MC34063 - StepUP from 3.8v to 7v - Page 1
Forum section: electronics / beginners

author: svlad1983
author_rank: Newbie
timestamp: June 06, 2014, 11:53:37 am
content: Hi guys,hope you can help me out. Trying to make a step up converter from LiIon battery 3.7V to get 7V output.With the schematics attached I get no change at all. Does it mean I already killed my IC?Diode Im using is this one:http://nl.mouser.com/Search/ProductDetail.aspx?R=B0540WS-7virtualkey62110000virtualkey621-B0540WS-7Max current is 90mAThanks!

author: c4757p
author_rank: Super Contributor
timestamp: June 06, 2014, 12:26:32 pm
content: Should work. You left out the current sense resistor, which is supposed to stop it from exceeding max current (which it is otherwise almost guaranteed to do at startup) so it has probably died. They've no other real protection - not a forgiving chip.

author: firewalker
author_rank: Super Contributor
timestamp: June 06, 2014, 12:28:04 pm
content: Did you made the calculations your self? Did you placed the ic correctly. A common mistake with MC34063. Because the schematic symbol is usually inverted "inverted" (pin 1 on the left side).You are also missing the current limiting resistor.Alexander.

author: svlad1983
author_rank: Newbie
timestamp: June 06, 2014, 12:35:17 pm
content: I used this calculator:http://dics.voicecontrol.ro/tutorials/mc34063/calculate.phpCurrent limiting resistor is 0.63 Ohm, so i thought it would be safe to ignore it?I have a SO-8 package, pin 1 should be on the side of the chip with the cut edge? correct?

author: c4757p
author_rank: Super Contributor
timestamp: June 06, 2014, 12:37:13 pm
content: 0.63 ohm is alargesense resistor for that chip. The datasheet gives an absolute minimum of 0.2.Yes, pin 1 is on the cut edge.

author: firewalker
author_rank: Super Contributor
timestamp: June 06, 2014, 12:38:28 pm
content: No, it is not safe. The output transistor has it's limits.Alexander.

author: svlad1983
author_rank: Newbie
timestamp: June 06, 2014, 12:40:57 pm
content: Thanks guys, ill add a resistor and try again!

author: firewalker
author_rank: Super Contributor
timestamp: June 06, 2014, 12:45:14 pm
content: Also try, a larger frequency value (50 kHz or something). It will reduce you inductor and Ct values.Alexander.

author: mariush
author_rank: Super Contributor
timestamp: June 06, 2014, 02:18:35 pm
content: Use the freeware application that's on sourceforge, it will give you more data :http://sourceforge.net/projects/mc34063uc/MC34063 is not that great for this project, at best you're only going to get about 250mA out if it.. well, since you say max will be 90mA, it's going to be fine.Here's an example using that software (below, picture). You can increase the inductor and output capacitor, you can change R1 and R2 to other values as long as you keep the ratio the same.. the current sense resistor needs to be there. Doesn't have to be exactly 0.47 ohm as in the picture, you could for example just put 2 x 1 ohm resistors in parallel to get 0.5 ohm and that will be close enough.The IC you have now is probably dead though. Like others said, the resistor needs to be there.

author: rob77
author_rank: Super Contributor
timestamp: June 06, 2014, 02:36:01 pm
content: your iC is definitely dead, the current sense resistor MUST be there, otherwise the chip will pass away during the very first pulse.0R22 is a safe value - if you use bigger one e.g. 0R5 , then you'll have a limited "ON" current through the inductor. that shouldn't be a problem in step-down configuration, but for step-up you need more current - so go for a smaller sense resistor.the chip will work well even with different than calculated inductors (+- 50%), but your max output current and voltage ripple on the output will be altered significantly.

author: svlad1983
author_rank: Newbie
timestamp: June 12, 2014, 05:47:20 pm
content: Guys, did a second try according to the values provided by mariush. (only different values Rsc 0.22R and L 68uH)The result was that when i had a scope connected on the output i was getting stable 3.66V with 3.89 V input.however when i just tried to connect voltmeter when the device was on I was getting sometimes 6.9V and sometimes same 3.7V, which is weird?It also was making some noise when i was connecting voltmeter. After playing with it for sometime (connecting and disconnecting voltmeter) it finally burned again.The question is why sometimes it was giving out correct voltage??I apologise for the rookie question:S

author: rob77
author_rank: Super Contributor
timestamp: June 12, 2014, 09:29:52 pm
content: did you checked the timing capacitor with scope ? was the frequency correct ? was the waveform a nice triangle/sawtooth (don't remember which one it is - but definitely one of those 2 ). ?

author: svlad1983
author_rank: Newbie
timestamp: June 12, 2014, 09:32:35 pm
content: No didnt check it. Gonna do another replacement tomorrow and try again. What else should I check?

author: mariush
author_rank: Super Contributor
timestamp: June 12, 2014, 09:36:52 pm
content: Could you post some pictures of the circuit (both sides), just to see that you connected everything correctly?Are you working on it on a surface that's insulating (a few sheets of paper or some plastic for example) ? You don't want to have shorts while testing.Some noise is normal, as it turns on and off thousands of times a second so you have pulses in the inductor.

author: rob77
author_rank: Super Contributor
timestamp: June 12, 2014, 09:42:20 pm
content: Quote from: svlad1983 on June 12, 2014, 09:32:35 pmmainly the signal on the timing capacitor - to check if it oscillates at the expected frequency and whether the waveform is correct.make sure your inductor have no shorts between turnscheck the current t through the inductor with scope - hook the probe across the 0R22 current sensing resistor.