How to drop a voltage using a series resistor?

Guest
Hello.

I'm looking at a schematic here that drops a 5V TTL output through a
series resistor into a 3.3V LVTTL input.

I cannot figure out how the value of the resistor is chosen, and would
like someone in the know to explain it (step by step, in small words,
like you would to a software engineer).

Thanks,
Paul.
 
In article <1123607455.005671.318380@o13g2000cwo.googlegroups.com>,
<pm940@yahoo.com> wrote:

Hello.
Click to expand...
Hi.


I'm looking at a schematic here that drops a 5V TTL output through a
series resistor into a 3.3V LVTTL input.
Click to expand...
A single series resistor isn't the way to go. Too many variables.
Also I don't think that a 5V TTL output can be guaranteed to be 5V.
In my trivial PIC programmer (http://www.finitesite.com/d3jsys) I
use HCT parts because the CMOS output is pretty much guaranteed to
be near the 5V rail.

I cannot figure out how the value of the resistor is chosen, and would
like someone in the know to explain it (step by step, in small words,
like you would to a software engineer).
Click to expand...
To be sure it would be better to use two resistors in a voltage divider
arrangement. Then you won't have to depend on the 3.3V LVTTL input
to present a set impedance to match up with the series resistor.

Check out the theory here: http://www.doctronics.co.uk/voltage.htm

Since 3.3 is 2/3 of 5V you want to generate a 2 to 3 ratio for your
resistors. Input current for LVTTL seems to be in the 10's of microamps
so using 10k and 20k resistors should provide more than enough current.
Create the voltage divider by tying one end of a 10k resistor to the
5V TTL output. Tie one end of the 20K resistor to ground. Finally tie
the two loose ends of the two resistors together. At that point the voltage
will be 3.3V when the TTL output is high. Tie the junction of the two
resistors to the LVTTL input.

When the TTL output is high, the resistors will consume 5V/30k = 160 uA of
current.

Hope this helps,

BAJ
 
Byron A Jeff wrote:
A single series resistor isn't the way to go. Too many variables.

To be sure it would be better to use two resistors in a voltage divider
arrangement. Then you won't have to depend on the 3.3V LVTTL input
to present a set impedance to match up with the series resistor.
Click to expand...
Thanks for the reply Byron. I understand how to do it with two
resistors, but that wasn't the question. I am not designing a circuit,
rather I am trying to understand how an existing circuit was derived.

Someone mentioned to me needing to look at the IBIS models of the part
being driven. This flew so far over my head it didn't move a hair.
They also said series isn't the way to go, but didn't have time to
elaborate on how R was determined.

So I'm still at a loss, and seeking knowledge.

Regards,
Paul.
 
A single series resistor isn't the way to go. Too many variables.
Byron A Jeff

I am trying to understand how an existing circuit was derived.
Paul (pm940 @ yahoo.com)
Click to expand...
They looked at the threshold spec and the typical input current spec
and hoped that their guess would hold for all iterations.
LOUSY ENGINEERING.

The repeatability of the 1-resistor technique
from one manufacturer to the next will not be predictable.
Even chips from different batches from the same manufacturer
may not work.

A series-resistor/shunt-resistor pair should be chosen
to swamp out the variables.
 
<pm940@yahoo.com> wrote in message
news:1123607455.005671.318380@o13g2000cwo.googlegroups.com...
Hello.

I'm looking at a schematic here that drops a 5V TTL output through a
series resistor into a 3.3V LVTTL input.

I cannot figure out how the value of the resistor is chosen, and would
like someone in the know to explain it (step by step, in small words,
like you would to a software engineer).

Thanks,
Paul.
Click to expand...
I'm not sure if I understand the question exactly but it seems to me if
your 5V source is being dropped to 3.3V then 1.7V is being dropped across
the resistor in question. The value of R would be determined via Ohms law.
R=E/I.



The device which needs the 3.3V source will draw a certain amount of
current and for this to work that current must be constant.



I am not familiar with LVTTL but just for the sake of discussion say it
draws 10mA at 3.3V. You can then think of it as a 330 Ohm resistor. 3.3V /
..010A = 330 Ohm.



To solve for your R, You know you need to drop 1.7V and you know the
current is .010A so, 1.7 / .010 = 170 Ohms





As I said, your LVTTL device must draw a constant current for this to
work. If it does not your 3.3V will fluctuate as the current draw
fluctuates.



Ohms law, E=I*R, I=E/R & R=E/I





To know what you device really draws for current 1.7V/R1 = I



I hope this answers your question
 
DBLEXPOSURE wrote:
I'm not sure if I understand the question exactly but it seems to me if
your 5V source is being dropped to 3.3V then 1.7V is being dropped across
the resistor in question. The value of R would be determined via Ohms law.
R=E/I.
Click to expand...
Sounds like you understood perfectly :)

I am not familiar with LVTTL but just for the sake of discussion say it
draws 10mA at 3.3V. You can then think of it as a 330 Ohm resistor. 3.3V /
.010A = 330 Ohm.
Click to expand...
It was the 10mA at 3.3V bit that confused me. JeffM suggests this is
unreliable, and my friend's comment about looking up IBIS models made
me think so too.


Any other notes, besides "lousy engineering"?

Regards,
Paul.
 
<pm940@yahoo.com> wrote in message
news:1123623538.197942.155080@g44g2000cwa.googlegroups.com...
DBLEXPOSURE wrote:
I'm not sure if I understand the question exactly but it seems to me
if
your 5V source is being dropped to 3.3V then 1.7V is being dropped across
the resistor in question. The value of R would be determined via Ohms
law.
R=E/I.

Sounds like you understood perfectly :)

I am not familiar with LVTTL but just for the sake of discussion say
it
draws 10mA at 3.3V. You can then think of it as a 330 Ohm resistor.
3.3V /
.010A = 330 Ohm.

It was the 10mA at 3.3V bit that confused me. JeffM suggests this is
unreliable, and my friend's comment about looking up IBIS models made
me think so too.


Any other notes, besides "lousy engineering"?

Regards,
Paul.
Click to expand...
So, now you understand how the size of the resistor was determind? I agree
about the lousy engineering and un relable part, Just want to make certain
your question has been answered.
 
On 9 Aug 2005 10:10:55 -0700, pm940@yahoo.com wrote:

Hello.

I'm looking at a schematic here that drops a 5V TTL output through a
series resistor into a 3.3V LVTTL input.

I cannot figure out how the value of the resistor is chosen, and would
like someone in the know to explain it (step by step, in small words,
like you would to a software engineer).
Click to expand...
---
There - is - no - resistor - needed.

http://focus.ti.com/lit/an/scba011/scba011.pdf

--
John Fields
Professional Circuit Designer
 
DBLEXPOSURE wrote:
So, now you understand how the size of the resistor was determind? I agree
about the lousy engineering and un relable part, Just want to make certain
your question has been answered.
Click to expand...
Yes, the only outstanding question is where the input LVTTL current is
documented.

Specifically for this question I'm looking at the Xilinx Spartan-3 FPGA
with LVTTL inputs. I don't see the input current documented in the
data sheet, but I may simply not be seeing it.

http://direct.xilinx.com/bvdocs/publications/ds099-3.pdf

The schematic I'm looking at is Digilent's Parallel I/O board for their
Spartan-3 starter kit. PC Parallel is TTL, and Digilent use 100 ohm
resistors to drop a TTL output down to something suitable for the 3.3V
Spartan-3. (http://www.digilentinc.com/Data/Products/PIO1/PIO1-sch.pdf)


Can you see the data that led them to 100 ohms?

Thanks,
Paul.
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:eek:abif19q5dc1qvfimb8p1qkfdtvose8a2r@4ax.com...
On 9 Aug 2005 10:10:55 -0700, pm940@yahoo.com wrote:


Hello.

I'm looking at a schematic here that drops a 5V TTL output through a
series resistor into a 3.3V LVTTL input.

I cannot figure out how the value of the resistor is chosen, and would
like someone in the know to explain it (step by step, in small words,
like you would to a software engineer).

---
There - is - no - resistor - needed.

http://focus.ti.com/lit/an/scba011/scba011.pdf

--
John Fields
Professional Circuit Designer
Click to expand...

I see what you are saying



----

Case 1 addresses a 5-V TTL device driving a 3.3-V TTL device. As shown in
Figure 1, the switching levels for 5-V TTL and

3.3-V LVC are the same. Since 5-V tolerant devices can withstand a dc input
of 6.5 V, interfacing these two devices does not

require additional components or further design efforts.

----

But earlier in the text,



----



The Case for Low Voltage

LVL, or low-voltage logic, in the context of this application report, refers
to devices designed specifically to operate from a

3.3-V power supply. Initially, an alternative method of achieving
low-voltage operation was to use a device designed for 5-V

operation, but power it with a 3.3-V supply. Although this resulted in 3.3-V
characteristics, this method resulted in significantly

slower propagation time. Subsequently, parts were designed to operate using
a 3.3-V power supply.

A primary benefit of using a 3.3-V power supply as opposed to the
traditional 5-V power supply is the reduced power

consumption. Because power consumption is a function of the load
capacitance, the frequency of operation, and the supply

voltage, a reduction in any one of these is beneficial. Supply voltage has a
square relationship in the reduction of power

consumed, whereas load capacitance and frequency of operation have a linear
effect. As a result, a small decrease in the supply

voltage yields significant reduction in the power consumption. Equation 1
provides the dynamic component of the power

calculation.

-----

Is this telling us, not need but preferred?
 
On Tue, 9 Aug 2005 18:43:07 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:


Is this telling us, not need but preferred?
Click to expand...
Apples and oranges.

One is saying that there is no problem with getting LVTTL inputs
powered from a 3.3V supply to switch when driven by TTL outputs
powered from a 5V supply, while the other is saying that that
combination will cause LVTTL to consume more power than if its
inputs were driven by LVTTL powered from a 3.3V supply.

--
John Fields
Professional Circuit Designer
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:gjiif1tj5ov96g3vl697a0tveiqv497pou@4ax.com...
On Tue, 9 Aug 2005 18:43:07 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


Is this telling us, not need but preferred?

Apples and oranges.

One is saying that there is no problem with getting LVTTL inputs
powered from a 3.3V supply to switch when driven by TTL outputs
powered from a 5V supply, while the other is saying that that
combination will cause LVTTL to consume more power than if its
inputs were driven by LVTTL powered from a 3.3V supply.

--
John Fields
Professional Circuit Designer
Click to expand...
So the OP's mystery resistor, (he was wondering how the designer chose it's
value) Serves what purpose? Current limiting? Seems to me there must be a
reason for it. Maybe not...
 
On Tue, 9 Aug 2005 20:11:15 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:


So the OP's mystery resistor, (he was wondering how the designer chose it's
value) Serves what purpose? Current limiting? Seems to me there must be a
reason for it. Maybe not...
Click to expand...
---
Here's what the OP's circuit looks like connected to a typical TTL
chip.


+5V---------------------+--->>---+
| |
[130R] |
| |
C |
B NPN |
E [1K2]R1
| |
[DIODE] |
| | R2
+--->>---+--[100R]--->LVTTL
|
C
B NPN
E
|
GND>--------------------+--->>--------------->GND


TTL, only being able to source 400ľA or so and guarantee 2.4V out,
(with a 5V supply) is pretty weak, and that's what the 1.2k resistor
in parallel with it is for, to provide extra current into the load.
How much? assume the load needs to see 2V and we have:

Vcc - Vil 5V - 2.4V
I = ------------ = -------------- = 0.002 ~ 2.0mA
R1 + R2 1200R + 100R

Also, those resistors will keep any downstream circuitry from
floating if there's nothing plugged into the connector and will
provide everything downstream with known logical staes if nothing is
plugged in

The resistance of the 100 ohm resistor is only [really] important
when the load is being pulled down, and since TTL can sink 1mA and
stay below 0.1V, that current through 100 ohms will cause an
additional 0.1V drop, resulting in a total drop of 200mV with 1mA
out of the load. With a maximum current of 75ľA out of the LVTTL
load and a 0.1V drop across the bottem totem-pole transistor of the
TTL sink that comes to a drop of:


Et = (Il Rs) + Vol = (7.5E-5A * 100R) + 0.1V = 0.1075V ~ 0.1V

Which is well below LVTTL Vil(min) of 0.8V.

Bottom line? the 100 ohm resistor isn't really needed but is in
there in case someone does something stupid like connect 5V with no
current limiting to the connector.

To find out, check the maximum input current allowed into LVTTL and
check if that's exceeded with raw 5VCD on its inputs when it's
running with a 3V supply.









--
John Fields
Professional Circuit Designer
 
On Tue, 09 Aug 2005 15:43:38 -0500, DBLEXPOSURE wrote:
pm940@yahoo.com> wrote in message
news:1123607455.005671.318380@o13g2000cwo.googlegroups.com...

I'm looking at a schematic here that drops a 5V TTL output through a
series resistor into a 3.3V LVTTL input.

I cannot figure out how the value of the resistor is chosen, and would
like someone in the know to explain it (step by step, in small words,
like you would to a software engineer).

I'm not sure if I understand the question exactly but it seems to me if
your 5V source is being dropped to 3.3V then 1.7V is being dropped across
the resistor in question. The value of R would be determined via Ohms law.
R=E/I.
The device which needs the 3.3V source will draw a certain amount of
current and for this to work that current must be constant.
I am not familiar with LVTTL but just for the sake of discussion say it
draws 10mA at 3.3V. You can then think of it as a 330 Ohm resistor. 3.3V /
.010A = 330 Ohm.
To solve for your R, You know you need to drop 1.7V and you know the
current is .010A so, 1.7 / .010 = 170 Ohms
As I said, your LVTTL device must draw a constant current for this to
work. If it does not your 3.3V will fluctuate as the current draw
fluctuates.
Ohms law, E=I*R, I=E/R & R=E/I
To know what you device really draws for current 1.7V/R1 = I

I hope this answers your question
Click to expand...
Probably not. According to the data sheet, the input leakage current of
an LV chip is 1.0 uA - that's microamps. So to drop 1.7V at 1 uA, you'd
need 1.7M. And depending on what kind of Voh your TTL is putting out,
(It's typically 2.7-3.3), the value was probably picked to give a worst-
case amount of current that can be shoved into the pin.

Yup. It's a bad design.

Hope This Helps!
Rich
 
On Tue, 09 Aug 2005 15:30:56 -0700, pm940 wrote:

DBLEXPOSURE wrote:
So, now you understand how the size of the resistor was determind? I agree
about the lousy engineering and un relable part, Just want to make certain
your question has been answered.

Yes, the only outstanding question is where the input LVTTL current is
documented.
Click to expand...
In the LVTTL data sheet. Ii, input leakage current. For the 74LV00, it's
1.0 uA (microamp).

Specifically for this question I'm looking at the Xilinx Spartan-3 FPGA
with LVTTL inputs. I don't see the input current documented in the
data sheet, but I may simply not be seeing it.

http://direct.xilinx.com/bvdocs/publications/ds099-3.pdf
Click to expand...
Page 4, table 6: Il, "Leakage current at user I/O, Dual-Purpose, and
Dedicated pins."

The schematic I'm looking at is Digilent's Parallel I/O board for their
Spartan-3 starter kit. PC Parallel is TTL, and Digilent use 100 ohm
resistors to drop a TTL output down to something suitable for the 3.3V
Spartan-3. (http://www.digilentinc.com/Data/Products/PIO1/PIO1-sch.pdf)

Can you see the data that led them to 100 ohms?
Click to expand...
For some reason, I'm not able to open PIO1-sch.pdf - but I can tell you
what led them to use 100 ohms - they pulled it out of the air. There
is no calculation involved at all - it's more of a rule-of-thumb, seat-
of-the-pants kind of thing. ("Oh, got a parallel port output? Stick a
100R resistor in series, 'cuz that's what everybody else does." :) )

Cheers!
Rich
 
On Wed, 10 Aug 2005 09:51:48 -0500, John Fields wrote:

To find out, check the maximum input current allowed into LVTTL and
check if that's exceeded with raw 5VCD on its inputs when it's
running with a 3V supply.
Click to expand...
Just a generic Q on this, since I'm too lazy to look it up - are there
LVTTL chips that _can't_ handle a 'regular' TTL level at its input?

Thanks,
Rich
 
On Fri, 12 Aug 2005 19:12:08 GMT, Rich Grise <richgrise@example.net>
wrote:

On Wed, 10 Aug 2005 09:51:48 -0500, John Fields wrote:

To find out, check the maximum input current allowed into LVTTL and
check if that's exceeded with raw 5VCD on its inputs when it's
running with a 3V supply.

Just a generic Q on this, since I'm too lazy to look it up - are there
LVTTL chips that _can't_ handle a 'regular' TTL level at its input?
Click to expand...
Dunno. I only checked into it for the OP's sake and got my info
from TI's 1994 Low-Voltage Logic databook, so since it's not
something I need to pursue for my own ends, I'm too lazy to look
into it any deeper. :)

--
John Fields
Professional Circuit Designer
 
Thanks for all the responses - more help than I expected.

Regards,
Paul.
 
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