Diode circuit does not

Hi all!

I'm having a problem getting my simulated circuit to translate in real life.

Please see image of the circuit or circuit attached. I'm just checking the voltage at the point of 'test '. When I run a simulation, I get a voltage square wave at this point that alernates between the 24V and (basically) 10V. When I build this circuit on a printed circuit, the voltage at this point varies from 24V and 0V. In addition, in the simulation, the tension in 'test' relies heavily on the load capacitor (10pF), but for my real-life circuit, the capacitor does not seem to affect the voltage.

Can someone help me understand this?

Thank you!

I can't run your simulation, but I can offer a few suggestions based on the images.

Simulation and analysis of the circuit I expect that on the first impulse, load capacitor voltage source (or very close to him, according to the correct diode model used). After that the Test voltage is not defined for ideal diodes (conduction perfect when skewed towards the front and no conduction when reverse biased) because the Test is isloated. However the Test will always be in the range of 0.24 V because outside that range one of the diodes would become before biased at a State of source.

In a real circuit impedance of entry of your measuring instrument becomes part of the circuit. If you use a digital multimeter or an oscillscope with a X 10 probe, you probably have an impedance to the ground resistance ohms 1E7 and a few picofarads of capacity.  The resistance could easily be much lower and much higher capacity with other equipment. With this impedance you currents flowing on the ground through D1 when the source is 24 V and current through D2 leakeage when the source is 0 v. As the capacity of the measuring system is comparable to the C1, the effect of the C1 will be difficult to observe.

Lynn

Tags: NI Software

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