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NCP7662 Voltage doubler calculate current output — Parallax Forums

NCP7662 Voltage doubler calculate current output

metron9metron9 Posts: 1,100
edited 2006-11-24 16:12 in General Discussion
I am not sure how much current this chip can output.

The datasheet:

www.datasheetcatalog.com/datasheets_pdf/N/C/P/7/NCP7662P.shtml

Says Power Dissipation (Note 2.) 720mW

So using it as a voltage doubler


The NCP7662 may be employed to achieve positive
voltage doubling using the circuit shown in Figure 9.


at 5.5V input and 11V output I would calculate A=W/V or A=.720/11 = 0.065A

Is that correct? I only need 9.1V. At 45ma, Right now we are using a voltage booster from Wall electronics but they are $15.00 each. The chip I am looking at is only $1.04 in 500 Qty

Its funny too I can't find that part on the On simi site but I got a quote on them from Rochester Electronics, LLC for 50 and 100 qty.
Think Inside the box first and if that doesn't work..
Think outside the BOX!

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Think Inside the box first and if that doesn't work..
Think outside the BOX!

Comments

  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-11-22 22:56
    metron9 said...
    Says Power Dissipation (Note 2.) 720mW
    That is total package dissipation, so it'll be more like VIN * IIN .·
    Remember, POUT will be·< PIN.·
  • metron9metron9 Posts: 1,100
    edited 2006-11-22 23:17
    Sorry PJ I am still confused. I understand it says package dissipation but that is the only power rating I can see on the datasheet, why can't thay just put a maximum mA output figure on it. All I want to know is can I run a 9.1 V 45mA load using 5.5V input.

    So if I take a stab at what I think you mean.

    If I need 11V at 65 Ma i would then need 130ma at 5.5V so 5.5V * .130A = .715W within the parameters of the chip.

    So the answer is yes I can run a 9.1V circuit that draws 45mA ?

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Think Inside the box first and if that doesn't work..
    Think outside the BOX!
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2006-11-23 03:22
    · The datasheet doesn't provide much info on the "voltage multiplier" application, it's mostly on about the "negative converter", especially the device efficiency aspect.·
    metron9 said... why can't thay just put a maximum mA output figure on it.
    · There is specified, in a graph anyway (the document's page 8), output current/s possible with the negative converter (I wouldn't infer that's valid data for the voltage multiplier.)
    ··I wonder what you're trying to operate with this IC.· LEDs or something?· That could be done with the negative converter easily enough.· The cathode has only to be more negative than the anode.

    · Sometimes you have to re-arrange what's in the box.· lol.gif·
    7662.JPG
    227 x 255 - 5K
  • metron9metron9 Posts: 1,100
    edited 2006-11-23 04:53
    No actually it's the gauge project I was working on where we tried to get a tachometer gauge to run at 6V instead of 12V Remember it has a zener regulator fixed at 9.1V so that is the voltage it needs, it draws maximum 45mA when I tested it, actually it matters not if the gauge reads 0 or 8000 it draws the same so the circuit is regulating the current by shunting it to ground.

    Anyway He found some 6V to 12V DC-DC converters and in small quantities his price was 25 bucks ea. per hundred 15 each.
    1.5W, 12V 125mA part PK1.5-D5-S12
    http://www.v-infinity.com/adtemplate.asp?invky=94723&catky=328060&subcatky1=248699&subcatky2=285565&subcatky3=

    They are out 3 weeks to get any so i continued to look.

    I found some at a company called Wall something, I cant find it now on this computer, but they were a little smaller, like I said I know the current won't exceed 45ma and the ones he is going to test are 85ma I think. But they are 15.00 each and they had 5 in stock so he bought them.

    So I continue to search for a dc=dc converter that perhaps we can build on a small circuit board for less.

    The converters we have found have pins so to do it right we still should put it on a board for wiring as it is in automotive and a solder joint bouncing around unconnected to a board directly soldered to a pin on the converter simply is not a good idea. I think a small board and some tiny pc snap standoffs would work great.

    Since a board is necessary and soldering anyway, I thought a chip like this one might just do it, if not enough power they could be doubled up I would think.

    Suggestions on DC-DC converters welcomed...

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Think Inside the box first and if that doesn't work..
    Re-arrange what's inside the box then...
    Think outside the BOX!

    Post Edited (metron9) : 11/23/2006 4:58:22 AM GMT
  • Chris SavageChris Savage Parallax Engineering Posts: 14,406
    edited 2006-11-23 20:17
    metron9 said...(trimmed)
    Sorry PJ I am still confused. I understand it says package dissipation but that is the only power rating I can see on the datasheet, why can't thay just put a maximum mA output figure on it. All I want to know is can I run a 9.1 V 45mA load using 5.5V input.
    Metron,
    ·
    · Usually when the Power Dissipation is listed rather than current it is because there is a variable voltage range.· When that happens the current rating will change with the voltage.· Ohm’s law will help you here.· Divide the Power Dissipation by your voltage and you’ll get the max current.· Take care.

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    Chris Savage
    Parallax Tech Support
  • Tracy AllenTracy Allen Posts: 6,666
    edited 2006-11-24 16:12
    This device is a switched capacitor voltage doubler. Charge is transferred at a rate equal to the switching frequency times the capacitance times the input voltage. Since current is charge transfered per unit time, you can't draw more than that at the output. Some of these chips have a means to increase the switching frequency to get more current.

    The output resistance is the parameter to looki at in the data sheets, and the data sheet will also have information on the ripple at the output as a function of current. For example, an output voltage of 10 volts and an output resistance of 80 ohms, will drop to 10 - (80 * 0.4) = 6.8 volts when the output current is 40 ma.

    That same chip is available from Maxim and Intersil, I think with the ICL7662 prefix and also from NJU7662. There is a high efficiency version available (lower internal switch resistance) with a D suffix.

    TI makes some very nice little voltage converters in DIP packages capable of one watt output, that cost around $5.

    Is it possible to operate your tach at a lower voltage without using the zener, if the voltage is regulated? That is possible with the LM2917LM2907 tachometer chip.

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    Tracy Allen
    www.emesystems.com
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