I am designing a DC-DC boost converter for Vin = 5V to Vout = 15V. To generate gate pulses, I will be using a NE555P Timer by TI - http://pdf1.alldatasheet.com/datasheet-pdf/view/161277/TI/NE555P.html. For astable operation, from the data sheet I realised the following data [Pg-7]:

Frequency (f) = 10 KHz, Capacitance (C) = 0.1uF, Ra + 2Rb = 1K, Duty Cycle (D) = 1 - (Vin/Vout) = 0.66 or 66.6%,

Priod = th + tl = 0.1ms,

Now, how do I calculate the resistance values of Ra and Rb for the desired duty cycle?. And what is the difference between Output driver duty cycle and Output waveform duty cycle?. Which one of these formulae can I use to find Ra and Rb satisfying Ra + 2Rb = 1K?

              //Some Calculations I did until now to find Ra and Rb//

D = Rb/(Ra+2Rb)), 0.66 = Rb/1K, Rb = 660 ohms,

Ra + 2Rb = 1K, Ra + 2(660) = 1K, Ra = 1000 - 1320, Ra = -0.32K????-----------> Getting a Wrong value since it should not come negative!!,

                     //My next try to calculate Ra and Rb// 

D = 1-(Rb/(Ra+2Rb)), 0.66 = 1-(Rb/1K), Rb = 340 ohms,

Ra+2Rb = 1k, Ra+2(340) = 1K, Ra = 320 ohms,

So, 320 +2(340) = 1000, satisfies Ra + 2Rb = 1K( But not getting 66% duty cycle).

So which one is correct? or my approach and understanding itself to finding the values of Ra and Rb is wrong?. Please help.

  • \$\begingroup\$ The difference is that the output amplifier is inverting, therefore its duty cycle is 1 - the astable duty cycle. Use whichever equation is physically realisable, and if necessary, add an inverter after the output to get the waveform you need. \$\endgroup\$ – Brian Drummond Jun 15 '15 at 9:58
  • \$\begingroup\$ Are you saying that, D = 1-0.66, so the output duty cycle I see on the oscilloscope is actually 34% (ON) and not 66% (ON)?. So, to get the 66% duty cycle I need to use an inverter at the output at any case? \$\endgroup\$ – PsychedGuy Jun 15 '15 at 11:30
  • 1
    \$\begingroup\$ Yes, with the standard application schematic, the smallest dutycycle you can get is just below 50%. This is because charging uses Ra + Rb but discharging only Rb. Therefore discharging always takes less time and ducy < 50%! You can work around that as you said with an inverter but there is another trick using a diode, see: electronics.stackexchange.com/questions/57226/… The diode takes care Ra is for charging only and Rb for discharing only, solving your problem ! \$\endgroup\$ – Bimpelrekkie Jun 15 '15 at 14:35

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