Phantom from a boost converter

[ruffrecords]

I can’t picture why or how one side gets to start first. Being both identical, it simply cooks my noggin that they oscillate.

If both sides were identical it would not start. But component tolerances ensure this never happens.

Cheers

Ian

 

[Rochey]

I figured as much. one always charges faster than the other i guess. the design assumption is that one will charge and after  that it'll oscillate.

Nice

 

[abbey road d enfer]

Being both identical, it simply cooks my noggin that they oscillate.

But they are not identical. No transistors are being created equal, same for resistors, and the xfmr has some unbalance to it. That what makes all multivibratot-type oscillators start.

 

[Rochey]

Abbey,

i;m back in the us, jetlagged to hell... its 5am and i've been thinking about short circuit protection. i found a nice circuit on  https://daycounter.com/Circuits/Short-Circuit-Protection/Short-Circuit-Protection.phtml

(after spending an hour on circuit sims designing my own more complex version)

this should work nicely, but there's a massive gotcha.
if i take the feedback point after the pass transistor, when it's output shorts, the dcdc will try and kick up its output to compensate, no?

 

[abbey road d enfer]

Abbey,

i;m back in the us, jetlagged to hell... its 5am and i've been thinking about short circuit protection. i found a nice circuit on  https://daycounter.com/Circuits/Short-Circuit-Protection/Short-Circuit-Protection.phtml

(after spending an hour on circuit sims designing my own more complex version)

It's a basic current limiter; very often used as CCS (Constant-Current Source). A CCS presented with a load that draws less current than nominal acts more or less like a low-value resistor, until the current draw reaches the nominal value of the CCS, where the current cannot increase anymore.
More clever short-circuit protection circuits actually mintor the output voltage and reduce current in a way that protects the series-pass transistor. Too complicated IMO for the intended purpose.

if i take the feedback point after the pass transistor, when it's output shorts, the dcdc will try and kick up its output to compensate, no?

That's correct, but may be of no consequence. It depends very much on your design choices.
Please note that I mentioned short-circuit protection in te context of capacitance multiplier.
The cause of failure is that the output goes instantly from V+ to zero, but the base stays at V+ plus 0.7V due the the capacitor, which blows the b-e junction (I erroneously mentioned zener effect before). The cure is very simple, adding a resistor between the cap and the base of the transistor. Indeed it slightly increases HF noise, but it's very easy to filter out with a capacitor at the output.

 

[JohnRoberts]

It's a basic current limiter; very often used as CCS (Constant-Current Source). A CCS presented with a load that draws less current than nominal acts more or less like a low-value resistor, until the current draw reaches the nominal value of the CCS, where the current cannot increase anymore.
More clever short-circuit protection circuits actually mintor the output voltage and reduce current in a way that protects the series-pass transistor. Too complicated IMO for the intended purpose.

"Foldback" current limiting, reduces the current available proportionate with the output voltage, appropriate for resistive loads.  This is commonly done with audio power amplifiers to protect output stages.

JR

That's correct, but may be of no consequence. It depends very much on your design choices.
Please note that I mentioned short-circuit protection in te context of capacitance multiplier.
The cause of failure is that the output goes instantly from V+ to zero, but the base stays at V+ plus 0.7V due the the capacitor, which blows the b-e junction (I erroneously mentioned zener effect before). The cure is very simple, adding a resistor between the cap and the base of the transistor. Indeed it slightly increases HF noise, but it's very easy to filter out with a capacitor at the output.

 

[Rochey]

Here is where i’m at so far.
Pass element is an A42 - as i’m Not pulling too much currentZ

 

[JohnRoberts]

You could add foldback to that simple current limit by adding another resistor from the base of the current limit npn to ground. This would allow it to deliver more current at full voltage than when severely loaded down. Of course you have to keep the foldback within reason or it will never start up.

I'm not sure there is much benefit from taking feedback from after the filter... this will also add lag to the NF that can cause stability issues. 

JR

 

[abbey road d enfer]

Here is where i’m at so far.
Pass element is an A42 - as i’m Not pulling too much currentZ

You need a capacitor after the choke.

 

[Rochey]

You need a capacitor after the choke.

Forgive me for appearing dumb -- it's that what the cap multiplier is doing?

 

[JohnRoberts]

Forgive me for appearing dumb -- it's that what the cap multiplier is doing?

The cap multiplier "looks" like a larger capacitance at its output, not it's input. It will smooth out perturbations caused by varying current draw from the load, but won't filter noise from source.

JR

 

[PRR]

> “which one starts?”

Put it in SPICE, you will "prove" it can not oscillate. All those millions of practical examples must be wrong.

Build it, one "13K" may be 13,100 and the other 12,900. Also two "hFE=100" BJTs will really be 98 and 105.

Even if you could build "true match", the universe's random noise would push it off-balance. (Nearly all RF oscillators start this way, and statistically might never start.)

 

[Rochey]

Genuinely wonderful! (And another evening, sleep deprived)... i’m Even worse than trump on twitter when i’m Sleep deprived.

 

[saint gillis]

New thread inside the thread :
Here's a boost converter from MC34063 datasheet :


Here's the circuit recommended to provide more current with an external transistor :


How would you do if you wanted to use an external transistor to avoid more than 40V going into pin1 (as the MC34063 doesn' accept more than 40V) AND getting 48V at Vout ?

 

[abbey road d enfer]

I would try something like that. With a 12-20V zener, making sure the reverse voltage is high enough. No gurantee here.

 

[Hubbub]

I would suggest just connect pin 1 to pin 8:
The pin 8 resistor sets the maximum base current for the external transistor.

With any boost converter be aware of what happens if feedback is lost. The output typically shoots up until the switching transistor reaches its actual breakdown voltage (more than rated voltage!). At that point the transistor can short out, and Rsc sees the full supply voltage causing it to heat significantly and emit smoke.

 

[saint gillis]

I would try something like that. With a 12-20V zener, making sure the reverse voltage is high enough. No gurantee here.
View attachment 112030

I've tried that but it didn't work

 

[saint gillis]

I would suggest just connect pin 1 to pin 8:View attachment 112037
The pin 8 resistor sets the maximum base current for the external transistor.

With any boost converter be aware of what happens if feedback is lost. The output typically shoots up until the switching transistor reaches its actual breakdown voltage (more than rated voltage!). At that point the transistor can short out, and Rsc sees the full supply voltage causing it to heat significantly and emit smoke.

Thank you, I'll try it

 

[abbey road d enfer]

I've tried that but it didn't work

So maybe Hubbub's suggestion is better.

 

[saint gillis]

I've run tests,
The principle works but the power efficiency is very bad (28%) and the transistor is very very hot, I've tried an MPSA42 and a 2SC2240 (same results more or less)

With the regular circuit (without the additional transistor), I obtain a 76% efficiency (without even fine-tuning the components values too much), and the chip stays cold and steady..

I've tested the principle with a 33V output voltage