voltage booster is providing 60V instead of 18.

Started by Boner, August 27, 2020, 02:14:36 PM

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Boner

I'm using a 555 timer to double a 9v input voltage, but instead of 18 I'm getting 60



555 timer is providing a 50kHz square wave and the duty cycle is 50%, I've verified it best I can with an old tektronix 2215 oscilloscope

I'm using a brourns 100uH inductor rated at 1amp
https://www.mouser.com/datasheet/2/54/rlb-777833.pdf

an IRF740PBF power mosfet
https://www.mouser.com/datasheet/2/427/91054-1768370.pdf

and a 1N5817
https://www.mouser.com/datasheet/2/308/1N5817-D-1801388.pdf

I've tried swapping out the 1n517 for 1n4004 after the 1n517 got too hot (1n518 I think would be a better bet but I dont have any)

seems like no mater what I do, the switch/inductor/diode when connected pulls about 200mA from the power supply and I get an output of 60V
Is my timer messed up or? Other than bringing up the frequency, which I'm gonna try and do now, I can't think of anything else  :icon_redface: :icon_redface: :icon_redface:

PRR

#1
That's not a "doubler", it does not naturally give 2X the voltage.

It is a "flyback". L1 kicks-up, ideally to infinite voltage. This is part of how car ignitions sparks are made. On iron-core coils the kick-up is often 5X to 20X the input voltage, limited by iron losses.
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Boner

#2
oh wow

I've been wrong before, but never "thats a completely different industry" wrong.


*edit*

maybe something like this?



wouldnt that be a charge pump and not a voltage booster? Or is what I had originally WAS a voltage booster but I was wrong to think a voltage booster limits the amount of voltage?

ElectricDruid

Quote from: Boner on August 27, 2020, 02:14:36 PM
I'm using a 555 timer to double a 9v input voltage, but instead of 18 I'm getting 60

Call it "super-boosted headroom" and mark it down as a feature?!? ;)

R.G.

The thing about flyback/boost power stages is that the don't have ANY well defined output voltage at all. The voltage is entirely determined by the load after the diode. That can be the arc-over voltage of a couple of inches of air, or the stray capacitance, or a load resistor/zener/dying IC, whatever.
Boost-mode voltage converters (modern fancy-speak for flyback) are nearly always used with a biggish output cap compared to the amount of energy in one cycle of the load-dump of the inductor so the inductor can't over-volt the cap very fast. This gives a regulator circuit time to decide not to put that next wibbet of energy out for a while.
This is the evil twin of a capacitor load-dump regulator like a charge pump. Capacitors have well defined voltages, but they don't have any well-defined current. Capacitor charge-dump devices like charge pumps have BIG spikes of current into and out of the bucket-cap. Currents are just harder to see than voltage spikes.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

PRR

> I'm getting 60

Yes, about in there.

It's not "infinite" due to coil losses, stray C, and probably because the current cut-off is not infinitely fast. In fact the '555 seems to struggle to shut-off my antique IRF150 model.

The diode model has been edited to 1,000V breakdown so that's not a factor.

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Rob Strand

#6
If the inductor discharges during the off time the effective off time is much reduced and the effective (discontinuous) duty is increased.   That will push the output voltage way up.    I haven't tried to simulate the ckt to see if that's occurring.

You might find different behaviours with different loads.
--------------
I'm pretty sure it's because there's not enough load.   With a 100uH inductor you need a minimum load of 1A to 1.5A output current, depending on the circuit losses.    The inductor current isn't allowed to go to zero otherwise the circuit operates in a discontinuous mode and the output voltage will rise due to the effectively higher duty cycle.

If you want to run a lower current you need a larger valued inductor.   For each minimum load current there is a minimum inductor value.


Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.


duck_arse

is no-one going to offer feedback? all you need is to read the output voltage, and use it to control the switcher - Q2 in this inst. like it was big volts, except small.

" I will say no more "

Rob Strand

This shows for an open loop boost converter that the voltage starts to increase when the inductor current falls to zero.
The critical minimum load is about 25 ohms, about 0.75A.

If feedback is added then the output can be regulated by controlling the duty, even if the inductor current falls to zero (DCM).

The load starts at 15 ohm and increases in 5 ohm steps.

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

R.G.

Quote from: duck_arse on August 28, 2020, 10:08:28 AM
is no-one going to offer feedback? all you need is to read the output voltage, and use it to control the switcher - Q2 in this inst. like it was big volts, except small.
Yep. Offered feedback in my post, above. Noted how it's done. Adding feedback also introduces loop stability to be dealt with, hence the slowing down issues with a capacitive load/filter. It's worth noting that the loop gain changes significantly when the inductor goes from return to zero (current) to non-return-to-zero.

It's also worth noting that there are single-chip flyback controllers that do all this, including feedback, in one package. Some of them include the switching transistor.

None of this is as simple to execute as a charge pump.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Rob Strand

#11
QuoteNone of this is as simple to execute as a charge pump.
The OP already knows about charge pumps as he has posted circuits in the past.    I'm assuming that he's using a switcher for more output in this case.     The charge pumps might go to 100mA or so.

For higher currents feedback is probably a good thing since inductor losses are going to cause an uncertain drop in output voltage- although you could tweak the duty.  The problem with the open loop case is you need a relatively large value inductor.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Rob Strand

Quote555 timer is providing a 50kHz square wave and the duty cycle is 50%, I've verified it best I can with an old tektronix 2215 oscilloscop
When I calculate the frequency from your part values I get 15kHz.

If you are getting 50kHz then the minimum currents are lower than my last post by a factor of 3.

Quoteseems like no mater what I do, the switch/inductor/diode when connected pulls about 200mA from the power supply and I get an output of 60V
Is my timer messed up or? Other than bringing up the frequency, which I'm gonna try and do now, I can't think of anything else
The inductor is going to start saturating at 1A so you need at least 50kHz switch frequency.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

PRR

> I get 15kHz.

My sim seemed to be 14.3kHz.

But I won't defend its honor. It lies; also I may have mis-copied a value.
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Rob Strand

QuoteMy sim seemed to be 14.3kHz.

But I won't defend its honor. It lies; also I may have mis-copied a value.
It's odd the sim and the hand calculations are close.   NE555's aren't normally 3 times off the calculations.  The calcs do go a bit wonky above 100kHz but not 3 times.

We will have to wait for Boner's (the OP) response to fill in some of the blanks.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.