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Glad to found an "electrical engineer"-version of stackoverflow, I greet you all.

I am working on a MPPT projects, using 5V solar panel, and MCU ATTiny24 (for its ability working on a low voltage).

The solar panel output is used for MCU voltage source and load. The problem is, PWM signal generated is not square if its on low duty cycle (see pics)

I also attach my schematics (sorry for the hand drawing, but I cant access the schematics right now) Now I am using PNP transistor C9012 for switching. My first guess is error caused by the transistor, anyone knows a P channel FET to switch low voltage? or maybe a P-channel substitution equal to MPF102?

When the duty cycle is higher than 60%, the signal is square.
I use load ranging from 1 ohm to 50k ohm.
Thanks in advance :) Ps: sorry i can only post 2 links because of lack of reputation Schematic

EDITED--------------------------------------------------------
I have try to modify the circuit as the suggestion, but still got the same result (and I'm sure the oscilloscope works). *sorry have to remove the former circuit, cant upload more than 2

enter image description here

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  • \$\begingroup\$ Post whatever you need \$\endgroup\$
    – user76844
    Commented Nov 4, 2016 at 10:18
  • \$\begingroup\$ For pmos look for bss84. Ah, load may matter. In steady state inductor current is at average like load current. \$\endgroup\$
    – user76844
    Commented Nov 4, 2016 at 10:21
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    \$\begingroup\$ Hi Gregory, thanks for joining. The load is just a resistor. no feedback from load (i think). It drops mostly when the resistance is too small, making the current for load is too big and decreasing current for avr. I'm terribly sorry if I am wrong, please let me try to keep up \$\endgroup\$
    – duck
    Commented Nov 4, 2016 at 13:58
  • \$\begingroup\$ Then it's inductance of pv or wires. Try adding some wires in parallel, see it there will be any effect. In any case it should not affect efficiency, i guess. \$\endgroup\$
    – user76844
    Commented Nov 4, 2016 at 17:03
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    \$\begingroup\$ I will make the input capacitance into 100uF and try using bss84 instead of PNP transistor. Sorry but I am a little confused in "add two wires coming to VCC and GND", I am sure you does not mean to short the circuit right? My English is not so good and so does my electronics comprehensive \$\endgroup\$
    – duck
    Commented Nov 7, 2016 at 6:15

2 Answers 2

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So now we know it's input capacitor. By the way, the MOSFET on your drawing is reversed. You should connect the source to VCC. It will conduct when the gate is pulled to zero.

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  • \$\begingroup\$ yeah sorry I didn't notice about the MOSFET drawing. Thanks! \$\endgroup\$
    – duck
    Commented Nov 9, 2016 at 6:05
  • \$\begingroup\$ I still have a question. Why not using integrated chip? \$\endgroup\$
    – user76844
    Commented Nov 9, 2016 at 6:40
  • \$\begingroup\$ which part do you mean? \$\endgroup\$
    – duck
    Commented Nov 9, 2016 at 6:44
  • \$\begingroup\$ At least the switch should be inside. Depending on your choice the diode may be inside either. \$\endgroup\$
    – user76844
    Commented Nov 9, 2016 at 6:46
  • \$\begingroup\$ Well since the purpose of this is educational, I try to build the circuit "as raw as it can" \$\endgroup\$
    – duck
    Commented Nov 9, 2016 at 7:18
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To answers you question completely, some missing information is needed. It depends if the load is connected when this waveform is generated and what value is. Is it resistive. The value of the base resistor is also needed to understand what is going on. The first thing you need to verify is the transistor turn on isn't causing the the solar cell voltage to drop. If this is the case, no transistor choice is going to make a difference. If the solar sell is not dropping when the transistor is turned on, try more base current. Do this by reducing the size of the base resistor. If you can't reduce the size of the base resistor, try adding a little capacitance across it.

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  • \$\begingroup\$ Hello, and thanks for the reply. The load is resistive, it occurs mostly when the load is lower than 10 ohm. About the transistor, I'm pretty sure its causing the voltage from solar cell dropping, because the transistor is used to drive a buck converter, which is also using voltage from solar cell. Don't you think adding capacitor will make it worse? because the capacitive traits will make the signal "less square"? \$\endgroup\$
    – duck
    Commented Nov 4, 2016 at 14:04
  • \$\begingroup\$ If the voltage at the solar sell is dropping similar what you see in your scope photo, no different device will fix this. The capacitor on the base resistor will also not help. You could try more capacitance across the solar cell. Under heavy load, more current will be taken out of the solar cell when the transistor takes over form the diode. This current will cause a drop across the solar cell source resistance causing the drop. The drop will appear at both ends of the switch. More capacitance will help. Otherwise you need a bigger badder solar cell. \$\endgroup\$
    – owg60
    Commented Nov 4, 2016 at 14:19

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