Can a Foster Seeley FM detector work if one of its diodes breaks down?

As the attachment shows, it's a simplified model of Foster-Seeley detector, just use for this discussion, if there's anything wrong with this figure, please point it out , I'll correct it as soon as possible.

if the detector works just as expected, the output voltage should be Vout=K(|Va|-|Vb|) where Va=V1+V2/2, Vb=V1-V2/2, and the phase difference between V1 and V2 is determined by the frequency f of received wave, roughly when f=f0, Vout=0, f>f0, Vout>0 and f<f0, Vout<0.

My question is, if D2 breaks down for some reason, does the detector work? Say it doesn't work well as usual, but once it still outputs a voltage around a reference voltage, may I say it's also working as a detector? I think even Vb is absent, if f=f0, Vout is Vref as a reference, and f>f0 Vout>Vref , f<f0 Vout<Vref, is this OK? Like...the shape of the envelope remains similar to normal status?

Try running a simulation in LTSpice.

alec_t said:

Try running a simulation in LTSpice.

Click to expand...

Thanks a lot for the reply~ Actually I ran the simulation on Multisim last night for a simple BFSK signal, but the resonant part(the LC tank) doesn't work well as expected(like when f!=f0, it's not a symmetric characteristic on each side), besides I can't distinguish the likelyhood between curves by my eyes >_< So, I still want to wait for a theoretical answer ^_^