DC Boost Converter Problems

[vne147]

3. The fact that the initial voltage is above 6v is a danger-point.

You said The output voltage starts at around 20V but within a second, everything stabilizes and I get a nice steady 5.68V
The circuit can "lock-up" when the voltage of the supply drops and this can either drain the cells completely or produce very high output voltages.

Your best choice is to use a 3.6v cell It has twice the energy-density of AAA cells and comes in rechargeable. Or even 3 x AAA cells

I guess that's what was happening before I switched inductors from 47µH to 180µH. It was locking up at around 28V. I thought it was normal for there to be a brief overvoltage condition though. Is it not?

Also, I was reading up on the Li-ion batteries and most places say that they require a protection circuit to prevent over discharging. I didn't build that into my circuit but I think I should be OK if I use these batteries or something similar. The charger should take care of over charging protection and the monitoring circuit built into the battery should take care of over discharging, right?

 

[Mr RB]

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Do you have any thoughts on the boot strapping idea? I have a couple DIP 34063s. If I have time tonight I'll test it out.
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The problem is that the 34063 drops about 1.5v in the output stage even with good base drive. With a 3v input it won't get good base drive to it's internal power switch transistor.

So with a 3v input it's trying to convert less than 1.5v to >6v, even with a perfect inductor you are looking at 4:1 current ratio so your 80mA output will require 320mA input. Now you also have losses from the 0.22 ohm current sense resistor, which drops 0.68 volts(!) at 320mA.
Now it's trying to convert 1.5v-0.68v = 0.82v to >6v, it will need something like 600mA input to try to get 5.7v 80mA out! Nasty.

If you MUST use the 34063 you could replace the 0.22 ohm resistor with a wire link, then use the output of the 34063 to drive a logic level NFET with a low Rds. After all that you might get the efficiency to 70% and get 5.7v 80mA out from about 3v 150mA input.

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I need about 80 mA at 5.7V. I'm powering an 18F1320 PIC, an SCA61T inclinometer IC, 2 LEDs, and an Xbee 1 mW transceiver.
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Why do you need 80mA? High efficiency LEDs will light up from 1mA each, the new PICs will run from maybe 2mA if you don't use a high freq xtal, so how much do the inclinometer IC and Xbee require? And why 5.7v? Were you going to add a low dropout linear 5v post regulator?

 

[MrAl]

I guess that's what was happening before I switched inductors from 47µH to 180µH. It was locking up at around 28V. I thought it was normal for there to be a brief overvoltage condition though. Is it not?

Also, I was reading up on the Li-ion batteries and most places say that they require a protection circuit to prevent over discharging. I didn't build that into my circuit but I think I should be OK if I use these batteries or something similar. The charger should take care of over charging protection and the monitoring circuit built into the battery should take care of over discharging, right?

Hi,

The Li-ion batteries are great if used right. Yes, there is a lower limit on the discharge, but many applications wont draw it too low anyway. For example, a high power LED and resistor, as the battery voltage drops the LED draws less current so it's harder to discharge below 2.5v.

What i found on the data sheet seems to differ with what MrRB must have found, because i found 1v drop in the transistor at 1amp. Thus, the design should work with 180uH, but then again we are working with very low voltages here. The peak current in the switch should be less than 0.5 amp for your required output from a source voltage of 2.7 volts. That's with 0.77 ohms ESR and 0.22 ohms sense resistor.

What i noticed too is that you seem to be using a 1.5nf cap for the timing cap. What made you choose that over say 1nf? 1nf offers a little higher frequency which might be better because that means it works better overall in most cases.

 

[Mr RB]

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What i found on the data sheet seems to differ with what MrRB must have found, because i found 1v drop in the transistor at 1amp.
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I was going from memory. The Vf of the chip's output stage does depend on how it is configured. Checking the datasheet you are right it is about 1v at 1A when configured as darlington, BUT with 5v Vcc AND with no DRI resistor! It says a forced beta of about 20.

At 3v Vcc it will be worse, and worse still as the drive comes AFTER the 0.22 ohm current sense resistor which drops Vcc, and much worse again becuse the OP is using a 180ohm DRI resistor.

So given a 3v Vcc and say 400mA on-current there will be 2.9v after the current sense resistor, through a 180ohm resistor into a 1v Vbe darlington (approximated) giving 10.5mA into the base of Q1 which has a 80 ohm resistor which soaks up about 7.5mA! So Q1 is only getting about 3mA base drive, it's Vce is going to be much worse than 1v...

Changing DRI to 100 ohms or less should really improve the circuit.

 

[RCinFLA]

With pin 8 tied before coil, the output device is not configured as darlington so you should have less then 0.5 v Vce sat on output npn for current you are running. With a 3 vdc input supply you should have a 51 ohm for the pin 8 resistor to give you about 32 mA (650 mA / 20 for forced beta base drive current) of output device base drive current. With 47-51 ohm to pin 8 with 3v supply, the Vce sat should be less then 0.4 v.

(3v - 0.7v Vce of driver - 0.65v Vbe of output device = 1.65v. 1.65v/32 mA output npn base drive = 51 ohms for pin 8 resistor.)

The pin 8 resistor needs to change based on input supply voltage. You can trade off the base drive overhead loss based on your operational sweet spot, taking a little higher Vce sat (and lower efficiency) at higher output current but saving base drive overhead at lower load current.

 

[Blueteeth]

For lower power, and lower voltage apps, I found the MC34063 to be rather disappointing in terms of effiency :/ But thats not a criticism, since the datasheet highlights the efficiencies of the example circuits, all of which have a moderate current output.

If you're aginst using proprietary chips (national semi, texas instruments, linear tech, zetex etc..) then, for <100mA output, and not tight frequency requirements, you might find a discrete boost converter to be more efficient. They're generally much less versatile, limited input voltage range etc. but for a one-trick pony they can work surprisingly well

 

[Mr RB]

With pin 8 tied before coil, the output device is not configured as darlington ...

Thanks RCinFLA, you are correct the driver is not darlington, I made a hurried mistake there.

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With a 3 vdc input supply you should have a 51 ohm for the pin 8 resistor to give you about 32 mA (650 mA / 20 for forced beta base drive current) of output device base drive current.
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(3v - 0.7v Vce of driver - 0.65v Vbe of output device = 1.65v. 1.65v/32 mA output npn base drive = 51 ohms for pin 8 resistor.)
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Close, the Vce sat of the driver will be less than 0.7v, more like 0.3v so total across Rdri will be about 1.95v, and you left out the 8mA base current lost into the 80 ohm Q1 base resistor so it needs approx 32+8mA across 1.95v = 49 ohms?

The beta is likely not quite as bad as 20 in actual use, so the most efficienct value of Rdri may be between 50 and 100.

And when the battery is low at 2.4v it's going to struggle.

 

[vne147]

Wow, thanks everyone for your help and input. I can report that over the weekend I tested out the bootstrapping idea I described in post #12 of this thread and it did not work. I tried the configuration both running on batteries and running on my bench supply. Instead of the expected 5.7V, the output was roughly .4V below the input voltage. I tried several different input voltages and the output was always about .4V below that. I'm not sure why the idea didn't work but I'm not surprised because the datasheet makes no mention of doing this. Oh well, I thought it was worth a shot.

If you MUST use the 34063 you could replace the 0.22 ohm resistor with a wire link, then use the output of the 34063 to drive a logic level NFET with a low Rds. After all that you might get the efficiency to 70% and get 5.7v 80mA out from about 3v 150mA input.

Can you elaborate on this idea please?

Why do you need 80mA? High efficiency LEDs will light up from 1mA each, the new PICs will run from maybe 2mA if you don't use a high freq xtal, so how much do the inclinometer IC and Xbee require?

The 80 mA requirement comes from:

Xbee = 50 mA (Peak RX current)
PIC = 1.8 mA (Internal 4 MHz oscillator)
Inclinometer = 4 mA (Maximum current consumption @ 5V VDD from the datasheet)
2 x LED = 2*10 = 20 mA

The LED is not high efficiency (although in hindsight it probably should have been), it is a bi-color red/green LED. I sized their resistors so they are only drawing 10 mA each.

All that adds up to 75.8 mA. I just upped it to 80 mA to be safe.

And why 5.7v? Were you going to add a low dropout linear 5v post regulator?

The 5.7V will supply a 5V LDO linear regulator which will supply the PIC, inclinometer IC, and LEDs. I chose the LP2985-50DBVT. The max dropout voltage will probably only be about 150 mV. I'm struggling to remember why I decided on the 5.7V. Maybe I just wanted a buffer but that still seems like a bit much.

What i noticed too is that you seem to be using a 1.5nf cap for the timing cap. What made you choose that over say 1nf? 1nf offers a little higher frequency which might be better because that means it works better overall in most cases.

I am using the 1.5nF cap because that was what was used in the datasheet for the example step up converter circuit. I did try a 1nF cap this weekend but unfortunately, it didn't work. Not sure why.

At 3v Vcc it will be worse, and worse still as the drive comes AFTER the 0.22 ohm current sense resistor which drops Vcc, and much worse again becuse the OP is using a 180ohm DRI resistor.

Similarly to the 1.5nF cap, I chose 180Ω because that was what was shown in the datasheet for the step up converter example circuit. I suppose that is one of the pit falls of my paint by numbers design methodology. I am not an EE and don't always understand how all the individual pieces work but I am usually pretty good at following directions and using them as building blocks. Usually being the operative word here.

If you're aginst using proprietary chips (national semi, texas instruments, linear tech, zetex etc..) then, for <100mA output, and not tight frequency requirements, you might find a discrete boost converter to be more efficient. They're generally much less versatile, limited input voltage range etc. but for a one-trick pony they can work surprisingly well

I'm not against using any proprietary chips. I just had a bunch of 34063s and inappropriately chose it for this project. Now I'm just trying to avoid starting over. For future projects though, I'll definitely consider the alternatives that you and others have identified for me.


So, the consensus so far seems to be that I should change the 180Ω resistor to 47Ω one and that I should replace the .22Ω resistor with a wire link. Unless anyone has any objections, I'll try that this afternoon.

Also, I did order some 3.7V AAA Li-ion batteries from China. If all else fails, I'll just bypass the 34063 on my PCB completely and power the two LDO linear regulators directly from the batteries. I think I'll keep trying to kick this dead horse until the batteries arrive though.

 

[RCinFLA]

If you MUST use the 34063 you could replace the 0.22 ohm resistor with a wire link, then use the output of the 34063 to drive a logic level NFET with a low Rds. After all that you might get the efficiency to 70% and get 5.7v 80mA out from about 3v 150mA input.

It would be a bad idea to eliminate the current limit resistor even if an external MOSFET switch is used. This would 'put the pedal to the metal' on startup and for battery operation would collapse battery voltage on startup potentially causing a continuous oscillation on startup attempts.

 

[Mr RB]

I agree it's generally not a good idea to remove the current limiting, but the 34063 has fixed on/off times so it still oscillates even when the output voltage is too low to operate the regulation cutoff.

It's probably safe enough to remove the current limiting, I've done it in 34063 buck circuits and relied on the fixed oscillation for startup. If the OP is worried he can try the 47 to 100 ohm resistor first, and see if it starts to work good enough.

Another suggestion would be to use a higher inductance L1 but with low ohms, say a 470uH <0.5 ohm. That will reduce current ripple and peak current.
And removing the LDO 5v regulator and setting the 34063 to 5.0v will reduce input current maybe 15% too.