# What happens if the Forward voltage of a diode is not met?

Ihave been playing around with transistors and diodes lately, and thinking about switching methods and techniques.

I'm not that educated on the matter of "Forward Voltage", but I illustrated a concept I am trying to grasp in the photo below:

In this circuit concept, I have a 1.5V battery hoked up across the terminals of a capacitor, but...there is a cathode of a diode hooked up between the negative terminal of the battery and the negative terminal of the capicitor.

The Diode, attached to a current limiting resistor,can provide a path to the other terminal of the capacitor, but only if the forward voltage is met???

Is that so? Would the diode act as some type of "flood-gate", allowing the battery to store electrical potential in the capacitor, until the pressure is adequate to provide the forward voltage to the diode?

My goal, or what I envision, is that I could hook up a relay or transistor the other end of the capacitor/ battery, and the temporary discharging of the capacitor would momentarily "power" the switch, allowing me to momentarily connect another power source to make a light blink or something.

If the diode funcctions in tth way I assume, then it would probably do this in a constant cycle, and could possibly be timed in accordance to functionality.

I'm just not sure if this is the actual science...

• Ignoring the schematic itself and focusing on your goals, I think you imagine that a diode could be arranged to wait until a capacitor charged up sufficiently and that when there was enough voltage present, the diode would "open" and let stored charge in the capacitor flow through the diode and turning something on until the capacitor lost enough charge that things stopped. Is that what you are hoping for? – jonk Oct 29 '17 at 7:22
• @jonk Yes, that is what I’m hoping for. – Iam Pyre Oct 29 '17 at 7:23
• Well, it won't work like that. First, you'd need to limit the current to the capacitor with a resistor (so that it takes time.) Next, the diode actually starts conducting earlier (at any forward voltage more than zero volts) and it's very likely an equilibrium state would result, instead. (In fact, it's certain to do so.) So this really won't "pulse." What you want is called a "relaxation oscillator." You can look those up. Or I could post up a schematic of one. But it involves a few transistors (and a diode or two) and I'm not sure you are ready to consider that much, just yet. – jonk Oct 29 '17 at 7:29
• A PUJT (programmable unijunction transistor) can make a very easy relaxation oscillator, though. You might look those up. That thing works almost exactly like you are imagining it should. – jonk Oct 29 '17 at 7:31
• @jonk well, transitory is what this all evolved from. I understand how to saturate a transitor, but I can only control it with a switch. I wanted a more “automated” way to control a transistor. I’m not good at programming, so using my Radpberry Pi isn’t a solution . I was hoping this method could be an alternative. Guess I got a lot more reading to do. – Iam Pyre Oct 29 '17 at 7:34

In this circuit, $R_4$ and $C_1$ are the timing pair. I think these are doing what you hoped for. Current flows through $R_4$ and charges up $C_1$, over time. $R_2$ and $R_3$ form a voltage divider that sets the "trigger voltage" that $C_1$ needs to reach before "things happen." As $C_1$ charges up, it eventually reaches that trigger voltage and when it does the PUJT "fires" and dumps the charge sitting on $C_1$ into $R_1$ as fast as it can. In doing so, the current caused by this charge-dump causes a voltage to develop across $R_1$, which is then used to turn $Q_1$ on and saturate it -- just as you'd hoped.
No Arduino. No microcontroller. Just a basic timer. You can change the timing of the pulses by adjusting the value of $C_1$. At it is, it runs somewhere between $3\:\textrm{Hz}$ and $4\:\textrm{Hz}$. I've set $R_{LOAD}$ to a particular value. If you discuss what your load actually is, I might be able to help adjust the above circuit. But at least this gives you an idea.