# Is a humble diode an Voltage to Current converting device?

I want to use a push-pull solenoid, so I did some study of a fly-back diodes etc.

Than it downed on me, that if you apply say 30V to diode, you still get only 1.6V drop. The rest of the energy is converted into the current.

So basically, whatever voltage spike you throw at a diode, it returns just a current with little voltage drop.

Practically, is it right to say that diode is an voltage to current converter?

P.S.
I want to protect not just Arduino, but my power supply as well. So I thought to use reversed diode to stop any spike coming out of the solenoid into power supply. Would that work?

DROBNJAK:
Than it downed on me, that if you apply say 30V to diode, you still get only 1.6V drop. The rest of the energy is converted into the current.

That dawn is a sun that will never rise in this universe. If you apply 30V to a diode, you get a 30V drop. Assuming a voltage source with zero Thevinin resistance.

Think about a resistor, it's much closer to a Voltage to Current converter, since I=E/R.   You’re talking a setup like this?
Yes, this would help protect your power supply.
No, the diode does not convert voltage to current - it only allows current to flow while dropping the available voltage some.
Supply voltage = (Vf of reverse protection diode) + (current x impedance of coil) + Vce of transistor.
If the coil needs 30V to fully turn on, and you are supplying (Vf + Vce) less than that, than perhaps other switching methods should be looked at.
The coil diode allows any back EMF voltage to be dissipated in the motor, and the power supply is assumed to be sturdy enough to not be bothered by that.

It's really the coil... As you may know, a transformer is two coils around a single core and a step-down transformer converts high-voltage low-current to low-voltage high-current.

A coil is a "current device". Inductors "resist changes in current".

When the voltage/current source to an inductor is removed, the electromagnetic field collapses and this collapsing field "tries" to keep the same current flowing.

If the circuit is suddenly broken you have infinite resistance so the inductor "tries" to generate infinite voltage to keep the same current flowing. Usually, you get an arc or some component breaks-down and fries so you don't instantly get infinite resistance, and of course you can't get infinite voltage.

That's how the "coil" for a spark plug works. (It's actually a transformer.) Contact is made and current flows. The contacts are broken and high voltage is generated.

Since the current is still flowing in the same direction, but the inductor is now "generating" energy, current through the inductor now flows from negative to positive (like through a battery) and the voltage is reversed.

Since the voltage is reversed, a diode can be used to conduct current and prevent a high-voltage spike.

Practically, is it right to say that diode is an voltage to current converter?

Since we are dealing with a fixed amount of energy there's actually one more factor in addition to voltage, current, and resistance... That's time.

Because there is a fixed amount of energy stored in the magnetic field, the field will collapse more slowly (current will decay over a longer period of time) with the relatively low 0.7V drop across the diode (with low effective resistance) compared to a higher-value resistor.

Ideal components(*) have a response curve - relating voltage to current, and so does
the circuit "driving" that component. The values of voltage and current you see are
those that fall on both response curves (ie where they meet), because that is the solution
to the equations governing electric circuits. In other words both the component and
the circuit co-operate.

30V forwards across a diode will lead to a current determined mainly by the series
resistance of the diode (this will usually vaporize the diode extremely fast).

(*) More correctly components that don't store energy and which have no time-dependent
behaviour (not inductors / capacitors - for these you have to solve differential equations).
The technique of finding where curves cross over can actually be useful when you have
several non-linear elements involved.

DROBNJAK:
Practically, is it right to say that diode is an voltage to current converter?

No.

That simply makes no sense. Please simply forget about it.

You are misunderstanding what a “flyback” diode does. Here is the normal current flow in the coil: And here is what happens when you open the switch or turn off your transistor/ FET: Note clearly that in both cases, the current is flowing in the same direction.

Now in the first case, the diode (albeit not shown) is not conducting because it is reverse biased. It conducts in the second case because the voltage across the coil has reversed. Note that - the voltage is reversed, not the current. In other words, as the switch opens, the voltage goes from positive to negative. It is as it goes negative that the diode - if present - “catches” it and limits it to the forward voltage drop characteristic of the diode - generally about 0.7 to 1V for a silicon diode.

If the diode is functioning correctly, the voltage as it conducts never exceeds that forward voltage drop. The current in the coil, and thus the current that the diode conducts, is initially exactly the same as it was before the switch opened (it thereafter decays), and the diode is rated for this current (because you chose it so,) so there is no anomaly here. And the voltage across the diode never exceeds in its reverse direction, the voltage that is provided to the coil by the power supply. There are no “excess” currents or voltages applied to the diode at any stage.

And just for completeness, the rate at which the voltage across the coil goes from positive to negative, is in practice limited by the intrinsic capacitance of the coil (and diode).

Think of current as similar to water flow and voltage similar to water pressure. A diode is comparable to a one-way check valve that needs a little pressure to push open. A capacitor is something like a bladder that fills with water inside a pressure tank (and compresses the air within). An inductor is the most difficult with which to make an analogy, but the penstock of a hydroelectric system can develop inertia that acts like a flow inductance (it will even oscillate with a bladder capacitor).