How to drop a voltage using a series resistor?

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.

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.

Hi.

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

I use HCT parts because the CMOS output is pretty much guaranteed to be near the 5V rail.

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:

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

Sounds like you understood perfectly :-)

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.

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.

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.

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

Thanks, Paul.

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

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.

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

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

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.

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

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...

--- 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 400µA 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 75µA 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

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

In the LVTTL data sheet. Ii, input leakage current. For the 74LV00, it's

1.0 uA (microamp).

Page 4, table 6: Il, "Leakage current at user I/O, Dual-Purpose, and Dedicated pins."

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

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

Thanks for all the responses - more help than I expected.

Regards, Paul.

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