# voltage drop from 555 / 556

Discussion in 'Electronic Design' started by Lyle, Jun 3, 2005.

1. ### LyleGuest

I'm playing with designing an IR tracking system using IR LED's flashed from
a 556. I want to run it off of a 9 v battery and was trying to calculate
the right resistor for high output but I can't find specs for output voltage
drop under load. I may just have to manually graph it out but by chance,
does anyone know what kind of voltage drop below Vcc I can expect out of a
555 or 556 at 100 - 150 mA max load?
thanks
Lyle

2. ### BanGuest

http://www.national.com/ds.cgi/LM/LM555.pdf
page 2,3
with 9V the droop will be typical around 1V with 100mA, but can be as high
as 2V worst case. You can see the 4 lower diagrams on page 3 which refer to
a supply voltage of 5V, 10V and 15V. These are the typical values, the max
values are specd for 15V only on page2.

3. ### Spehro PefhanyGuest

If you read the data sheet, it says Voh for sourcing 100mA on a 15V
supply, 13.3V typical, 12.75V minimum. At 200mA, 12.5V typical.

So, one guess would be 1.7V drop at 100mA and 2.1V at 150mA. Assuming
you're talking about the bipolar type. Be sure to bypass the battery
well.

If you use a transistor such as a 2N4401 wth 15mA of base drive, the
drop will likely be less than a couple hundred mV.

Best regards,
Spehro Pefhany

4. ### Spehro PefhanyGuest

I assume he's talking about sourcing current, rather than sinking,
since he mentions "voltage drop below Vcc", but I could be wroing
(sic).

Best regards,
Spehro Pefhany

5. ### MSCGuest

Build it, try it...

Sometimes we get so bogged down in designing things on paper and in
software that we forget that occasionally the best method is just to
get the old breadboard out and try it!

Mike

6. ### BanGuest

Well, the droop is pretty symmetrical sourcing or sinking, even if the
graphics and the tables are not giving the same values.

7. ### BanGuest

Well, the droop is pretty symmetrical sourcing or sinking, even if the
graphics and the tables are not giving the same values.

8. ### BanGuest

Well, the droop is pretty symmetrical sourcing or sinking, even if the
graphics and the tables are not giving the same values.

9. ### BanGuest

Well, the droop is pretty symmetrical sourcing or sinking, even if the
graphics and the tables are not giving the same values.

10. ### BanGuest

Well, the droop is pretty symmetrical sourcing or sinking, even if the
graphics and the tables are not giving the same values.

11. ### Spehro PefhanyGuest

Typical values @25°C:

Source Sink 2N4401 ZTX1051A
NE555 NE555 Ib=Ic/10 Ib=Ic/100
1mA 1.3V 0.015V 0.006V 0.05V
10mA 1.4V 0.1V 0.005V 0.05V
100mA 1.6V 1.5V 0.1V 0.05V
200mA 2.5V 2.5V 0.15V 0.06V

Best regards,
Spehro Pefhany

12. ### LyleGuest

thanks all! gives me a good place to start and I'll bread board it and test
from there.
thanks
Lyle

13. ### LyleGuest

LOL, yup, I just wanted to get a good starting idea as I'm trying to get as
much out of the IR LED's as is reasonable but their voltage drop is about
half that of a standard LED so its easy to fry them, very easy.
Lyle

gesundheit!

15. ### LyleGuest

I see it now, thanks for pointing it out. What do you mean bypass the
battery well? I was just going to use the output from the 556 in this
version until I need higher currents, do I need a transistor driver? (I'm
just a hobbyist)
Lyle

16. ### Spehro PefhanyGuest

The battery will have some internal resistance. It will get worse as
the battery runs down. You should put a capacitor across the battery
that will supply most of the current *during* your high-pulse with
little droop. The battery can recharge the capacitor between pulses.

For example, suppose you wanted to pulse at 150mA for 1msec and 200mV
of drop was considered okay.

You'd need C = 0.15A * 0.001 sec/0.2V = 750uF (use 1,000uF/10V).
Put 0.1uF ceramic in parallel.

Best regards,
Spehro Pefhany

17. ### Pooh BearGuest

That's why you use a series resistor. To set the current. The variation in Vf
for the diode ( or the margin in Voh for the 556 ) will make very little
difference to the current.

Best not to design right on the edge anyway.

Graham

18. ### Pooh BearGuest

Put a big capacitor across it ! The current drawn by the IR led(s) will see the
full battery impedance otherwise and the supply volts will sag.
A transistor driven by the 556 would give a better saturation voltage. On a 9V
supply that's not really a big issue. A bipolar transistor would also 'waste'
the base current thus reducing battery life.

Graham

19. ### Fred BloggsGuest

Do you have the data sheet for IRLED you're using? Some of them can
really take a wallop, go ohmic, and develop 5V drop. But you need the
datasheet to determine maximum current, maximum permissible pulse width,
and allowable average power, as well as heat sinking techniques using
short leads to large traces/planes. 9V Type N are not particularly
famous for high capacity, and in alkaline will spend most of their
useful life in the 7.2-8.5V range, so that the resistor drop method is
not the best way to drive the LED.