Anomaly with my negative trigger

[TaaviT]

Hello, i am having anomaly with my circuit. I designed a negative trigger circuit to detect car door switching.



Problem occours in example 2.
As i know current flows from higher potential to lower. Considering the 2V at the cathode, at the anode should be 2.3V, but in reality, voltage there is smaller, how is this possible?

As you can see in example 1, when my circuit is unconnected, in the circuit behind the question mark, which I dont know, is represented 1.5V, so when i connect my circuit to it like in example 2, it is being pulled up to 2V, and current is flowing in the direction shown, which is totally expected. Also, in example 3, when switch is closed, voltage behind the diode is 0.3V, like it should be.

Why is the voltage 1V at the anode of example 2?

 

[barry]

Without knowing what's behind that question mark makes this a little difficult.

Question: What makes you think current is flowing in case two? If you put 100v on the cathode of a diode, and 2v on the anode, how much current flows? (Hint: not much). I think you have a reverse-biased diode, in which case the voltage on the anode has nothing to do with voltage on the cathode.

How are you measuring voltage? Did you actually measure current flow, or are you just assuming?

 

[TaaviT]

Without knowing what's behind that question mark makes this a little difficult.

Yes, that is true, but still, shouldn't the rule, current is flowing from higher potential to lower be undependant of circuit behind question mark? Why is voltage on anode droping lower than cathode voltage?

If you put 100v on the cathode of a diode, and 2v on the anode, how much current flows? (Hint: not much). I think you have a reverse-biased diode, in which case the voltage on the anode has nothing to do with voltage on the cathode.

Indeed, but why inside my designed circuit on the anode i can see 1V, even though there's a pull-up to 5V?

How are you measuring voltage? Did you actually measure current flow, or are you just assuming?

Voltages were measured relative to ground. Current I measured with a multimeter and got 0.4mA.

 

[d123]

Hi,

Indeed, but why inside my designed circuit on the anode i can see 1V, even though there's a pull-up to 5V?

Even if the ratio is 10:1 and it doesn't divide down to 450mV if you get 1V, is that actually a pull-up or really a voltage divider?

 

[TaaviT]

Hi,

Even if the ratio is 10:1 and it doesn't divide down to 450mV if you get 1V, is that actually a pull-up or really a voltage divider?

It indeed is a voltage divider, but why is the voltage on anode divided below the voltage on cathode??

 

[barry]

That is only a voltage divider to AC; you're showing D.C, So that cap is basically an open-circuit.

I suspect there's something wrong with your measurement. If you truly have 0.4mA flowing through the diode, those voltages make no sense.

 

[TaaviT]

So i guess voltage on the anode is correct, considering the 0.4mA current? But on the cathode is false? Maybe the circuit behind question mark is using different ground or something?

 

[barry]

... or something?

Like maybe there's AC in there that you're not aware of?

 

[TaaviT]

I will check with oscilloscope tomorrow, but i doubt.

Thank you everybody for answering anyway.

 

[TaaviT]

Like maybe there's AC in there that you're not aware of?

Problem solved. There was actually voltage switching, checked with oscilloscope, couldn't believe that tbh.