The theory behind step up converter is very fascinating. It literally works like a voltage pump. If you would like to find out more about the theory I recommend the video of Prof. Sam Ben-Yaakov about them.
The basic idea is to switch the voltage of circuit on and off so that an inductor works as voltage pump and pumps voltage into a capacitor. This way you get a higher output voltage than the input voltage.
There are several IC available for that kind of circuit. I decided to go for a traditional solution: The MC34063
Depending on the peripherals the MC34063 can work as step down converter, voltage inverter or as step up converter. It is able to deliver up to 1.5A at a maximum switching frequency of 42 kHz. Its operating voltage is between 3V and 40V input.
This was my first project in which I tried to design my on circuit. And I was very surprised to learn that designing in this case means to stick to recommendations of the data sheet: There are circuits available for all three use cases. TI even provides an Application Report. So, here’s the basic circuit from TI’s data sheet of the MC34063A:
TI advises to use am 1N5819 Schottky diode, but this type was not deliverable, when I ordered my parts. So I ordered a 1N5817 diode instead. They are pretty much the same, but do have a higher forward voltage (0.45V compared to 0.35V of 1N5819).
Next step is doing some Maths. TI delivers following table in their datasheet:
You need to start in the first row with the calculation. I was too lazy to do it by myself, so I just set up a Numbers sheet (Excel or similar is also suitable). This way I was able to play around with the values which helped me to understand the circuit a little better. There are some websites out there you can use, but be careful some of the sites I tried did not solve the equations correctly.
| Name | Value | Unit | |
| vin | 3,7 | V | Input Voltage |
| vinmin | 3,3 | V | Minimum input voltage (vin -10%) |
| vout | 5 | V | Output Voltage |
| Iout | 0,25 | A | Output Current |
| fmin | 33000 | Hz | Minimum output switching frequency |
| vripple | 0,1 | V | Desired peak-to-peak output ripple voltage |
| vF | 0,5 | V | Forward Voltage of Schottky Diode (datasheet of 1N5817) |
| vsat | 0,5 | V | Saturation voltage of output switch (datasheet of mc34063) |
| R1 | 2200 | Ohm | |
| ton/toff | 0,7857 | ||
| ton+toff | 30,30 | us | |
| toff | 17 | us | |
| ton | 13,3 | us | |
| CT | 533 | pF | |
| Ipk_switch | 893 | mA | |
| RSC | 336 | mOhm | |
| Lmin | 42 | uH | |
| CO | 300 | uF | |
| R2 | 6600 | Ohm |
I was designing this circuit for the operation of a ESP32 Cam. The ESP32 can consume up to 140 mA if the CPU is running on 240 MHz. I am planing to run the camera with infrared LEDs attached to it. So I decided to use a maximum output current of 250 mA (Iout in the table above).
The typical switching frequency of the MC34063 is 33 kHz at 5V. The maximum is 42 kHz at 5V. As my input voltage will be lower I decided to not use the maximum frequency but instead use the typical frequency. With that frequency and my desired maximum output current I am still able to keep the limits of the Schottky diode: Its maximum current is 1A which is a little above the calculated value of 0.89 A. But I think that’s ok as I will not draw the maximum current for a long time.
As output ripple I defined 0.1V. I do not know if this is a good value or not, but somehow it seemed to be a value that allows to describe the output voltage as stable. Unfortunately I do not own a oscilloscope (yet), so I can not prove if my calculation worked out fine. At least my multimeter do not show a ripple. But I do not want to take this as proof.
Required Hardware
- 1 x MC34063A
- 1 x 8-pin IC socket
- Resistors
- 1 x 2.2 kOhm (R1)
- 1 x 6.8 kOhm (R2)
- 1 x 0.33 Ohm (RSC – I used 1 Ohm (1 Watt) instead)
- 1 x 180 Ohm (connected to pin 8 of MC34063)
- Capacitors
- 1 x 533 pF (CT – I used 470pF instead)
- 1 x 300 nF (CO)
- 1 x 100nF (Filter at output)
- Inductance
- 1 x 220 uH
- 2 x Screw Terminals
- 1 x Schottky diode 1N5817 (better: 1N5819)
Circuit
Soldering
After I calculated the values and ordered all necessary parts, I drew the circuit diagram in Fritzing. Fritzing also allows to layout the perfboard, but unfortunately the Fitzung part library did not have the right terminals and inductors, so the layout was just „kinda“ helpful. But at least I learned that my perfboard needs to be 7 rows high and 21 columns broad. After I finished soldering the first step up converter (I made three) I decided that 8 rows and 22 columns are better.
As Fritzing was not the perfect tool for me I decided to layout my perfboard in a classic manner: With paper and a pencil:
As you can see it took me several approaches before I was happy with the layout. There is a reason why I wrote „Front“: If you solder the lines like on the picture above, everything will be mirror-inverted. And unfortunately this will not work as the pins of MC34063 will not be mirror-inverted. Therefore you have to flip over the paper and redraw the lines on the other side. The backside will then be the master for soldering.
And to be able to decide whether it is the side to be soldered or not I just wrote „Front“ on the front side of the drawing. Unfortunately I did it to late: I had to redo the second version of the step up converter, because the paper was laying with front side on top and I did not realize.
But in the end everything turned out well. At the time of ordering I was very confident that I had 300 uF capacitors laying around. But I hadn’t. So I just used three 100 uF capacitors in parallel instead. As you can see in the picture, I tried different capacitors: 25V and 16V.
Next step: I will do some measurements. I am interested in the current consumption and the efficiency of my converters. And I am also very interested in the influence of the different capacitors on the behavior.
Update:
In the first version of the article I wrote that I put the three 100 uF capacitors in row to get a 300 uF capacitor (This is still shown in the picture of my PCB layout). But this is wrong.
Putting capacitors in row the resulting capacity calculates like
\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{(C_2C_3 + C_1C_3 + C_1C_2)}{C_1*C_2*C_3}Or:
C = \frac{C_1*C_2*C_3}{(C_2C_3 + C_1C_3 + C_1C_2)}Hence the capacitors I used are 100 uF this results in
= \frac{100*100*100}{100*100 + 100*100 + 100*100} = \frac{1.000.000}{10.000+10.000+10.000} = \frac{100}{3} = 33.3 [uF]That’s why the capacitors must be in parallel, not in row.