Re: Explanation of the whole 300 mA thing

On Dec 3, 2:09 am, Jim Granville <no.s...@xxxxxxxxxxxxxxxxxxxxxx>
wrote:
rickman wrote:

The feedback is due to the fact that the load is in the input voltage
loop as well as the output. If the load voltage increases, the
voltage across the B-E decreases lowering the current to the load. A
lower voltage on the load likewise increases the B-E current and
increases the load current. A common emitter arrangement has no such
feedback and changes in the load will affect the output voltage much
more significantly.

You would need to draw this, as I cannot picture a circuit from that
description.

Look up any emitter follower diagram. I shouldn't need to draw this.


The Vbe slope, gives 600-700mV @ 300mA and 1.0-1.2V at 2.1A, so that's
a change in drive-drop of ~400mV from that alone.

I don't see this data in the *** for the ZDT6753T. Regardless, the
current variation from the changes you describe are not large. There
is significant variation between LEDs and this should be within that
typical variation in brightness.

LEDs are quite good within a batch. I've measured under 10mV of
variation, on a sample of 100. There IS, of course, variation between
Red / Green, but the issue is trying to minimise crosstalk or brightness
modulation, caused by common mode resistance.

You are talking about the I-V curve being similar. But you are trying
to balance brightness. My understanding is that there can be
significant variation in brightness between units at a given current.
I have never tried to measure this, but this is what I have read.
Even so, doesn't the I-V curve vary between lots?


Then the base current also matters - 2.1A at Hfe of 100 is 21mA, which
is quite high for HCMOS, Even at 210Hfe, (rarer, but doable) it is 10mA

Yes, you are right, the ZDT6753T may not be the best choice for this
circuit dut to its low hfe. The ZDT6790T or a similar part should
work much better with an hfe of 150 min at 2 amps requiring only 13.3
mA base current.

Even a high drive SN74LV595 specs 16mA @ .55V max Nch, and 0.8V Pch
- that drop is current linear, so adds to the modulation effects.

I'm not sure why you are discussing this part since it is not in the
circuit.

See my other notes, the HC165 is a poor choice, as it lacks a Latch, and
OE control. The LV595 is the highest drive, 5V latched shift register
I could find quickly.

I still don't know why are you talking about output drivers on MSI
chips. None of the LEDs are being driven by MSI chips! The MCU
outputs drive the high current, tristate row drivers. The low current
drivers are common emitter, so their output voltage do not depend on
what is driving them.


The high current drivers are controlled directly from the
MCU outputs. Many MCUs can directly drive 20 mA LEDs. I don't know
the specs on the OP's MCU, but 13 mA should be doable without
excessive voltage drop.

with a high drive LV595, and your numbers, that's 675mV, which is
significant in the voltage budget. It also varies from 96mV to 675mV
as the load current (number of LEDs ON) changes.
To me, that is intolerable to have in the voltage budget - so
a circuit change is needed.

I think we are losing traction here. I don't know which numbers you
are using or how you are using them. On top of that I have no idea
why you are using the LV595 instead of the parts I have described.


Then the LED drop at 300mA needs to be included (esp non-red), so the
whole thing looks tight at 5V, if using ths simple emitter-followers in
both paths.

I don't think you understand the circuit I am describing (I may have
mistyped getting common collector/emitter follower confused with
common emitter). The emitter follower push-pull circuit is only used
in the row drivers. The column drivers are standard common emitter
push-pull drivers. The load resistor should be duplicated from each
common emitter collector output to the load to help minimize shoot
through, rather than tying the collectors together with a single
resistor to the load.

A drawing would help - a picture is worth 100 words :)

The voltage budget is getting tight.
Few Display LEDs specs to 300mA, but slope of 500mv/50mA is not
uncommon, for Green.
Most matrix displays expect to be driven at lower than 16:1 duty cycle.

The dominant resistive voltage drop, should be the current limiting
resistor, not any drivers, and especially not any column drivers.

I have never seen a diode that was resistive in the forward
direction. I think you will find the forward voltage drop to be
logarithmic. Yes, the dominant resistive element is the current
limiting resistor, that is why it controls the current.


That starts to dictate N fet and Pfet column drivers.
vanilla 100mOhm fets are 210mV at 2.1A, with no base-drive factors.
50mOhm fets are ~100mV

This is not so much an issue of FET vs bipolar as common collector vs.
common emitter. Common emitter allows the device to be in saturation
with a lower voltage drop across the transistor output. But that
circuit requires more control circuitry. The point of the emitter
follower stage is that it duplicates a high impedance on the MCU
control pin to the driver output. This is not doable with a common
emitter stage without two control lines per driver. This is also not
doable with FETs in a source follower configuration since the
threshold voltage is so much higher for a FET than for a bipolar
transistor.

If you use the correct voltage levels for the drivers, you will see
that there is adequate drive from a 5 volt power rail.

Depends on the LED. The specs I have for 5x7 matrix displays show none
rated to 300mA, and suggest ~3V drop at 100mA drive (green).
That can barely tolerate one follower, so it does need a low saturation
column driver. If we lower to 100mA, for the *** I have, and allow
0.8V for emitter follower + base effects, that leaves 1.2V for column
driver, and limiting resistor. Not enough room for a second follower,
so it needs a open-collector saturating driver. At 5V targets, FETs
are better than bipolars.
[also not nearly enough voltage budget for the OPs original Emitter
follower/darlington follower/LED/resistor chain ]

I am confused. Are you saying that using an row emitter follower with
an open collector column driver will work? That is what I am
describing!

The problem with the FET is that you need two control lines per driver
in order to get a tristate driver output. How do you connect this to
the MCU?


Std 5mm leds got nowhere near 300mA ratings (even peak) but I found
a 'traffic light' green led, specs 3.8V max Vf, at 350mA. (typ 3.3V)

You need to ask the OP what is being used in this circuit.
.

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