Wednesday, February 25, 2015

3D printer hot end temperature control system Version 2 [unfinished]

Almost a years ago, I post the first version of a simple temperature control system ultilizing only a LM324 quard op-amp with some resistors and a MOSFET. The system is for 3D printer's hot end.

The controller has two major functions: (a) turning a power resistor on and off automatically to maintain a constant temperature (adjustable manually); (b) showing the real time temperature in the range of 120-260 C. It is simple, low price, and efficient. The circuit board is shown below.
Temperature control system Version 1. See this post for detail. The values of the resistors here are R0=600 ohm, R1=10k ohm, R2=3.9k ohm, R3=6.19k ohm, R4=2.4k ohm, R5=10k ohm.

However, I realized later, also as reader epineh pointed out, the second function, displaying real time temperature, requires a constant +12V power supply. This could be problematic. Because most time people use low price ATX computer powersupply to power their 3D printer. And those powersupply DOES NOT output constant voltage. When there is a large current drain change, say once the power resistor is switched on and off, the power supply's output voltage can jump between 10 and 12V. As a matter of fact, the displayed voltage can be off by 20%!! The control system, on the other hand, works just fine because it relies only on the ratio of the resistance of thermistor and R0.

One solution is to use a logic power supply that output constant 12V for the LM324 and ATX powersupply for the hot resistor R_hot. Or another way is to add a voltage regulator, which turns the unstable 12V to a stable 5V as the voltage reference for LM324. Only one power supply is required. The circuit diagram is shown below.

Temperature control system Version 2. The values of the resistors here are R0=600 ohm, R1=10k ohm, R2=3.9k ohm, R3=6.19k ohm, R4=12.4k ohm, R5=10k ohm.

There are two differences between version 1 and version 2.
(a) In version 2, a LM340T5 +5V voltage regulator is used to convert 10~12V to a constant +5V. So the display will not be affected by the status of the hot resistor.
(b) R4 is changed from 2.4k ohm to 12.4k ohm. This is important in order to correctly display the temperature. I will show why later.

[this post is unfinished yet]

Friday, September 19, 2014

Change car charger's output voltage

Car charger is a must-have today, thanks to smartphones' larger and larger screen. It is very cheap to get a car charger adapter, which converts 12V, voltage of car's socket, into 5V, voltage of USB standard. I have several of these too.

However, Now I am in need of 9V 1A power while driving, and I decide to convert one of my 5 V car charger into 9 V.

Initially I was thinking about using a 9 V voltage regulator, say, LM7809, to do this. I believe it will work, but I found the solution can be even simpler after I actually opened one 5 V car charger.

This is what it looks like inside the 5 V car charger.

On the top is a 2 A fuse, which connects to the anode of the car charger (12V). The two thin metal sheets connect to the ground. The circuit at the bottom does the 12V-5V converting job.

Although those larger aluminum capacitors at the bottom occupy most of the space, the most crucial part is the IC (integrated circuit) under the black wire.

Look carefully into the circuit I found the IC part number is MC34063A, which is a 1.5A step up/down/invert DC-DC voltage regulator. I found its specs online

Sounds promising. The regulator can theoretically output voltage from 1.25V to 40V, and it can output up to 1.5A. So it is possible to achieve my goal, 9V 1A, by simply modifying the circuit a little bit.

The datasheet also provide a step-down circuit example, which outputs fixed 5V.

The output pin of MC34063A is pin 2. Resistor R1 and R2 form a voltage divider so that the voltage of pin 5 is
V5 = Vout*R1/(R1+R2),
which gives
Vout = V5*(R1+R2)/R1 = V5*(1+R2/R1).

From the circuit we see that V5 is tight to a 1.25V reference by a voltage comparator, meaning that V5 must equal to 1.25V. Hence
Vout = 1.25V*(1+R2/R1).

Given R1 = 1.2k and R2 = 3.6k = 3R1, there is
Vout = 1.25V*(1+3) =  5V.


Look carefully into the circuit I found that the blue resister on the right side of the IC is labeled R1, and the yellow resistor hidden beneath the large black capacitor is labeled R2. R1 links pin 5 with GND, and R2 links Pin 5 and output (the red wire). They match the above schematics perfectly!

Using a multimeter I found that R1 = 1k ohm and R2 = 3k ohm. Hence Vout = 1.25*(1+3) = 5V.d

Now the task is super easy! I removed R2 and replace it with a 6.2k ohm resistor so that
Vout = 1.25V*(1+6.2k/1k) = 1.25V*7.2=9V.

Guess what, after this simple modification, the voltage output of the car charger is 9V!

I opened several other car chargers I have and they all use MC34063A or identical chips. So looks like it is a pretty standard way to make inexpensive car chargers. Therefore the method I used here would work for those car charges as well.