current reducer 1A to 250 mA

Hello,

I am using a PIC18F4550 for a circuit. I have a DC-DC converter that has an output of 5V and 1A.
I would like to reduce this for the PIC to a max of 250mA.

I am also using an LCD and a couple of sensors that each require 5V. I had considered just using a resistor to reduce the current, but I cannot afford the voltage drop.

What would be a good circuit for this? Would a combination of transistors be a good way to reduce this output current?

Would anyone have any good links to follow on this?

Thank you as always.

you do understand that a power supply RATED 1A does not HAVE TO output 1A, right?

i do understand... I have a few more things on this circuit that can draw upto 0.8 A however and I do not want the PIC to see this much current since it does funny things above .25A.

i do understand... I have a few more things on this circuit that can draw upto 0.8 A however and I do not want the PIC to see this much current since it does funny things above .25A.

Click to expand...

from your answer, it is clear that you do NOT understand.

I am using an EC4A21H from CINCON...
so you are saying that I could connect this straight to the PIC and I would be just fine?

Thanks btw.

- do you want to supply power to the PIC with this DC/DC converter?

or

- do you want to measure the current supplied by the DC/DC converter?

so you are saying that I could connect this straight to the PIC and I would be just fine?

Click to expand...

I said none of that.

what I did say is that you simply do not understand this whole thing. reading the pic datasheet or a good textbook on electronics 101 should help you tremendously.

a pic will draw whatever it is going to draw from a voltage source (like a power supply), regardless of how much current the power supply is CAPABLE of supplying. it is like you can drive a car rated at 300HP at less than 300HP. the current rating on that power supply is simply an indication of its capability. when you connect a load to it, the power supply may or may not supply the current at its rated capacity.

now, mcus in general consume very little power. that can be an issue for some power supplies if its specified minimum load current is greater than what the mcus can consume. whether or not it is an issue with your combination of power supply and mcus is only known to you as you have to read the datasheets and look at your application.

good luck with that.

Kurenai_ryu - yes I am trying to power the pic with the DC-DC converter.

Millwood - I get what you keep saying, and the datasheet does not give this info, unless I am missing something on it (you are welcome to take a look). I have read the datasheets, I DO understand, and this is why I am asking for ADVICE on a current reducing circuit to use.
Dont get me wrong I appreciate the reply, but I tend to get "go read" way too often and I have read. Please quit assuming everyone is a 4 year old.

The forum is meant to discuss these issues and I am asking for a recommended current reducing circuit or some specific links that point me in that direction.
I am not sure if using dual transistors is the best choice or if just a resistor to ground would be adequate.

It would be possible to make a current limiter but you will lose some voltage and voltage stability due to the current sensing resistor and series pass transistor. You could put it on the input side of the DC-DC if the input voltage range was not too wide.

However, I cannot understand why you need a current limiter. Is there some circuitry round the PIC that may demand a high current that you want to prevent? In general, if a device needs a high peak current then you supply it rather than limit it. I guess you have had some experience that has made you want this limiter so it would be useful to understand that to help.

Keith

I am asking for a recommended current reducing circuit

Click to expand...

you are asking that because you still don't understand the "problem" you think you are facing. You are essentially trying to solve that problem that may not exist in the first place, even 4-yr olds don't do that much often.

I am using an EC4A21H from CINCON...
so you are saying that I could connect this straight to the PIC and I would be just fine?

Click to expand...

The answer is simply yes. I don't know this particular device, but all similar types have proved to supply 5V logic and processors safely. If you fear transient overvoltages, a TVS diode in parallel to the converter output can be used, but it's not necessary generally. Other circuits shouldn't drive out current to the 5V bus, however.

The other prerequisite is to keep the maximum ratings of the PIC's I/O pins. Not to cause short circuit at the outputs, not do drive input current above or below the supply rails. Otherwise, it could be easily damaged with or without 250 mA current limit. There are a few cases where a VCC current limit can be expected to protect the device, e.g. if you manage to cause a device latch-up by seriously overdriving a pin out of the rails. But it's rather unlikely with modern chips.

don't forget, a DC power supply of 5v/1A, can supply at most 1A but also can supply at lower currents too! (it should be in the datasheet-the lower current.) the DC power Supply, 'tries' to keep the output voltage, from low current to a maximum current - always keeping the output voltage (5v) ofcourse it's not perfect and near or over 1A this voltage drops...

most microcontrollers works with a DC power supply, so, they need a constant 5v (commonly) to work, but depending the load you put on the micro or the complete circuit, it will drawn more and more current, (mostly by the i/o ports, and internal peripherals...)

a common micro, can drawn around 1mA to 100 mA (or more depending your complete circuit) so a 5v/1A power supply will keep the 5v over that range...



...

Maybe when you set more and more loads to the same power supply (800mA? near 1A?) the voltage get distorted and that makes your PIC 'see' funny things, as the pic NEVER sees all this current (or more appropriately, never should drawn that much current... in that case, limiting the current will not do anything more that make your problem worse as it will drop the voltage considerably...)

if you want a more accurate answer, post your circuit. (including the power supply) and where do you measure the 250mA or the 800mA or what current and voltage do you measure at certain points...

Hello!

Ok, let's try to have another run.
Your DC/DC converter can supply 5V / 1A.

Your PIC needs, say, 10 mA, which is 0.01A.

If you plug your PIC directly to your power supply, what will happen? Nothing.
It should work fine. Your PIC will draw what it needs, and your DC/DC converter
will continue to feed it with 10 mA, not more, not less.
And since it's capable of providing 1 A, it just means that you can still provide
0.99 A to other devices.
You say that your "few more things" draw upto 0.8 A. Therefore adding these few
more things and the PIC, the whole stuff will draw 0.81 A, nor more, not less.
No need to reduce the current.

Millwood used an image of a 300 hp car that you don't HAVE TO use at
300 hp, and he's right.

Another analogy. You enter an elevator that can handle 8 persons / 600 kg.
You're alone. But it will work fine, you don't need to reduce its power
before boarding, although you are far (I hope) from weighting 600 kg.

Now a last analogy which is closer to the subject. You have a plug in your
kitchen, that can provide 15 A at 220 V. If your plug an oven, you will likely
be close to the limits. Now if you plug a fan that uses about 1A, you don't
need to reduce the current of the plug and it works fine. I hope you agree
on this one.

Dora.

Unless you are using a load which require a constant current source like high power LED's, you neednt control the current for any microcontroller. But point to remember is that these devices require a regulated voltage source. When you connect it to any current with a regulated voltage, the PIC will draw only what it is required by it to work at full efficiency. Stiil if you like to be on the safer side, you may connect a 1E 0.25 watt resistor in series to the Vcc ternimal which will drop the current to certain extent depending in the PIC's consumption. But i personally dont see any need for it.
Cheers

I like the elevator analogy from @doraemon !

hope all of this helps

Thank you all. Yes those are good analogies.
I will try to make a schematic and post what I am doing.

I have tried so far to just connect the DC-DC to the PIC and LCD.
(previously I had been powering them both with the programmer...)

I do notice that the LCD is brighter (although I am measuring 5.016V from the converter {i will measure the current here shortly}).
The pic 'boots' up just fine, but then after it gets to the main screen it seems to lock up (which it had never done before).

I will post the schematic.

Thanks again.

---------- Post added at 06:15 PM ---------- Previous post was at 04:26 PM ----------

With connecting the uC to the DC/DC converter I have placed a jumper inline and measured the current through it at ~11mA. which is great...
but even with everything removed except the PIC and LCD I am getting the PIC to constantly reset.
I have specified in the code to disable the brownout however.

any other reason this could be resetting like this?


also, FvM, could you clearify this for me: "keep the maximum ratings of the PIC's I/O pins."
thank you...

So far, after connecting the DC/DC I am getting either the pic to Lockup at the first screen, or I get it to constantly reset.

Again, I will post a schematic soon.

Here is a very quick schematic for the parts that i have wired currently.

**broken link removed**
**broken link removed**

the only component that I did not show are two 22pf caps on the main clock to the pic.

I am using PWM to control the brightness of the LCD and the motor speed. The motor has an internal controller and I just PWM to the speed pin. It is a 48V DC motor (hence the 48v circuit)

On the main screen I am displaying time, pressure, (ultimately CFM), temperature.

The buttons allow me to change menus, fan speed, etc.

Like I say, this all works fine until I removed the programmer and replace the power with the DC/DC converter.

The current does not seem to be the issue however.

Could a 100KHz switching frequency be affecting anything?

Also, I am supspicious of the connection for the brightness to a 5V supply...

Thanks again everyone.

Could a 100KHz switching frequency be affecting anything?

Click to expand...

that's a fair easy thing to confirm, isn't it?

in my experience, it hasn't been if you do a good job decoupling the supply, as suggested by the OEM.

---------- Post added at 09:13 PM ---------- Previous post was at 08:46 PM ----------

wow, did you really connect the ccp pin to the 5v supply?

AND there is no diode on the motor?

AND there is no pull-up / pull-down resistor on the tech line?

did you really do that?

I am decoupling...

I do have a resistor on CCP2 that I left off of the schem...

No I do not have a diode on the motor, thanks for pointing that out.

Yes my 'tech line' does have that... I actually use a secondary board which is not shown... i never put that on the primary board just as a preference.

I know that you are trying to be helpful millwood. And i do appreciate that, but i have one word for you: Lexipro!

here is what I see as your fundamental problem. You cannot post a schematic saying that it reflects your system, and then turn around saying that many items are left out of that schematic so that it does not reflect your system.