Darlington Line driver

[lilzz]

ULN2803.PNG (34.27 KiB) Viewed 989 times

I have some question about their device.
1)The Common signal which connected to 12V and with R4 and D5 zener diode and C5 cap. why there's zero ohm R4 running parallel to the zener diode?

2)on the output, why there are 2 diodes versus one diode on the input? Doesn't the diode clamp it at 0.7V so regardless of how many diodes they are? Those diodes are preventing if input and output drop below 0V right?

3)Why there's a reverse diode connecting to the common pin? preventing the Output goes above 12.75V right?

[Burngate]

Where did you get that diagram from?

From the data sheet (I'm not sure where I got it from, but it's originally TI)

ULN2803A.png (16.5 KiB) Viewed 961 times

there's no sign of R4 or D5, so they must have been added by whoever created your diagram, for no reason that I can see
Also, there's only one diode on the TI data sheet, and only one should be necessary, as you say

The diode behind the COMMON pin is, as you say, to clamp the outputs to the supply - when driving inductive loads, quickly switching the current off can give massive voltage spikes, from V=L.di/dt

[lilzz]

Where did you get that diagram from?

From the data sheet (I'm not sure where I got it from, but it's originally TI)

there's no sign of R4 or D5, so they must have been added by whoever created your diagram, for no reason that I can see
Also, there's only one diode on the TI data sheet, and only one should be necessary, as you say

The diode behind the COMMON pin is, as you say, to clamp the outputs to the supply - when driving inductive loads, quickly switching the current off can give massive voltage spikes, from V=L.di/dt

What would you connect the COMMON pin with?

My diagram is from a design.

[FLYFISH TECHNOLOGIES]

Hi,

What would you connect the COMMON pin with?

Connect it to 12V.


Best wishes, Ivan Zilic.

[lilzz]

ULN2803A.png

The diode behind the COMMON pin is, as you say, to clamp the outputs to the supply - when driving inductive loads, quickly switching the current off can give massive voltage spikes, from V=L.di/dt

Is the diode behind the COMMON pin, if not conducting, considered to be open circuit?

And if it's open circuit, how would the 12V going to the circuit?

[Burngate]

That diode is only going to conduct if the voltage on the output rises above the 12v supply - and it's only going to do that if the transistor is off, and what it's feeding is trying to push current into it

When power is first applied to an inductor, current slowly builds up; energy is being stored in the magnetic field
di/dt = v/L so the greater the inductance the slower it'll grow. But energy E = ½Li²
Switch off the source of power, by turning off the transistor, and suddenly there's nowhere for the current to go, other than into stray capacitances.
Since those will be small, their voltage will rise quite high, with the risk of breaking something
But put a diode across, and as soon as the voltage rises above 0.7v the diode will carry the current and dissipate the energy

You are always advised to put a diode across motors or relays, and for your convenience Texas Instruments builds it into their device