An led at 11.4V at 232mA = 2.65 Amps
a high power led driver, the narrow pulse, stored energy and all that ...
Since the IRFZ44 is popular, a search for high power led driver finds those mosfet circuits
https://www.edn.com/power-an-led-driver-using-off-the-shelf-components/

3W led strobe triggered by a 4017

 

Amps?

 

Thank you for that correction, not amps.
Watts
by searchng for driving 2 or 3 Watt led using mosfet found the IRFZ44 circuit.

 

Ok, can anybody advise whether there would be any advatage/disadvantage (other than cost) in ordering lower voltage mosfets online as opposed to using one of three I can source locally (post #55)?

AnalogKid, is it then just a case of substituting the selected mosfets and CD4049 into the circuit you posted at post #35? You also mentioned 2 decoupling capacitors. My original circuit had a 47uF in front of the CD4017, would I keep this and add another one in front of the inverter or something else?

 

In round numbers, for the same drain current rating, devices with lower voltage ratings usually have lower Rdson values, lower cost, or both. For the same drain current in the circuit, a lower Rdson in one part versus another means lower power dissipation, which goes directly to reliability.

General rule of thumb for long-term reliability in power switching circuits: to reduce electrical stress, double everything. If the peak voltage across a transistor is 12 V, use a 25 to 30 V transistor minimum. If the peak current is 2 A, use a part with a minimum 4 A current rating. Etc. Same for voltage rating of electrolytic capacitors, power rating of resistors. Etc. Next, heat. For a set of circuit conditions, the lower the Rdson rating, the lower the heat dissipation. Extra-low Rdson probably will increase the cost, but could eliminate a heatsink.

Rule of thumb for decoupling signal-level ICs (TTL and CMOS logic chips, "normal" opamps, etc.): one 0.1 uF (min) ceramic capacitor per power pin, as close as possible to the power pin with the shortest possible leads. If the device has a GND pin like most logic chips, or multiple ground pins like CPLD's and FPGA's, each cap goes to the closest GND pin. The cap value is flexible. For example, if the only other capacitors in the circuit are 1.5 uF, it's fine to use one of them as a decoupling cap rather than add another component value to the design. BOM (bill of materials) optimization is not a big deal for a one-off home project, but is serious money when you're building 1,000,000 cell phones every day. No harm in thinking like the pros.

For opamps and other bipolar devices, decouple separately the positive and negative power input pins to ground. If the device is moving extra current, like an LED driver or an opamp driving a 600 ohm load, add a 10-100 uF electrolytic cap in parallel with the ceramic. Always look at the sample circuits on the datasheet. If they show a decoupling cap bridging the + and - power inputs, do that.

ak

 

Thanks AK, I have had a look at various through hole p-channel mosfet datasheets and it seems that mosfets in the 30-50V range allow for 10's of amps drain current which is way more than required. I have no idea how to choose an appropriate mosfet, can you possibly recommend one that will work in the circuit you posted?

 

Given I don't know which mosfet is appropriate to use in the circuit at post#35 or if using mosfets would require gate resistors or voltage dividers to be included in the circuit, how about trying PNP transistors.

AK mentioned a TIP115, could somebody please explain how to calculate the base resistor value if I use CD4017, CD4049 and TIP115 with collector current of 231mA?

 

if using mosfets would require gate resistors or voltage dividers to be included in the circuit,

No gate resistors or voltage dividers. The gate inputs of most MOSFETs are rated for +/-20 V, so your 12 V circuit is plenty safe. That's the big advantage of MOSFETs for something like this - less clutter.

The TIP115 is a power darlington bipolar transistor, so it requires approx 100x *less* base current than a regular bipolar transistor. 1 mA is a safe value. The datasheet charts show this to be well into the saturation region at 500 mA, more than twice your collector current.

The tradeoff is that the "on" voltage for a darlington is greater than with a MOSFET. A darlington will dissipate around 200 mW per transistor - enough to feel warm. Any MOSFET with an Rdson rating of 3.3 ohms or less produce less heat.

TIP115 datasheet saturation voltage curves, Vcesat @ Ic = 250 mA = ~0.8 V. 0.8 x 0.24 = 0.192 W

For a MOSFET to produce the same heat, its Rdson can be calculated with a permutation of Ohm's Law:

R = E / I

Rdson = 0.8 / 0.24 = 3.33 ohms >> This is the Rdson needed to produce the same amount of heat as the TIP115 example above. A MOSFET with a lower Rdson rating will produce less heat. It also will have a lower voltage drop across it when on, which will increase the LED current and brightness (for the same LED resistors).

ak

 

Thanks AK I have learnt alot from your posts and appreciate you putting up with my basic questions. Mosfets sound like the way to go but as I mentioned before I am having difficulty selecting one which will be appropriate. It seems that a mosfet in the right voltage range allows way more current than I require, there must be a happy medium but where? If you were making this circuit what mosfet would you use?

Thanks again,
John.

 

Thanks AK I have learnt alot from your posts and appreciate you putting up with my basic questions. Mosfets sound like the way to go but as I mentioned before I am having difficulty selecting one which will be appropriate. It seems that a mosfet in the right voltage range allows way more current than I require, there must be a happy medium but where? If you were making this circuit what mosfet would you use?

Just in case there’s a tiny bit of confusion, there are RATINGS and SPECIFICATIONS.

A specification is a value that specified what a component will run with. A rating is a maximum operating value. Thus, a resistor is rated in ohms. A 100 ohm resistor with 5V across it will use 50mA. A power supply might have a rating of 5A. If it produced 5V into a 100 ohm resistor, the current will NOT be 5A; it will still be 50mA. If you attached the power supply to a device that drew 10A, either the voltage would be reduced or the power supply would fail. It’s only rated for 5A.

When picking a MOSFET, the current rating is a maximum. A 10A MOSFET can switch up to 10A (maybe 8A practically). So as long as the MOSFET is rated for more current than your LEDs, it will work.

Given that you’ve chosen the correct type (N or P channel), the choice comes down to price and availability.

 

Thanks djsfantasi, I am aware that it is the device or load which determines the current draw in the circuit. My concern was whether having a mosfet rated at say 10A was overkill for a circuit drawing only 231 mA. If this isn't considered outside norms then it makes my mosfet selection alot easier.

 

AK mentions at post#35 that the circuit needs two power supply decoupling capacitors. Wouldn't it need three power supply decoupling capacitors when I include the 555 timer?

 

AK mentions at post#35 that the circuit needs two power supply decoupling capacitors. Wouldn't it need three power supply decoupling capacitors when I include the 555 timer?

Yes, but the circuit I posted is independent of the clock source, so I didn't address that.

ak

 

My concern was whether having a mosfet rated at say 10A was overkill for a circuit drawing only 231 mA. If this isn't considered outside norms then it makes my mosfet selection alot easier.

It is overkill. But in today's design environment, after decades of switching power supply development, low voltage FETs with unusually high currents (compared to bipolars) are everywhere and cheap.

Also, that 10 A rating gets you a very low Rdson, which is directly proportional to the heat dissipated in the device. Lower is good.

ak

 

Great thank you AK.

Regarding decoupling capacitors. On the 555 TI datasheet that they say that the minimum recommended capacitor value is 0.1 μF in parallel with a 1-μF electrolytic capacitor. Presumably the 0.1μf is a ceramic?

The datasheet for the CD4049 says for devices with a single supply, TI recommends a 0.1-μF capacitor but it is acceptable to parallel multiple bypass capacitors to reject different frequencies of noise. 0.1-μF and 1-μF capacitors are commonly used in parallel.

The datasheet for the CD4017 doesn't appear to mention decoupling or bypass capacitors at all.

Do you think it best to use the parallel capacitor setup on the 555 and separate 0.1-μF on both the CD4017 & CD4049?

 

Many datasheets assume that you know that a decoupling capacitor close to an IC is very important.
One reason it is important on a 555 is that it was designed using an old TTL type of circuit and when its output switches it draws a supply current spike of 400mA!
A modern Cmos 555 (LMC555, TLC555 and ICM7555) does not do that.

 

I just wanted to say a big thank you to everyone who offered advice and made suggestions, especially AnalogKid and Audioguru again. I really appreciate your expertize and patience.

I got all my parts yesterday and built the circuit on a breadboard and it worked first attempt. The IC's and Mosfets don't even get warm.

Thanks again,
John.