LM2623 Design Problem

As you discovered, you can't help yourself by trying to enhance the LM6263's n-mos switch to ground with a p-channel source follower. That's because the follower needs gate voltage to turn on, and therefore can't act as a switch to ground. E.g., the Si4435 needs about Vgs=3V to conduct 4A at Vds=Vgs, according to its characteristic curves. So if you pull the gate to ground, and push 4A into the source, it'll be sitting at about 3V above ground.

Your best bet would be to select a switching-controller IC that is meant to drive external mosfets.

But if you really want to use a part like the LM2623, and interface it to an external larger n-channel mosfet, then you need an inverting mosfet driver. Something like TI's tps2828, which can drive a medium-sized mosfet.

Delay times are an issue, so you'll need to use a low-value pullup resistor to create the logic signal. Usually when attempting tricks like this, lower-speed controllers are used. 2MHz is pushing it.

Hi All,

I've tried the above method to no avail.

I've got a question to ask and hope someone would be able to help. The datasheet may be downloaded at National semi website. It is stated teh LM2623's output voltage is programmble within certain range. But the dsheet only lists the electrical specs (eg. load current for given Vout=5V and various efficiency vs. Vin graphs etc) at a given Vout. Since its Vout is programmble within certain range via resistor divider, how can one have an intuitive feel what is the max load current a step up dc-dc converter chip (such as the LM2623) can deliever for a certain Vout other than that stated in the dsheet?

I am not asking for detailed calculation, just ball parking figure to give gut feel whether certain step up dc-dc converter part is worth pursuing which may just meet my electrical requirement.

Many thanks in advance!

> I've tried the above method to no avail.

If you want to continue with the LM2623A part, you could give us more detail about what you tried (sketches would help) and what problems you ran into.

> intuitive feel what is the max load current a step up dc-dc converter

SMPS 101. We'll use the O.P.'s specs: convert 3V in to 9V out, at up to 700mA load. He's selected a converter IC with an internal switch rated at 2.2A.

A step up, or boost converter switches an inductor to ground to charge it, I_SW, then opens the switch so the inductor can "flyback" to a higher voltage, and discharge its current through a diode into an output capacitor. If D is the switch-ON time, defined as a fraction of the switching period, then

V_OUT = V_IN (1 / 1-D).

A boost converter with a high voltage step-up ratio has a long switch-ON time, D, compared to a short output-charge time, 1-D. For example, to step-up 3V to 9V, a factor of three, 1-D must be 0.33, so D = 0.67, or 2/3 of the time.

I_OUT = I_IN * V_IN / V_OUT.

We can see that only a small portion of the input current, 1/3, ends up, on average, at the output. Assuming a constant current in the inductor, if you have a switch capable of 2A conduction, the best you can get at the output is 2/3 of that, or 0.67A, for our 3V to 9V example.

In reality, the situation is worst than that, because we can't have truly constant current in the inductor. Even in continuous-conduction mode (right), it'll be ramping up and down, and I_L-AVG will have to be some fraction of I_SW peak, even as poor as 1/2 I_SW(max), which could mean in this case that I_OUT = 1/6 I_SW max, or 330mA. In discontinuous mode (above), it would be even worse than that.

Additional losses are in the switch's ON resistance and in the output diode.

OK, now turning to our O.P's situation. First, I wonder if he's using the LM2623A parts, rated at I_SW(max) = 2.2A min, as opposed to the LM2623, rated at 1.2A min, that could be an issue.

He states that he's verified continuous-conduction mode, but hasn't said how much his current varies. To approach I_OUT average = 1/3 I_SW, he needs a large enough inductor so that the inductor current approaches a steady I_SW = 3 x I_OUT. And of course the inductor must not saturate at I_SW max.

Things look on the hairy edge for our O.P., but there is one silver lining in the cloud: he's driving a solenoid coil. When he first turns on the coil his desired voltage will be available, and the solenoid should actuate. Thereafter the excessively-high current may cause his output voltage to sag, and the current will fall from its maximum. But if the solenoid remains activated at the lower current, perhaps he can live with it.

It's common for pull-in currents to exceed steady operating currents. In fact it's not unusual to intentionally lower relay and solenoid currents after activation to avoid overheating their coils.

Things are also on the hairy edge for the LM2623A, but if the solenoid activation times are short, it may be OK. However, we'd really like to see a better safety margin.

The O.P. asked about better ICs. There are parts like NSC's LM2588 that can easily handle higher currents, I_SW(max) = 5A (min), and provide the safety margin we'd like to see. But the LM2588 can only work down to 4V, so that's out. It also only runs up to 200kHz, which means a larger inductor.

Aha, LTC has a part, the LT1370, rated at I_SW = 6A max, and it runs at 500kHz! It'll work down to 2.7 volts, and looks pretty good. Both of these parts are considerably more expensive than the LM2623A, but hey, that's life!

Hi Win, All,

I will post the findings for the above 1-2 days later. There is some discrepancies I could not understand but I hope get to learn more from that.

Meanwhile, am bus with other stuff but got time to look for an alternative solution to the more (MC34063E).

The datasheet is available at: http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00001232.pdf

Application Note is available at

http://www.intusoft.com/onsemipdfs/AN920-D.pdf

I am using the DC-DC converter to drive coil which is needed to vibrate at certain frequency. The On/off control to be done by MCU. I cannot get away with lower drive voltage else the vibration will get weaker (as vibrating force will be used to drive something). The coil resistance is about 8ohms + so the current drawn will be about close to 0.7A to 0.8A. I have attached the scanned images of my calculated design with MC34063E and am posting them here so I can get some comments before proceeding to build the circuit. Pls take a look and give me your comments.

I am particularly uncomfortable with peak switching current Ipk(sw). How does one visualise this? Does it means the battery source have to supply this instantaneous current? Also, is an inductor with current rating higher than this needed? Or can I get away with one with a lower DC current rating? Cause this will affect the physical size of my circuit(components).

The worksheets are located at:

https://www.yousendit.com/download/MEtRN3RTZ2dCSWRjR0E9PQ (pg1)

https://www.yousendit.com/download/MEtRN3RkUnFlcEpFQlE9PQ (pg2)

Thanks!