dc dc converter step-up output load issue

[fuji]

Hi Ron,

Well could you show your circuit too? My calculations tell me about 150ma peak during start up but dropping down significantly later. I am using 36 ohms total series resistance and shooting for 11ma output at 11 volts (1k load), with capacitance.

I guess at some point we could try simulating the more exact circuit. The feedback mechanism of this chip is a little different than most switchers but should work to some degree.
One guess would be that it cant start properly with load so it never gets up to the full operating conditions.

My circuit above as I checked also has around 150mA-160mA peak on start up. I had a current drop right after Rsc at around 8mA on Rd resistor.

Do you suspect that this could be a cycling issue within the IC because of the voltage drop on the output load? Perhaps the current I want is too low for it to properly load the inductor.

Nokia used this IC in their products but I'll assume they had higher currents for step-up and step-down with their products which made it suitable for their applications.

What boost converter do you recommend but not too expensive? Maybe a successor to the mc34063.

 

[ronsimpson]

Think about the TS5171 from TI. I have not used it. They had some nice parts but they cut off below 7V. I think that is a voltage to stop pulling from the battery. LV5636 and LV56351 cut out at 8V.

 

[ronsimpson]

I have used several of this family for "flash light LED bulbs".

 

[fuji]

I have used several of this family for "flash light LED bulbs".
View attachment 91811

Thanks for showing me that schematic. Does this boost converter become accurate at very low current outputs as well? I'll need to check the datasheet.

For this schematic, where is the power source coming from? A 9v battery or a power box? I see the Vmin is 5.

My power source is a 9v battery, so I'm assuming the minimum for 9v battery is +-10, which is I think 8.5v as min.

 

[ronsimpson]

Thanks for showing me that schematic. Does this boost converter become accurate at very low current outputs as well?

You have the voltage min there at 5. Since there is 9V max, shouldn't their be a +-10 of 9, which is 8.5v as min?

The software I used asked for Vin min and Vin max. I just entered numbers. Probably 8 to 9V would be fine. If you let your battery discharge too much it will likely leak with some time.

 

[MrAl]

My circuit above as I checked also has around 150mA-160mA peak on start up. I had a current drop right after Rsc at around 8mA on Rd resistor.

Do you suspect that this could be a cycling issue within the IC because of the voltage drop on the output load? Perhaps the current I want is too low for it to properly load the inductor.

Nokia used this IC in their products but I'll assume they had higher currents for step-up and step-down with their products which made it suitable for their applications.

What boost converter do you recommend but not too expensive? Maybe a successor to the mc34063.

Hi,

Theoretically i think this should work, but yes there are practical aspects that come into play. The BH curve for inductors is like a stretched out lazy S, so yes the inductance can do down with extremely low current. It would be hard to believe here though with such a low current inductor and the current is not the same as the load current it can be higher when operating with some max and min.

What do you mean by "current drop"?
Try to describe the problem as well as you can.
It's not unusual for a new design to have a problem but usually there is a solution and the solution is found by making measurements and comparing those measurements to theory. I have the theory but you have to supply the measurements.

We could look for a spice model and do a few closer to real life simulations. I thought i tried this a few years back but ended up making my own model anyway. I'll see if i can dig it up.

LATER

Ok dug it up, tried it out, but first i had to modify it because i had it set up for buck. Now that it is boosting, i have trouble getting it to fail with anything but an input voltage that is too low.
This tells me that there is a chance that the battery and wiring can not deliver the current when needed, so the dynamics of the input supply cant meet the dynamics of the boost circuit. To possibly help this situation, you could try an electrolytic cap in parallel with the battery, right at the PC board. This may allow for a higher peak current reserve which could get it working. So measuring the input voltage with or without load with a DC volt meter is not good enough if it does show normal voltage but the circuit still doest work, it would have to be a scope, or just tack a 10uf cap in parallel.
This is just one thing to try though.

 

[ronsimpson]

The inductor in the MC34063 design must be able to handle a much larger current then you would think.
Most PWM will deliver a little power with every cycle. The MC34063 often is in a mode where it grabs a huge amount of power, one out of 20 cycles. Then it skips the next 19 cycles. So the inductor must be sized for a much higher current then you would think.

 

[fuji]

This tells me that there is a chance that the battery and wiring can not deliver the current when needed, so the dynamics of the input supply cant meet the dynamics of the boost circuit. To possibly help this situation, you could try an electrolytic cap in parallel with the battery, right at the PC board. This may allow for a higher peak current reserve which could get it working. So measuring the input voltage with or without load with a DC volt meter is not good enough if it does show normal voltage but the circuit still doest work, it would have to be a scope, or just tack a 10uf cap in parallel.
This is just one thing to try though.

Thanks for the reply. Was a little busy with family and couldn't respond to both of you sooner.

I'd like to let you both know that their are two types of 9v batteries I am testing with. Alkaline (500-600mAh) and Carbon-Zinc (200-250mAh).

I'll get on with the electrolytic cap in parallel and get back to you on the result. Maybe then the inductor can have enough current to power up the output.

As for the battery and wiring that can not deliver the current on your side, I am wondering if my breadboard is causing a similar issue. Are you using the regular pc board with the througholes? I do have the pc boards to use and might as well keep the mc34063 circuit off of the bigger breadboard I'm using instead. I'll solder an 8 DIP socket into the board to prevent heat damage to the mc34063. Maybe their will be a better connections with the components.

I'll turn on my oscilloscope and check the peak to peak current right after the battery with the 10uF cap. My measurements and calculations for the mc34063 (step-up) are based off the equations in the attachment. Theres also my older calculations for the schematic I first posted. Not sure if that calculator is accurate or not.

 

[fuji]

Ok dug it up, tried it out, but first i had to modify it because i had it set up for buck. Now that it is boosting, i have trouble getting it to fail with anything but an input voltage that is too low.

Quick question. Did you also test (with a power box) the output voltage if it was a constant 11volts by lowering the input voltage of the IC? The inductor should still conduct enough to keep 11v going out despite lowering the voltage from the input, even at around 3.5 volts as I've seen myself with my older mc34063 circuit. Most likely until the power source is dead.

When I have a constant 11v output using a 9v battery from the input of the IC, the current on the output is what it should be (6.2-7mA). The problem is, when I lower the input voltage (8v, 6.5v, 4v, 3.5v etc) of the mc34063, it would stay at 11v on the output, but the current goes down on the output. Is this normal based on Ohms law or should the current output still be the same (7mA) despite lowering the input voltage of the converter? What I do know is, the additional instruction of the converter is to never exceed 7mA, but it doesn't have the instruction to keep the current the same (7mA) when the power gets low. Or should it also be doing that as well?

 

[MrAl]

Hello again,

With a resistive load, if you have 11v output constant then the current should be constant too. If it is 7ma then it should stay at 7ma unless the voltage changes from 11v to something else.
If you measure 11v and the output current drops, that would indicate that the circuit could be in discontinuous mode which may make the voltage meter look constant (or nearly so) while the current changes (depending on what current meter you use too). So check the output with a scope.
Normally a scope is used to check many of these problems. A voltmeter is used after the basic operation has been established.

In my simulation i was able to get down to about 5v before the output starts to drop. This could be because i am using only one transistor as the driver. Depending on the voltage drop, it may take a little more to keep the output switching higher.

Id be very interested to hear how the input cap works, if it helps, and try to use a scope. The output voltage should be fairly constant with just a little sawtooth. If the sawtooth grows too much that could indicate a problem with low input line (low input line is the lowest input voltage ever expected during normal every day operation) or with the dynamic input reserve (input capacitor again).

To test the entire dynamic input capability, you could look at the voltage at the left or top side of the inductor (terminal not connected to the transistor). The most demand comes near the end of the 'on' cycle, where the inductor is passing max current. If the voltage at the inductor drops too much at that time, it limits the output, and this could be because of the input capacitor or any other impedance that happens to be present between the input capacitor and the inductor. You could also make sure the transistor does not drop too much voltage during the end of the 'on' cycle (scope again).
Since the ground is an important current passing part of the circuit too, you could check for any voltage drops in the ground line from input to the bottom of the output cap, again using the scope so you can catch the activity near the end of the 'on' cycle.

 

[Tony Stewart]

The series resistance of the coil must be less than the Current Sense R to get any accuracy on predictions. This drives the inductance down from 1mH to 10uH~100 uH for sub 1 Ohm Rs

Here I simulated the essence of the circuit using a Hysteretic oscillator and same Darlington and a 10uH inductor.

The Pull up-down cap stores the initial and final voltage which are the same (V/2) to eliminate under/overshoot on startup. ( Press Reset)

allow Java here

Feel free to edit the values. ( enlarge window, change Load R and current, ESR included in Cap for realistic effects. Change anything you want) ( rt click edit)

 

[MrAl]

Hi,

After reading and contributing to this thread i am almost tempted to run one of these circuits up myself
Not sure if i will be able to do that though as other things are on the table right now.
But this is quite interesting, and with my home made model i see 25uH might work, but i am being cautious here because usually we see 1mH inductors with these circuits, or maybe that is just with the buck configuration. I'll have to go back and check my model against the data sheet again.

 

[Tony Stewart]

Looks like it is possible with 1mH 10R in my design.

Vin 9 V 10 mA
Vo 11.1 V 7 mA

Adjust Rfb for Vo and selected R load for 7mA.

But series R in coil and battery will cause Vo variation with load due to load regulation limits of high ESR design.