On the oscilloscope continuous operation. Intermittent simulation. Apparently the model MC34063 is not entirely accurate.

 

Don't you think connecting an oscilloscope can distort the signal timing diagrams?

 

Note that I have reduced the 180 Ohm resistor and even connected it in the wrong way for more effect. It is possible not to change the connection method, but also to reduce the resistor to 100 Ohm. Don't be afraid, at current through the choke 1 A it is quite linear and will not go into saturation. I had to look for the core material on the Internet more than once. And it's never been difficult to find what kind of material. The truth is, I did ask GooGle in Russian. Using Schottky's diode will further increase the output voltage by at least 0.5V. In this scheme, using 1N4148 is not successful. On it, with such a high current for it, a lot of voltage drops (1.5 - 2 V). The diode you use causes high losses because it is slow.

Thanks for your advice, and the reply!


I do some changes to my circuit, and I'll show the result in this comment.
Oscilloscope connection:
Yellow: Channel 1, connect to the pin1 of MC34063, which means the switching signal.
Blue: Channel 2, connect to the output terminal.

DVM connection:
The red probe connects to the output of boost-converter
The black probe connects to the load positive side

1.Diode
I bought 1N5819 from the local component store, though it has more forward voltage then 1N5817, the store doesn't sell 1N5817 so I can't get it.
Thus, I replace the FR107 diode in my original circuit, rebuilt the whole circuit on the breadboard using 0.6mm single core wire instead of DuPont line, and connect the component as close as the IC if possible to reduce the influence of the resistance of connecting wires.
Then I use the component which design at the beginning, Ct = 250pF, L=100uH (toroid core mixture-26), and still using 180-ohm resistor at pin 8 and connect it series to the 0.5-ohm current limiting resistor.
The result is fine that the output current can up to 100mA as the original design, awesome!!
Of course, the switching frequency is 128kHz @100mA load, but the inductor seems fine when I attach a 110-ohm resistor.
BTW, I found that the control method of MC34063 is changing the switching frequency instead of changing the duty cycle to maintain the output voltage.


2.Timing Capacitor
As @Bordodynov said, the high switching frequency may cause the inductor into saturation.
So I use a 470pF ceramic capacitor as Ct then test the circuit, and we can see the frequency goes down to 90.29kHz.
And here is something confuse me:
Q1: I think the frequency should be half of 128kHz but it is still around 90kHz and I don't know why.



3.Inductor
Q2: I bought this inductor I saw at boost module like this, is this an iron powder inductor?
I replace my toroid inductor into this power inductor there is nothing change when I compared these two inductors.

4.Peak current
Now I can output 100mA current at 12V for real, but I still can't drive the nixie clock which is the purpose I build this circuit.
I attach the nixie clock, the DVM and oscilloscope show that the output voltage drop to 6 volts and the current becomes 311mA.

I decided to change the input voltage to 10 volts, and the circuit drives the nixie tube perfectly, then I decrease the input voltage, the threshold seems is 6.4 volts. Sorry that I'm not an English speaker, so let me show the phenomenon through video.

P.S Sorry for my bad pronunciation
I try to add an external NPN transistor which is TIP41C in the P.6 of MC34063 datasheet, but there are adding more problems when I use the transistor and do no advantage to the circuit.

Thanks for your reading,
best regard.

 

What kind of current do you measure? I don't speak English, either. If you measure the current from the power supply, it can drastically damage the transducer. It can be caused by an ammeter's resistance. I recommend that you measure the power supply current at an increased current measurement range, e.g. 10 Amps.

 

Hi,Bordodynov, thanks for your fast reply.
I add an explanation to the previous comment, thanks.
I have measured the output current of my boost converter in my video and in the picture, the red one connects to the output, and the black one connects to the load positive side.

This time I measure the 5V power source current using 10A range of DVM, and they have the same result like this:

The nixie clock can not light up and the output voltage is only 5.5v.
Thanks for your help.
Best regard

 

Judging by your messages, your indicator consumes more (significantly more) with a lower power supply. I will continue to think how to help you.
Is there a built-in voltage pulse converter in your indicator?

 

Judging by your messages, your indicator consumes more (significantly more) with a lower power supply. I will continue to think how to help you.
Is there a built-in voltage pulse converter in your indicator?

Thank you very much!

Yes, the tubes (indicator) are driven by another MC34063 boost converter, that PSU circuit inside of the clock is like this:
(12V input / 180V output)
This circuit provides a 180V DC source for the nixie tubes, and I use some MUX and high voltage transistors such as KSP42 and KSP92 to be a scanning circuit to light up the tubes.
And because this circuit was be founded on the internet, this is why I used the wrong diode which not fast enough for the boost converter.
I also did some search from google, I think the problem may be caused by the output capacitor in this 180v circuit or the inrush current.
Thanks for your help,
best regard

 

I have not been able to increase the load capacity of your pulsed stabilizer in a simple way without greatly complicating the circuitry. The only chance is to make the high-voltage stabilizer run smoothly over time. I've done this before by bypassing the upper divider resistor with a capacitor. It'll make the consumption current rise smoother. I propose to parallel the R51 resistor with an capacitor of 1 µF (200 V) or more. We'll have to experiment.

 

I suggest if possible replacing FR107 with UF4007. The proposed diode is eight times faster. This will reduce the current consumption of your indicators.

 

Thanks for your reply, I'll try to find the UF4007 diode and also add an additional capacitor in the 180v circuit tomorrow.
I hope the electronic store sells the diode.
If those methods don't work, maybe I need to find another controller IC to do this job
Best regard, and very much appreciate your help!

 

@Bordodynov
Instead of stabbing around, we start with the facts, the datasheets, the calculations and work from there. Always debug by going from knowns.

What nixie are you using? IN-14?

To use this BOOST regulator to go from 5V to 12V, delivering up to 100mA, with 50kHz switching; here are your component values:



I made this spreadsheet to make calcs easy for VFD and NIXIE clock projects I have done.

 

The problem is that the 12-volt drive is loaded on another overdrive. As we know, they have negative input resistance. In other words, it is more consumption with less input resistance. Therefore, an inverter with a load current of up to 500 mA is required. But I suggested another bypass, to add a capacitor to reduce the inrush current of the high voltage converter.

 

I've calculated how a high-voltage converter behaves when it's powered by a current-limited source. It turns out that this source in the original circuit produces a voltage slightly higher than 4 volts. The charts for this are shown in green. The second case (blue) of the source voltage is close to 12 V.

 

So I've calculated the established process. With the additional capacitor the consumption current is 83 mA, and without the capacitor 350 mA AND the voltage has not reached the required value.

 

What nixie are you using? IN-14?

Yes, they are IN-14.

So I've calculated the established process. With the additional capacitor, the consumption current is 83 mA, and without the capacitor 350 mA AND the voltage has not reached the required value.

Thanks for your reply and simulate, the result of the simulation is close to the true circuit.
It is really useful and very awesome!

I have not been able to increase the load capacity of your pulsed stabilizer in a simple way without greatly complicating the circuitry. The only chance is to make the high-voltage stabilizer run smoothly over time. I've done this before by bypassing the upper divider resistor with a capacitor. It'll make the consumption current rise smoother. I propose to parallel the R51 resistor with a capacitor of 1 µF (200 V) or more. We'll have to experiment.

I add a 1u/250v capacitor parallel with R51, but I can't find UF4007 at three electronic stores in my city, so the diode I'll replace after the product delivery, maybe a week later.
I'll keep updating this topic.

After I add the capacitor, I observe that the time of peak current getting stable is shorter than it didn't be added.
Yellow: 12V power source
Blue: Nixie clock power in
Red: Voltage across the 1-ohm resistor, aka, the input current
The measurement method had been shown in the first discussion.

p.s: Two pictures are using a 12V/2A AC-DC adapter, and the left picture is the first post of this discussion, using Ct = 250pF, FR107 diode.

Then I connect the nixie clock to my 5-12V converter, the converter works very well!!
There is a small problem but not serious is after reconnecting the nixie clock less than 4 seconds, the output voltage and current will go back to the waveform before adding the capacitor (output voltage drop to 5.4V, and current increase to 300mA).

I guess it is because the time constant \[ \frac{1}{RC} \]of R51 (330k ohm) and the capacitor (1uF) is 3.3 seconds, so we need to wait for the additional capacitor to discharge so it can play its role (?
I think the actual principle should be explained by @Bordodynov
This method is amazing!

Thanks for everyone's help,
best regard.

 

The initial problem was the insufficient load characteristics of the 5 V --> 12 V inverter. This problem is due to the fact that the step-up transmitters draw more current from the power supply when the power supply is low. In our case, the step-up converter consumes approximately 350 mA when powered from a 4 - 5 V power supply. In this case, the output does not have the required voltage. There is a direct dependence of the consumption current on the output voltage. At smaller output voltages, the consumption current is lower. By adding my capacitor, I make the initial output voltage low and the average consumption current low. Then the output voltage increases smoothly. As a result, there is no high power consumption when the high voltage converter is switched on, and this allows the 5 V --> 12 V converter to operate at its rated load at switch-on.
And yes. I apologize for not warning you about the restart problem. I knew this would happen, but I decided it wasn't significant.