For the following schematic I get the following currents - they are close enough to 20ma to be satisfied theoretically.
Practically should I reduce the forward amperage (in this case using 20ma) to something smaller say 15, 16 mas?

 

What would be the point of doing that?
ETA: If you could choose resistor values from a continuous spectrum of 0.0001% values, then you could make both currents closer to 20 mA if that was your goal.

 

Practically should I reduce the forward amperage

It depends. Will the LEDs be sufficiently bright at a lower current? If so, there's no reason to operate them at the maximum continuous current rating.

I often operate standard brightness LEDs used as indicators at a few mA. There's no benefit in dissipating power needlessly.

The flow in your schematic is mostly conventional, but why did you choose to draw it "sideways" from how it would typically be drawn?

 

For the following schematic I get the following currents - they are close enough to 20ma to be satisfied theoretically. Practically should I reduce the forward amperage (in this case using 20ma) to something smaller say 15, 16 mas?

If the max current of your LEDs is 20mA, then yes, you don't want to run at that current and 10-15mA is fine. Some highBrightness LEDs give off a lot of light at just 1mA.

 

The flow in your schematic is mostly conventional, but why did you choose to draw it "sideways" from how it would typically be drawn?

Sorry I don't know how to answer this question, must mean I have done something wrong : )

 

Sorry I don't know how to answer this question, must mean I have done something wrong : )

Note the use component designators and the lack of colors.

 

View attachment 278634
Note the use component designators and the lack of colors.

I'm sure he'll understand your version as well. Black and white is boring but still understandable.

 

Lots of people have silly rules for schematics. I don't believe "sideways" is wrong.

Generally, schematics flow from upper left to lower right. One rule to understandability is the grounds point down, positive Voltage connections up, negative voltage connections down.

 

Here is what I get for 20 ma and a forward voltage drop of 1.16 Volts. There still might be a smaller error in your calculations.
ETA: I fixed the forward voltage of the presumed LED and the right resistors to get exactly 20 mA. still don't exist.

 

Here is what I get for 20 ma and a forward voltage drop of 1.6 Volts. There might be an error in your calculations.

How does running a simulation with the LED Vf set to 1.6 V indicate that the TS might have made an error in their calculations for LEDs with a Vf of 1.16 V?

@clangray
We don't really need to throw problems like this at a simulator. A moment's worth of analysis will yield equations that we can use and reuse to our heart's content.

For a supply voltage of Vcc driving n LEDs each having a forward voltage drop of Vf at a current of If, we want a limiting resistance, R.

The voltage across R is just the supply voltage minus one LED Vf drop for each LED, so (Vcc - n*Vf). By Ohm's Law, then the resistance we need is given by:

R = (Vcc - n*Vf)/If

If possible, look up the device data sheet your are using. As an example, here is a figure from the data sheet from the HAN1002-1-TR infrared LED that seems like it might be close to what you are working with.


As you can see, at 20 mA it has a (typical) Vf of ~1.25 V, while at 10 mA that is reduced to 1.17 V.

Just decide what current you want, estimate the voltage drop at that current, as well as how many diodes in series, and then calculate the resistance.

You won't be able to find that resistor, so find the nearest one in a table of standard values, and then invert the formula to find the expected current.

If = (Vcc - n*Vf)/R

You will be using a voltage that is slightly wrong, but that is almost certainly washed out by the uncertainty in your particular LED's actual parameters compared to the manufacture's typical curves.

In fact, if you consider that the LED in my example has a typical forward voltage of 1.2 V at 20 mA, but a max voltage of 1.4 V (no min is given), that 200 mV (on the top side only, mind you) covers the change in typical forward voltage from 2 mA to 50 mA, so while looking at "better" values from the charts may give us a warm fuzzy, it is arguably not worth the effort. Just use the typical 1.2 V figure and use that to size your resistor regardless of current. That's one of the several reasons why it's always recommended to stay away from operating near the max device limits.

 

Lots of people have silly rules for schematics. I don't believe "sideways" is wrong.

Some people have silly rules for their drawing and some simply get paralyzed when they see something outside of the way some "expert" or "professor" said was "right". I say, let the newbies be creative - as long at the plus goes to the plus, the minus goes to the minus, and they connect to other things along the way so they don't creat a short circuit. I've worked with teenagers trying to learn how to build something. One has to understand that leaning to keep the bicycle on two wheels for a few laps around the track is way more important than proper racing position and not making a full lap.

 

Well yes, I got the forward voltage wrong. It was more about getting the current in the two legs equal at 20 mA with resistors that do not exist.

 

Vf 1.16 volts seems a bit low for a visible light LED. Are these IR (infrared)?

 

Vf 1.16 volts seems a bit low for a visible light LED. Are these IR (infrared)?

Yes. The TS had very similar thread awhile back and it was established that he was using IR LEDs.