voltage drop from 555 / 556

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

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

Typical values @25°C:

Source Sink 2N4401 ZTX1051A NE555 NE555 Ib=Ic/10 Ib=Ic/100

Best regards, Spehro Pefhany

gesundheit!

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

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

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

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

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

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

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

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

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

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

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

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

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.