Vcc & Vdd on the same chip (LB1205)

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laeeq
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Vcc & Vdd on the same chip (LB1205)

Post by laeeq » Thu Feb 21, 2008 11:50 am

Hi,

I want to use LB1205 (high current/voltage darlington drivers), in one of my projects.

Normally, I am familiar with seeing only one supply voltage in a chip (Vcc or Vdd). But on LB1205, I see both.

Vcc on pin 1, 7
Vdd on pin 10.

According to data sheet Vcc max is 62 Vots & Vdd max is 7 volts.

I want to use LB1205 to drive a stepper motor which requires 5 volts per coil at 1 amp.
My question is, what are right values for Vcc & Vdd in this case.

Thanks in advance.

Regards
Laeeq Khan

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haklesup
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Post by haklesup » Thu Feb 21, 2008 12:07 pm

7.0V and 62V are absolute max values. Those which if you exceed you will blow up the part. The recommended VDD is between 3V and 7V meaning 5V is just fine.

Looking at the schematic most of the power comes from VDD. VCC looks like it is just a reference voltage for the spark killer diodes and any value = or > than VDD should work. Not much is said about this pin in the datasheet though. No minimum or typical value is given for VCC

If the outputs are driving an inductive load, you could have a voltage spike when the output switches off. This spike will be clamped by the internal spark killer diodes (up to 1.5A) to the max level of VCC+Vf of the diode.

http://pdf1.alldatasheet.com/datasheet- ... B1205.html

Bigglez
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Re: Vcc & Vdd on the same chip (LB1205)

Post by Bigglez » Thu Feb 21, 2008 12:09 pm

Greetings Laeeq,
laeeq wrote:I want to use LB1205 (high current/voltage darlington drivers), in one of my projects.

Normally, I am familiar with seeing only one supply voltage in a chip (Vcc or Vdd). But on LB1205, I see both.

Vcc on pin 1, 7
Vdd on pin 10.

According to data sheet Vcc max is 62 Vots & Vdd max is 7 volts.

I want to use LB1205 to drive a stepper motor which requires 5 volts per coil at 1 amp.
My question is, what are right values for Vcc & Vdd in this case.
This device is intended to interface from 5V logic to
high power loads (such as your stepper motor) which
may have higher voltage ratings. The power supplies are
separated to allow different values, and reduce feedback
from the heavy load to the logic section.

Connect VDD to the logic 5V supply. Connect VCC to
the stepper motor supply. Or connect both together
if they are the same voltage.

It would be wise to add separate decoupling capacitors
between each supply and ground near the IC.

Read the datasheet for advice on adding a copper foil
heatsink to your PCB - the device will get hot under
heavy loads.

Comments Welcome!

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haklesup
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Post by haklesup » Thu Feb 21, 2008 12:14 pm

The power supplies are
separated to allow different values, and reduce feedback
from the heavy load to the logic section.
The diodes point in the wrong direction for VCC to be able to supply any current to the output. See the equivelent circuit.

Bigglez
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Post by Bigglez » Thu Feb 21, 2008 12:51 pm

Greetings hakeup,
haklesup wrote:
Bigglez wrote:The power supplies are
separated to allow different values, and reduce feedback
from the heavy load to the logic section.
The diodes point in the wrong direction for VCC to be able to supply any current to the output. See the equivelent circuit.
The internal diodes are to conduct current from inductive
loads back to the supply (aka "back-emf" or "free-wheeling
diodes").

As stated earlier, the VDD logic supply is separated
from the VCC load supply to allow either different
voltages and/or reduce cross-talk from the high power
load to the controlling logic.

VCC and VDD can be joined together if the load is also
5V, and load current is routed away from the logic.

Comments Welcome!

laeeq
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Post by laeeq » Thu Feb 21, 2008 1:22 pm

I agree with bigglez's view on function of internal diodes. I have seen similar diodes in ULN2003 chips and they work fine.

I am trying to understand this back emf, voltage spike, thing. I dont think I fully understand it yet. I am trying to do the transient analysis of a simple LR circuit and see the origin of this spike.

Thanks everybody for help.
Laeeq Khan

Bigglez
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Post by Bigglez » Thu Feb 21, 2008 2:23 pm

Greetings Laeeq,
laeeq wrote:I am trying to understand this back emf, voltage spike, thing. I dont think I fully understand it yet. I am trying to do the transient analysis of a simple LR circuit and see the origin of this spike.
The spike is the release of energy stored in the
magnetic field of the solenoid (inductor, coil,
motor winding, relay armature). The collapsing
field causes a voltage spike of opposite polarity
and "infinite speed and unlimited voltage" until
the energy is dissipated. In practice the voltage
is limited by stray capacitance but could damage
voltage sensitive devices in the circuit.

Adding the diode directs the spike to dump energy
back to the power supply, protecting the other
parts of the circuit.

Dumping large amounts of energy and depending
upon layout can cause feedback to the logic section.
Of less concern but still an issue is that when the
catch diode conducts the current flows through the
power supply and can cause "ground bounce" which
is a problem for some bipolar logic and may also
cause latch-up in CMOS logic (due to internal ESD
diodes on inputs).

If you model a circuit in SPICE or similar to view
the spike please publish your simulation results
here, it would be helpful to the next person asking
about back-emf protection.

Comments Welcome!

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haklesup
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Post by haklesup » Thu Feb 21, 2008 4:45 pm

Ok, now I see the outputs as an "Open Collector" type. As such the load would be biased directly by its source VCC. Likewise, the device should be connected to the same VCC source to clamp any voltage spikes due to back EMF.

The output pin in series with VCC and the motor provide a path to ground for the current to flow through the windings. When not conducting (turned off) the output pin will see the voltage of VCC and therefore has protection to that same voltage. The VCC pin of the device still does not provide power to the motor, that VCC node is connected directly to the motor.

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