Simple diode and transistor circuit

#12

Joined Nov 30, 2010
18,224
Base of Q2 needs a resistor to ground to bleed off the emitter to base current that will accumulate on C2. Time constant of that resistor and C2 must be faster than the highest switching speed so Q2 can be fully off before the next high pulse arrives.

I made the same mistake with my circuit, except the accumulated charge will be a lot smaller with only parasitic capacitance.

The only thing I tried to do was arrive at: When the uc voltage output is low, the current through the next transistor is zero. If I understood the request correctly, I think it will work, as in, "fail in the off condition".

I can not interpret what Scott Wang said a a sentence, so I did not try to explain how, "the input is feedback".

Bedtime now. Sorry guys. The last time I woke up was Saturday at 2 pm. (Probably why I thought this was still Saturday :D )
 
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inwo

Joined Nov 7, 2013
2,419
Base of Q2 needs a resistor to ground to bleed off the emitter to base current that will accumulate on C2. Time constant of that resistor and C2 must be faster than the highest switching speed so Q2 can be fully off before the next high pulse arrives. Thanks

I made the same mistake with my circuit, except the accumulated charge will be a lot smaller with only parasitic capacitance.

The only thing I tried to do was arrive at: When the uc voltage output is low, the voltage output from the current amplifier transistor is low. If I understood the request correctly, I think it will work, as in, "fail in the off condition".
I agree!..................................................................................
 

ScottWang

Joined Aug 23, 2012
7,397
He's using an opto-coupler. He asked for a way to have it fail in the off condition if the Uc output got stuck, "low". I considered that he wanted those two things and made a simple circuit that would do it.
I saw many people using the photocouple in many situations, the most common reasons is the interference, specially when the uC and Motor are using in the same project, sometimes we can just using RC filter from the Vcc to solve the problem, sometimes we need the LC parts to solve the problem, sometimes we just need two capacitors for the motor.

Many people like to using the photocouple, but does the case really need to that, I can't say what is the best way, just case by case to solve the problem, as what I knew was that we have to find out where is the real problem could be, and using the O'scope is a good way to do.
 

ScottWang

Joined Aug 23, 2012
7,397
I can not interpret what Scott Wang said a a sentence, so I did not try to explain how, "the input is feedback".
Scott Wang said:
You using the Q1 bjt as as switch, but the output pin as a input, what the output will be happening when the input is feedback?
Sorry,#12.
It just a short form to showing a long story.

What the circuit from inwo, that is a circuit similar as a SCR, using this kind of circuit should be careful when it is a high voltage or big current, it could damage the Vin side, because the output of Q1 is connected to the Vin, although it is not connected directlly.

Normally we will using the base of bjt as a input side, and the collector will be the output side, but in this circuit that the Q1_C output of bjt connected to the Vin side, that's what I said the "the output pin as a input", the Vin will get the feedback from C through B, and this circuit also easy to lockout as a real SCR, it will loss the original function.
 

ScottWang

Joined Aug 23, 2012
7,397
Thank you for looking and the question.

I'm sorry, being self taught, I often have trouble following others explanations and questions.:p

Possibly, a more detailed description would get thru to me.
Or maybe #12 can help me out.:confused:
Don't be sorry.
But what I trying to say is that if you still want to play with EE, you should using solder or breadboard to do the experiment not the simulation software, I didn't mean that the simulation software is bad, it is a tool when you know more the basic theory, and doing more the practical experiments, the tool will help you save the time and see the waveform.

You can find the problem from the real experiment, sometimes you can't get it from the software, and you will learn something from the experiment, even the experiment is wrong.

And you should continuing to attach and show your circuit, it will let the users to help you to find out the problems, and you will learn from that, so the circuit was wrong that's ok, if you don't show the circuit then you can't find the problem, that's a real problem.

when I attach the circuit that I also hope other user to find out the problem and tell me, even that is a small error, but I still felt that very grateful, and thanks for pointed out the problem.
 

AnalogKid

Joined Aug 1, 2013
10,987
We have a similar problem with brushless DC fans that can die with the tach pulse either high or low, and solve it with a single capacitor. AC couple the 20 KHz square wave with a 0.1 uF cap, adding around 1 ohm at that freq. The transistor base can stand 5V of reverse bias, but you can add a catch diode to the +5 V for extra protection. No matter which way the uC output fails, the cap will eventually charge to a steady state, current will stop flowing, and the opto will turn off.

ak
 

inwo

Joined Nov 7, 2013
2,419
We have a similar problem with brushless DC fans that can die with the tach pulse either high or low, and solve it with a single capacitor. AC couple the 20 KHz square wave with a 0.1 uF cap, adding around 1 ohm at that freq. The transistor base can stand 5V of reverse bias, but you can add a catch diode to the +5 V for extra protection. No matter which way the uC output fails, the cap will eventually charge to a steady state, current will stop flowing, and the opto will turn off.

ak

So, are you saying that you just capacitive couple the driver circuit somewhere?

Would that not adversely affect the square wave?
 

#12

Joined Nov 30, 2010
18,224
A capacitor too small will make the flat part of the square wave sag. The right size capacitor will not interfere at all.
 

inwo

Joined Nov 7, 2013
2,419
Did I just say something stupid?:confused:

Sorry. Wasn't paying attention very well.
No no no. :D

I'm serious!!!!!!!!!!!!

Driving thu a capacitor somewhere in the circuit is the logical answer.

If it works!

ps.
You must think I'm stuck in sarcasm mode! :)
 
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