I am trying to design a voltage inverter based on old/cheap/jelly bean parts. I wanted to use the MC34063. Mostly for learning purposes, so I don't look for easier parts to use for now.
I run the calculations, and I also derived the formulas on the datasheet to really understand the math behind. Then I went to practice, and it seems to work fine - except for a corner case with lower input voltage. FYI, from the following requirements, I have come to the same values which can be extracted from a website like http://www.nomad.ee/micros/mc34063a/index.shtml.
- Input voltage: 4.5V to 9V
- Output voltage: -25V
- Output current: 10mA (before goind DCM) to 50mA at Vin min
- Inductor current ripple: +/- 30% (not important now)
- Output voltage ripple: +/- 1% (not important at all now)
- Switching frequency: 50 kHz
And I used the inverting configuration:
With these parts (relative to above picture):
- R3 = 0.33 Ohm
- C2 = 47μF (electrolytic)
- R2 = 49.5 kOhm
- R1 = 2.49 kOhm
- C3 = 680 pF
- D1 = V4PAN50-M3/I
- L1 = SRN1060-221M
- C1 = 47μF (electrolytic)
Specifically, with these values, I get a peak inductor current of 900mA, for a maximum output of 55mA at the theoretical minimum of 4.5V input, considering a 1.3V of saturation voltage of the switch.
The calculations are made to be sure that the inductor ramp, with the maximum on-time of the switch, is enough to charge the output capacitor at the given voltage, and sustain the given load.
For example, with a Vf of the diode of 0.5V, Vin min of 4.5V, saturation voltage of 1.3V and output voltage of -25V, from (also provided in the datasheet):
Which brings to 18sμ of maximum on-time.
From my tests, at 50kHz and with an input of 4.5V I do have a -25V unloaded, but with a few mA drops immediately to -14V.
I can have a 25mA load at -25V only from 6.5V in input.
To be sure that the problem is the inductor that does not have enough time to load, I tried to lower the frequency to 35kHz. Indeed now there is more time, and I can sustain a 25mA load already at 5.9V.
So the question is, if is a problem of the on-time, why the calculations are not fitting the reality by a lot?
If it helps, here is the acquisition of the scope at pin 1 (in yellow) and 2 (in light blue), during the 25mA load, and also showing how we are having a correct maximum on-time:
As a side question on something else that is not clear, the "long negative slope" on pin 1 is the current increasing in the inductor, as I think increases the drop on R3. I am just clueless on the meaning of the yellow during the off-time (when the inductor is pushing the current into the capacitor), as I should not see this shorter slope on pin 1. Is this due to some coupled capacitance of the internal Darlington to the input?