What is the input impedance of the Netduino board?

Try reducing the value of the 82kΩ collector resistor at the output to about 10kΩ.

 

I'll give the 10k resistor a try. I'm not sure where to find out what the impedance of the board is. http://www.netduino.com/netduinoplus2/specs.htm

 

Crutschow, I tried a 10k resistor, but ended up with a 1k in order to get an output pulse voltage of about 2.5v. This was enough to trigger the Netduino interrupt.

If I went below 1k on the collector resistor, the steady state output voltage started creeping up above zero. Why would this happen?

 

I see that you have an 82k Ohm pull-up resistor on your output. Try changing it to something in the range of 510 to 330 Ohms.

[edit]
Whoops, I didn't see the other replies?
You don't have a resistor from the B out to the BC547's base; you will burn up the transistor and maybe the 556 very quickly. A 1k resistor should be OK.

[eta]
A caution; the Linear-supplied 555 timer is an idealized model that's optimized for speed. A component model is more accurate; as an actual 555 timer has a Darlington output, which won't go higher than about Vcc-1.3v even under light loading. This can be significant when interfacing with logic levels.

 

Sgt, I have a 10k resistor in front of the base on the output 547 on my physical circuit. I just forgot to add that to the circuit diagram. Thanks for your comments. I'm still curious to know why the steady state output voltage creeps up when I lower the resistance below 1k on that collector?

 

It's because you have the 10k resistor on the base that you didn't show in the schematic, and when you dropped the collector resistor, the transistor started to come out of saturation. Being saturated means that a further increase of the base current (Ib) will not lower the collector to emitter voltage (Vce).

The "rule of thumb" for "forced beta" (gain) when using a transistor as a saturated switch is to use 1/10 of the desired collector current for the base current. If you have 5v across a 470 Ohm resistor on the collector of the output transistor, that would be 5v/470 ~= 10.64mA current flowing through the resistor when the transistor is turned on, requiring ~ 1.07mA current through the b-e junction in order to saturate the transistor. About the most you'll get out of the 555 output is 5v-1.3v = 3.7v, and the b-e junction will have about a 0.7v drop, so 3.7v - 0.7v = 3.0v to drop across a base resistor. 3v/10k Ohms = 0.3mA, or about 1/4 the amount of base current (Ib) you needed for a 470 Ohm collector resistor in order to saturate the transistor. 3v/1.07mA ~= 2.8k Ohms; the closest standard E24 value would be 2.7k. You could use a 2.4k resistor also.

 

Thanks Sgt. I knew it had to be something that I did not understand about transistors and IC's. I'm a mechanical engineer, not a EE, so my knowledge in this area is pretty limited. Your explanation goes a long way to helping me understand your comment about the idealized 555 model too.

The world is a better place because of guys like you and crutschow who are willing to share your knowledge and wisdom with hacks like me!