I think this is a buck converter, but I'm not entirely sure.

Question

So I have this little chip here from a cigarette lighter psu, but there isn't a voltage regulator any where in it. If this is indeed a buck converter, that would explain the lack of a regulator, but I'm new to this kind of stuff, (I first got into it while creating a project for school last year) so I would appreciate any help I can get.

shabaz · Accepted Answer

Hi Jake,

That is a voltage regulator, also known as a DC-DC converter or a Buck converter when it uses an inductor and switching and steps down the voltage.

You could confirm by examining the part code on the integrated circuit, and obtain the datasheet for the part online. It is extremely likely to be MC34063 (or similar) because that is common in low-cost USB car chargers (and from the photo, the pinout looks about right for that).

rachaelp · Answer

Take a look at the datasheet for the controller here (assuming it is the one Shabaz is referring to): http://www.farnell.com/datasheets/1723944.pdf?_ga=1.226316314.1698039299.1446112257 If you look at Figure 15 on page 11 this is the circuit for a step down converter. In the circuit in this drawing R1 and R2 are the feedback resistors which are used to set the output voltage. The CII input feeds into a comparator inside the device, the other input being a 1.25V reference. The device will try and adjust the output such that the voltage on the CII input matches the reference voltage so if you want to change your output voltage then select values of R1 and R2 such that CII is at 1.25V for your desired output voltage. Best Regards, Rachael

rachaelp · Answer

I hope it's not 1.2 ohm. Do you mean Kohm? You can do as you said to calculate the two resistor values. You might need to replace both to get the output you need as you will be restricted to standard values which can be easily obtained.

shabaz · Answer

Hi Jake, You didn't specify the reason for 4.7V, but if you're happy with "about 4.7V" then you could put a diode in series with the output. The output won't be accurate any more, but will be lower than 5V slightly (could vary to 4V or lower). For any accuracy you need to take the approach Rachael mentions with the resistors,but needs the steps mentioned (soldering capability, resistors purchased). If you're interested in this as a hobby, see this URL for lots of information: I Want to Do Electronics! Where do I start?