic circuit called dc converter to increase the voltage to have higher value be 9 volt instead of the battery 6 pack, connected series.
It can also increase the voltage from the battery 3 volt to 9 volt as well be the original circuit, and it will allow more current.
Properties of the IC TL496
– A TL496 be control circuit of the power supply, by the circuit was designed, so that the output voltage is 9 volt. For the change of the input voltage, which comes from the battery 1 or 2 pack. The TL496 operates in a switching regulator, the power loss is less than the typical linear regulator.
3V to 9V dc converter
How it works.
– When connecting the battery to the input pin 2, which may be a one battery pack (voltage 1.1 -1.5 volt) or 2 pieces (pressure 2.3 – 3 volt). At low voltage circuit starts. This low voltage circuit starts, the feedback voltage to bias to transistors inside the integrated circuit. This means that a short circuit, L1 is connected to pin 6 to ground, resulting in the accumulation of energy in the form of a magnetic field on current in L1 increases until a maximum value at about 1 amp.
– The IC internal circuit will reset itself. As a result, the transistor stops (open circuit), so there is no current flow, the magnetic field in the coil L1 will collapsing. The induced current flows into At pin 6, lead to increased pressure within the IC be 9 volt out to pin 8(output pin)
How to build
– Since this project using very small amount, We may use the anchor cable replacement PCB time, or if you want a nice tidy it up a PCB. Which has copper pattern which is shown in Figure 3 is the position of the device.
Based on circuit in Figure 2, will seen that there is no point jumping legs 3 and L1., If a single battery. I used jumper cables to connect the two together, which will be have the DC 9 volt output voltage and current about 40 mA.
Then if insert battery two pack or 3 volt entered to input. to disconnect the jumper out. This will increase to current (about 80 mA) at the jumper cables are blanked. May will used to switch on / off. come to instead for ease in use.
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