1 MCU open-drain pin to drive 4 NPN

[banh]

hi,

i intend to use 1 MCU open-drain pin to drive 4 npn's as in the figure.

IoL of the pin is 10mA.

each npn needs to be injected 1mA into the base for saturation.

So, this is what i expect:
- when the NMOS of the port is off -> all the NPN are turned on and the leds will be lighted up.
- when the NMOS of the port is on -> all the NPN are off, and the leds are off.

i need your comments on this circuit.

thanks.

 

[artem]

instead of 560 Ohm resistor put pnp transistor with collector connected to npn's base resistors, base connected through resistor to open-drain mcu output
and emitter to Vcc

 

[IanP]

Should work well ..
You can easily increase the value of the pullup from 560Ω to, say, 1kΩ ..

Regards,
IanP

 

[Sceadwian]

The 560ohm resistor will give a higher gate bias value. At 560ohms the gate bias is about 2.5 volts (asumeing 5volt VCC) at 1k ohms the gate bias is about 1.8 volts. I'm not sure how this effects the responce of the mosfet, but the closer to 'on' you can bias a mosfet without it actually turning on the better. Considering that lower resistance draws 8ma's which is pretty close to the 10ma limit of the pin it's probably the best value to use, unless the rise/fall time of the I/O pin is too high with the large current draw.

 

[banh]

thanks all,

another concern of mine is that instead of the pin being configured as open-drain, it is accidentally configured as normal push-pull output.

in this case Ioh of the pin is only 2mA.

is this a problem? as the pin may source currents to the NPN's, and may exceed 2mA..

 

[IanP]

The "push-pull" output still will be comfortably drive these 4 transistors ..
Keep in mind, that common NPN transistors have β>>100, and that to drive, say, 10mA collector current of NPN transistor you need I(base) =Ic(ollector)/β = 10mA/100 = 100µA (or less, for β>100) ..
For 4 transistors, therefore, the driving current will be <400µA, so with 2mA Ioh of the micro's pin the driving capability is ≈20 of NPN transistors ..

Regards,
IanP

 

[banh]

the transistor i use has hFE (min) of 20.

hence i need to pump into the base around 1mA.

 

[IanP]

In this case, if you don't want to invert the phase of the signal, add another NPN transistor, and connect its base directly to the microcontroller pin, collector to +5 and emitter (through resistos) to bases of these 4 transistors ..
Obviously, you can do it in one-hundred-and-one other ways ..

Regards,
IanP

 

[banh]

why can you connect directly the base of the transistor to the MCU pin? shouldn't you have a resistor to limit current?

 

[IanP]

You can connect additional resistor between the base and the MCU's pin, but it is not needed there as you have resistors wired at the emitter side ..

Rgards,
IanP

 

[banh]

thanks IanP,

attached is the modified circuit according to your suggestion.

i'm not so clear about calculating:

- how the current flows in the npn connected to the MCU pin? how much current is pumped out from the mcu pin?
- the current flowing into the base of the other 4 NPN's? are they still (5-0.75)/3K9 for each NPN?

 

[IanP]

Base current of each of the 4 NPNs:
(5V-0.75-0.75)/3k9 ≈ 0.9mA (second 0.75 stands for Ube of the added NPN)

MCU pin current (assuming H=5V):
≈ (4*0.9mA)/β (of the added NPN) ≈ 3.6mA/20 ≈ 0.18mA

Regards,
IanP