Boost converter, MCP1640 output not regulate - Page 1
Forum section: electronics / beginners

author: chancs
author_rank: Contributor
timestamp: July 09, 2019, 08:53:34 am
content: MCP1640 Datasheethttp://ww1.microchip.com/downloads/en/DeviceDoc/20002234D.pdfI want to build a boost converter that can output 5V @5mA peak from a CR2450 coin cell battery. After some researches in Digi-Key, MCP1640D is chosen because of its low price.I followed the datasheet page 15 equation, calculated that with R_TOP = 680K and R_BOT = 220K, VOUT = 4.95V. Page 19 Figure 6-1 has an example circuit of how to use it with a button cell battery, a 3.3uF inductor is used instead of 4.7uF as I don't have one at the moment.I built the circuit on the breadboard and measure the input (CH1, yellow) and output (CH2, cyan) voltage. Figure 1 shows the output with no-load, and Figure 2 shows that output with 100 OHM load. Both output voltage is nowhere near the expected 4.95V, and someone points out what I have done wrong?Figure 1Figure 2

author: hli
author_rank: Frequent Contributor
timestamp: July 09, 2019, 02:26:58 pm
content: That connection between the two positive rails looks a little bit dodgy. Is EN (pin 3 ) correctly pulled high? Otherwise try to connect the caps as direct as possible to the IC. And maybe check that the resistors really have the value they are supposed to have...

author: mikerj
author_rank: Super Contributor
timestamp: July 09, 2019, 02:55:15 pm
content: Quote from: chancs on July 09, 2019, 08:53:34 amThat is very likely the cause of your problems, breadboards are simply not suitable for this type of circuit. Use Microchips recommended PCB layout, it's simple enough that you could prototype it with a piece of plain single sided PCB and a small dremel tool.

author: Pseudobyte
author_rank: Frequent Contributor
timestamp: July 09, 2019, 03:01:39 pm
content: Agreed, switching power supplies do not fair well on bread boards. You would see an improvement if you switched to a soldered prototype on perfboard, but even then you introduce some loop inductance and stray capacitance that can affect the stability of your circuit. Your best bet would be to lay it out with surface mount components as shown in figure 5-2 and 5-1 of the datasheet. Component selection is very important also, read each paragraph regarding each component and the consideration that needs to be taken. Going purely of component value can cause instability when you introduce high ESR caps into a switching design.

author: Pseudobyte
author_rank: Frequent Contributor
timestamp: July 09, 2019, 03:12:30 pm
content: AN1311 may help you learn a thing or two about this specific application.http://ww1.microchip.com/downloads/en/AppNotes/01311a.pdf

author: Buriedcode
author_rank: Super Contributor
timestamp: July 09, 2019, 03:18:40 pm
content: Adding to comments about breadboarding SMPS... that inductor is a signal inductor - not a power inductor. Whilst it will be 3.3uH, its saturation current will be very low and DCR possibly high enough to bring down efficiency.Also, a CR2450 won't be able to supply much instantaneous current. Whilst you're after 5v @5mA (= 25mW, so for 60% eff. thats 25/0.6 = ~42mW in, = 14mA input current) the peak current in the inductor will be much higher, which will cause the battery voltage to sage during on-time.You could scope the inductor current with a 1 ohm resistor in series with the inductor.

author: chancs
author_rank: Contributor
timestamp: July 10, 2019, 04:26:25 am
content: Thanks all. I will prototype the circuit in PCB, and see how it goes.

author: yeoman
author_rank: Newbie
timestamp: March 18, 2023, 03:26:01 am
content: Quote from: chancs on July 10, 2019, 04:26:25 amhello,chancs, do you prototype it in PCB successfully ? does it can output 3.3v in the beadboard ? I meet the same question, I'm very confused it.