><Digitm...@gmail.com> wrote in message >news:1161498877.726962.314130@e3g2000cwe.googlegroups.com... >> Hey I need some help calculating some resister values, I cant seem to >> find the formula anywhere. I'm trying to trigger a NPN transistor to >> switch on a set of leds when my IC's output goes HIGH. Heres my values:
>> NPN transistor: >> Typical Hfe - 200 >> Max Vce - 30v >> Ic - 800mA >> Power dissipation - 1.8w
>> What I need to happen is when the output goes high is to turns on 6 >> leds (parralle) operating at 20ma each from a 9 volts source. I already >> know I need a 39 ohm resistor for the leds, what I really need to know >> is how to calculate the resistor value I need between the IC and the >> transistor.
> It looks like you might be pushing it a bit from the figures you've >given. (6*20mA)[LEDs] / 200 [Hfe] = 0.6mA, but the IC can only supply 0.4mA >@ 3.5V. What voltage will the IC's output drop to when sourcing 0.6mA? > What is the IC and what is the transistor?
This thread should be in sci.electronics.basics, I think. Adding it there, with followups set to .basics.
Johnny, the transistor's beta is 200 only if it is kept out of saturation and even then only if not at too high currents, often. So the situation is probably still worse than just the 600uA versus 400uA would otherwise indicate.
I think this kind of thing does call for two transistors, to relieve the load on the IC output.
Q1 gets close to saturation in this case because Q1's collector probably has to get dragged down some 800-900mV. That places it very close to the Q1 base voltage, when on. So if the 3.5V drive is a little higher up than 3.5V (and it might very well be) then the base drive current will rise a lot. So really this isn't so good, even though it really limits the drive current -- if the Q1 base voltage is kept low enough to keep Q1 out of saturation, but not so well otherwise.
So perhaps a better circuit using another resistor is:
That will load the IC output more, as Q1 is also saturated now, but it will deal with an input voltage that spans the likely range and the worst case current is about half the IC output max spec of 400uA.
> On Sun, 22 Oct 2006 17:39:10 +1000, "Johnny Boy" <NoS...@Please.com> > wrote:
>> <Digitm...@gmail.com> wrote in message >> news:1161498877.726962.314130@e3g2000cwe.googlegroups.com... >>> Hey I need some help calculating some resister values, I cant seem to >>> find the formula anywhere. I'm trying to trigger a NPN transistor to >>> switch on a set of leds when my IC's output goes HIGH. Heres my values:
> That will load the IC output more, as Q1 is also saturated now, but it > will deal with an input voltage that spans the likely range and the > worst case current is about half the IC output max spec of 400uA.
Woaw... 2 transistors and 3 resistors.
Just use the right transistor for this: logic level mosfet.
<fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote: >Woaw... 2 transistors and 3 resistors.
>Just use the right transistor for this: logic level mosfet.
Makes sense. Particularly in the 120mA region.
I know that just about any cheap NPN and PNP will work in the circuit I gave. Junk box stuff. I'm not nearly as familiar with MOSFETs. Can you point me to a good example of any one that would work great with 3.5V on its gate? I'd like to learn more, myself.
> On Sun, 22 Oct 2006 11:48:33 +0200, Fred Bartoli > <fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote:
>> Woaw... 2 transistors and 3 resistors.
>> Just use the right transistor for this: logic level mosfet.
> Makes sense. Particularly in the 120mA region.
> I know that just about any cheap NPN and PNP will work in the circuit > I gave. Junk box stuff. I'm not nearly as familiar with MOSFETs. Can > you point me to a good example of any one that would work great with > 3.5V on its gate? I'd like to learn more, myself.
Sure. Just plug logic level mosfet or digital mosfet into any mos manufacturer website and you'll find plenty of.
Even the good old 2N7002, if the logic is 5V CMOS may just do it. I didn't checked at 120mA, but given the price, just use 2, or run the leds lower than 20mA each.
>> On Sun, 22 Oct 2006 11:48:33 +0200, Fred Bartoli >> <fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote:
>>> Woaw... 2 transistors and 3 resistors.
>>> Just use the right transistor for this: logic level mosfet.
>> Makes sense. Particularly in the 120mA region.
>> I know that just about any cheap NPN and PNP will work in the circuit >> I gave. Junk box stuff. I'm not nearly as familiar with MOSFETs. Can >> you point me to a good example of any one that would work great with >> 3.5V on its gate? I'd like to learn more, myself.
> Sure. Just plug logic level mosfet or digital mosfet into any mos > manufacturer website and you'll find plenty of.
> Even the good old 2N7002, if the logic is 5V CMOS may just do it. I > didn't checked at 120mA, but given the price, just use 2, or run the > leds lower than 20mA each.
The PMV45EN would easily handle this; SOT23, 5.4A ID(max), Vgs(th) 2.2V (max across temperature) for ID = 1mA.
I use this as a power switch from logic levels (3.3V system) and I've never had an issue with it.
Interestingly, the datasheet also specs the sub-threshold leakage currents, which is unusual but useful.
Thanks for your help and for the diagram, one small problem though. My second power source is 9 volts not 4.5 volts. The chip is running of 5 volts and leds are running off 9 volts operating at 4.5 volts. So is it better to use individual resistors to each led instead of one higher 1 watt resistor?
Digitm...@gmail.com wrote: > Thanks for your help and for the diagram, one small problem though. My > second power source is 9 volts not 4.5 volts. The chip is running of 5 > volts and leds are running off 9 volts operating at 4.5 volts. So is it > better to use individual resistors to each led instead of one higher 1 > watt resistor?
> Thanks again you guys rock.
It's usually best to use a resistor for each LED to prevent current hogging from the lowest resistance LED (they will have varying characteristics).
On Sun, 22 Oct 2006 10:15:01 -0700, Digitmode wrote: > Thanks for your help and for the diagram, one small problem though. My > second power source is 9 volts not 4.5 volts. The chip is running of 5 > volts and leds are running off 9 volts operating at 4.5 volts. So is it > better to use individual resistors to each led instead of one higher 1 > watt resistor?
> Thanks again you guys rock.
It's better to put LEDs in series, unless there's a serious variation in brightness at the same current, in the same batch.
Just add up the VF's or, say, three LEDs and calculate your resistor as (9V - (Vftotal)) / Iled, unless I've gotten it upside down. ;-)
That way your switch transistor doesn't have to handle all of that current, and you waste much less power in dropping resistors.
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