1N4148 datasheet question

i use the 1N4148... n this is the datasheet....


and i wanna know the maximum allowable forward current,, but i cannot find it...

bellow is the absolute maximum rating...

the "average rectified forward current " is 200mA.. can it go any futher?...

please explain... thanks...

regards,
sp

Here is the value from an ancient paper databook.

1 second surge 1 A

1 us surge 4 A

DC steady state 300 mA

recurrent steady state AC half cycles 400 mA peak

These values are for the leaded part in free air. If you use surface mount versions the heating will be greater and you will have to operate at lower values.

In general, for reliability you should always be at 80% or below maximum ratings.

if i use the 1N4148 in this cicuits for the motor controller as the pic show,,, is it ok if my current to the motor coil is about 1A?...

thank you...

1N4148 is a small signal diode and is to small for that job.
Go for something like UF5402, BYV27-200 or similar .. which have I≥2A and tr<50ns ..
Regards,
IanP

like flatulent mentioned... the diode is not used for continuous...


so the diode can cope wth it?....1N4148 in motor controller bridge...

regards,
sp

sp said:

so the diode can cope wth it?....1N4148 in motor controller bridge...

Click to expand...

???
Please rephrase your question.

You don't want tiny 1N4148 diodes in a motor controller.

could i ask what will happen if i dont use the diode n connect to the motor?...

it spoil the supply or the driver or the controller or everything?..


thank you...

regards,
sp

Maybe this schematic helps you. It is worth noting that the peak current is very high (as high as 1A), but only a peak. When a swith of the full bridge closes the inductance of the motor coil try to maintain the current. So if there is no diode the voltage can be very high and maybe destroy other chips.

actually the circuits u show is using the 1N5405

for 1N5xxx they nvr mention the trr(reverse recovery time)... why????

i need trr <= 200ns.... 1N5xxx fullfill this condition??...

the 1N5xxx mention is standard recovery rectifier.... what does tht mean???... what is the trr??...

can anyone clear my doubt?... ???

thank you....

regards,
sp

The schematic says "VF <= 1.2V at 2A". That immediately rules out small diodes such as 1N4148.

If a rectifier's data sheet doesn't mention trr, then you should assume it is horribly slow. That is common for a standard recovery rectifier.

Choose a fast recovery rectifier that meets the requirements stated on the schematic. Be conservative. Assume the VF requirement is continuous, not pulsed.

I assume you are using the same power supply voltage and stepper motor as described in the application note. Otherwise, all bets are off.

I'm puzzled why the schematic says "2 Fast Diodes". I wonder if each diode symbol is actually two diodes in parallel. It may be wise to use 4A rectifiers instead of 2A parts.

The 1N4148 is definitely NOT the diode for that job.
The peak current will be the motor peak current, so if the motor draws 1A, the peak current through the diode will be 1A. You need a diode that can handle peak currents well in excess of 1A.

Don't rely on those surge parameters. They generally assume that the initial junction temperature is 25°C and at the end of the surge it reaches the maximum specified junction temperature (150-175°C). So those ratings are for one-time occurrences, with enough time for the diode to cool down. Another surge like that with the junction at say 50°C and the diode may not survive.

Then there is the power dissipation issue. Assume the current through the diode is 1A 50% of the time, so the average is 0.5A (this is like your average rectified current; you should calculate it for worst-case).
Multiply that by the voltage drop at 1A and you get the diode conduction losses. To that you should add switching losses. If the diodes are ultrafast, you can neglect these losses in a first approximation.

Check the maximum junction temperature of the diode at the maximum intended ambient and see if it's reasonable. By reasonable I mean 100~110°C. Remember, this has to be at the maximum intended ambient.
Since you neglected the switching losses, make sure the junction temperature is actually less than 100°C, so you can accommodate the extra losses.

thanks for everyone help.... i pressed helped button

emmm... actually i use 1N4148 for the diode n the motor function well.... but i just use 10V for the motor.... and no prob so far....

i gonna buy the byt03 n L6210(ic)... but all is quite expensive....

anyway.. can anyone tell me y the 1N4148 working?... bcos i am not using large voltage?...

thanks everyone for help... grateful..

regards,
sp

Heating is the main problem that limits the current capabilities of a diode. If the duty cycle on your diode is low, you can put a little more current through it (you just can't all the time). If you put a heat sink on your diode, even better (no one ever does this for a 1N4148). Anyway, I've used a 1N4148 in this application before, but my motor currents were small (like a 5 ohm winding at 5 Volts or 1 A through the motor). The impulse currents through the diodes will be much higher, BUT of very short duration. The question is really how much power a diode can dissipate without frying. Keep in mind that the diode must dissipate the energy in the inductor everytime you switch (1/2*I^2*L).