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I am designing a battery charger circuit controlled by a microcontroller. Once the battery charges at constant current it switches to constant voltage and hence I need to design a switch that will be controlled by the microcontroller to switch between these two circuits.

I designed the following circuit, where an NPN transistor will act as a switch to control the PNP transistor connected to it. It will basically switch it on and off. However, the circuit works well for up to 0.35Amps. Increasing the current beyond 0.35A, will cause very inaccurate readings. What might bevthe problem? Am I biasing the transistors wrongly or taking incorrect readings/decisions in my circuit that will cause the switching circuit to function badly beyond that certain point? I want my circuit to work up to max around 0.8A.

Thanks in advanceenter image description here

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    \$\begingroup\$ Have you considered using a P-channel MOSFET for Q2? \$\endgroup\$
    – brhans
    Commented Nov 30, 2020 at 19:55
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    \$\begingroup\$ No, a P-channel MOSFET requires a Gate voltage 'more negative' than its Source voltage. Since you'll be connecting the Source to your +7V rail you'll have no trouble pulling the Gate below that (just as you're doing now using Q1 to pull Q2's base down). \$\endgroup\$
    – brhans
    Commented Nov 30, 2020 at 20:06
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    \$\begingroup\$ @brhans is right. All you need to do is to find a PMOS with low Vgs(th). The ones with -2V to -4V would be sufficient. Using a PMOS is a better option because saturating the TIP32C for 0.8A requires higher base current. By the way, Beta is meaningless for a saturated BJT. Here's a practical tip: Let Bx is the one tenth of the minimum Beta for desired collector current. Select a base current of \$I_B=I_C/B_x\$. This guarantees the saturation. Again, Beta is not used for a saturated BJT. \$\endgroup\$
    – Rohat Kılıç
    Commented Nov 30, 2020 at 20:16
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    \$\begingroup\$ No again. To switch the PMOS on requires that you pull the Gate lower than the Source by at least the PMOS's Vgs(th) value. To switch the PMOS off requires that you allow the Gate to rise so that the Gate-Source voltage is less than Vgs(th). So for example if the source is held at 7V and Vgs(th) is 2V, then pulling the gate below 5V will turn the PMOS on and letting the gate rise above 5V will turn the PMOS off. No negative voltages are required, nor are any voltages above your 7V. \$\endgroup\$
    – brhans
    Commented Nov 30, 2020 at 20:34
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    \$\begingroup\$ " Increasing the current beyond 0.35A, will cause very inaccurate readings." - what readings, where? Exactly what does the switching circuit power? \$\endgroup\$
    – Bruce Abbott
    Commented Dec 1, 2020 at 5:14

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