RF signal on wirebonds

I am curious to know how RF signal is transmitted through wirebonds. I thought that in microwave transmission lines, the energy was stored in electric and magnetic fields inside the dielectric material which stays between two conductors. I imagine that this is the case for coaxial cables and microstrip lines for example.

However, I can't see how wirebonds transfer the RF signal. Where would be the ground plane in this case ?

According to the microwave101, wirebond can be modelled as an inductance element whose impedance depends on the diameter and the length of the wire.

Yet, I still can't see where the energy is stored physically.

The energy associated with electric currents in conductors is stored in electromagnetic and electrostatic fields, concentrated mostly near the conductors, but also potentially extending or radiating to great distances.

All this can happen in the air or in vacuum, i.e., no dielectric material is required to store EM energy.

When a dielectric material is placed near conductors, it is polarized by the electric field and its polarized molecules store some additional energy and modify the field. For instance, dielectric material in coaxial cables and microstrip lines affects their characteristics, such as capacitance, impedance, etc.

• So I suppose that the only function of the insulator in transmission line is to physically separate the core from the shield. And the loss in dielectric material, is it due to the polarization then? – Krlngc Jul 25 '18 at 5:55
• @Kirlangic Yes, the main function of the insulator is structural, but, as mentioned, it affects the impedance of the cable, which is important for matching. There are some low loss coax cables with air as a dielectric and spacers, like a plastic spiral, for example, to maintain distance between the core and the shield. Yes, the loss in dielectric is due to the continuously changing polarization. – V.F. Jul 25 '18 at 11:54

The magnetic field surrounds the thin short wire just as if it was a steady state field, in fact it is just that in this approximation when we think of it as an inductor; no ground plane or other metal is needed.

• I see. Can we say that this approximation holds as long as the wire bond length << signal wavelength ? – Krlngc Jun 24 '18 at 22:30
• usually one assumes that the bond-length (or anything else) < $\lambda/10$ and then the quasi-static (steady-state) approximation is pretty good. – hyportnex Jun 24 '18 at 22:37