Convert Squarewave output to Always on.

[ADWSystems]

I need to convert a 50% duty cycle squarewave output of an existing circuit (open collector) to a steady signal with voltage the same as the peak voltage of the square wave. (ie., how to I go about filling in the low cycle of the squarewave without reducing the peak voltage?)

 

[3v0]

Is the peak voltage fixed ?

 

[birdman0_o]

A capacitor at the base of a transistor?

 

[ADWSystems]

Is the peak voltage fixed ?

Yes. I think (don't quite understand the question). Same peak voltage on each pulse.

 

[birdman0_o]

Wait, if you just want something that is always "high" just make it "high"?
Do you want it to be low when there is no square wave?

 

[MOSFET KILLER]

Use a capacitor to smooth the squarewave just like filter capacitors in power supplies. If you need a cleaner signal just use an NPN transistor, however, you may need to reduce the voltage off of the capacitor so you dont burn out the base of the transistor.

 

[ADWSystems]

Wait, if you just want something that is always "high" just make it "high"?
Do you want it to be low when there is no square wave?

I'm tying into an existing circuit. The output is pulse high when active and steady low when not. I think it would be easiest if I had a steady high and low. I'm not sure what the original intention was of the pulse-high/steady-low regime.

 

[ADWSystems]

Use a capacitor to smooth the squarewave just like filter capacitors in power supplies. If you need a cleaner signal just use an NPN transistor, however, you may need to reduce the voltage off of the capacitor so you dont burn out the base of the transistor.

OK. Open collector output, attach pull-up resistor to power, cap to ground, current limiting resistor to NPN base. NPN emitter to ground, resistor from power to collector, steady out should come off collector-resistor junction.

Dumb question time as the answer is completely eluding me. How do I figure the value of the current limiting base resistor? Assmuing a plan-jane 2N2222A NPN transistor, +5V power, and Ic=100mA (Rc=50 ohms).

 

[mneary]

Conservative: Assume a saturated gainof the 2N2222A of 10. The 2N2222a charts sho an assume beta of 10 for saturation. If your collector current is 100mA, you would want a base current of 10mA. If your supply is 5V, the sum of your pullup and your base resistor would be (5-0.65)/0.01 or about 435 ohms. You can safely increase the resistance by 20% or more.

Less conservative: The gain improves dramatically if you don't demand saturated Vce (0.2V) and tolerate something like 1V Vce instead. If you assume a beta of 20, then you can use resistors more like 1k (like 470 pullup and 470 on the base.)

 

[birdman0_o]

Why do you need to assume a beta? Doesnt the DS tell you what it is equal to?

 

[mneary]

Why do you need to assume a beta? Doesnt the DS tell you what it is equal to?

The data sheet gives saturated characteristics with β = 10. This appears to be a common figure in the industry, for a wide range of devices. The charts also show β figures for Vc=5V and 10V, a long way from saturation. β is not given, for example, for Vc=1V which would pull in the relay just fine with a supply voltage of 5V.

I was suggesting that he try designing to Vc=1V, therefore an assumption needs to be made that the β is greater than 10 (saturated chart) but less than 100 or so at Vc=5V.

 

[KMoffett]

ADWSystems,

I didn't see anywhere you mentioning the frequency or voltage of the pulsing signal, or how critical the timing is, but it seems that your need would be easily met by a "missing pulse detector" circuit. Google'ing that provides lots of circuits. Like the "Basic Missing Pulse Detector" here:
**broken link removed**

Ken

 

[audioguru]

Why do you need to assume a beta? Doesnt the DS tell you what it is equal to?

The datasheet for all transistors shows a wide range of beta when the collector to emitter voltage is fairly high so the transistor is not saturated.

But the datasheet for most little transistors shows a base current that is 1/10th the collector current when the transistor is saturated, not using beta.