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I built the following circuit:

enter image description here

The boost onverter works fine when applying 0.5 time ON PWM signal (Vout = 2.5V) but when I try to increase the voltage using 0.8* time PWM signal the Vout** stays about the same.

*Vout = 1/(1-tON) **Vout relative to ground

Why does that happen and how to fix it?

TECHNICAL DATA: Inductor current = 30mA. DCR = 16 ohm.

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    \$\begingroup\$ Q1 needs a base resistor or a very carefully controlled base voltage. \$\endgroup\$ – Andy aka Mar 6 at 12:08
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    \$\begingroup\$ you'll need a MUCH larger inductor if the PWM is only 1khz and the input is only 1.2v. you would also be much better off with a logic-level n-channel fet instead of a BJT (given the 1.2v input) \$\endgroup\$ – dandavis Mar 6 at 19:40
  • \$\begingroup\$ What is the DCR of L? and driver of BJT ( 1.2V CMOS? ) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 7 at 3:44
  • \$\begingroup\$ What is the drive current and Vbe? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 7 at 12:14
  • \$\begingroup\$ ok so I*DCR =ΔV= 30mA*16=480mV which is 40% of your supply voltage. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 7 at 12:15
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schematic

simulate this circuit – Schematic created using CircuitLab

Most probable cause is that inductor goes in saturation. If we ommit the resistance of the inductor, then the current trough inductor is $$ i_l=\dfrac{1}{L}\int u\cdot dt$$ where u is the voltage, for example 1.2V and time is 0.8*1/960 = 0.00083, so the peak current is: 1.2*0.83/100= 10mA.

Check the inductor data what's the saturation current.

EDIT:

It has to be noted, that this exaple is oversimplified. The main simplification is that current is trough inductor is discontinouos. For better explanation you should use some simulation software.

You can click on simulation.

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Lack of details for design data such as component specs and driver current will always lead to unexpected results.

Your load is pretty small, Vout/R= 2.5V/4k7=0.53mA and at 50% Ipk= I=V/Ldt= 1.1V/0.1H*0.5ms = 5.5mA pk

If the inductor is a small size the L/R ratios can be a wide range. This is important. As the duty cycle increases, the current can no longer rise at the rate of dI/dt=V/L if the voltage is reducing internally due to series resistance call DCR.

A 100mH is pretty large and so the DCR also tends to be large ranging from 0.6 to 700 Ohms in small parts. R/L=T needs to be small (<<2% ) compared to the load and also say 100R/100mH = T << 1/f=10ms. So lowering the frequency will increased the output voltage until the IR drop becomes significant or L saturation current is reached and L drops in value.

Conclusion

State your DCR and L Sat current, and datasheet link.

So lower f or change L to a smaller part and put forth some design specs next time.

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  • \$\begingroup\$ Please see update \$\endgroup\$ – Danaro Mar 7 at 11:32

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