[Amps] LDMOS Solid State Amplifiers

[Gary Schafer]

Ok I see where the 20 comes from converting to voltage. But why do you
negate A and how do you negate it?

Thanks
Gary  K4FMX

> -----Original Message-----
> From: amps-bounces at contesting.com [mailto:amps-bounces at contesting.com]
> On Behalf Of Steve Thompson
> Sent: Monday, May 07, 2012 3:29 AM
> To: amps at contesting.com
> Subject: Re: [Amps] LDMOS Solid State Amplifiers
> 
> I think the underlying point is that IMDs add together as voltages
> not powers so you have to work back to the levels in Volts, not
> power, in order to can see the relative contributions and the
> total overall level. IMD levels in dB are powers; power=20log(V)
> and V=10^(power/20) which is where the 20 comes from.
> 
> Steve
> 
> > Hi Gary,
> >
> > B = 20Log(1+10^(-A/20))
> >
> > Let's say you measured the input IMD3 at -44 dB and the output IMD3 at
> > -32 dB.  The difference is 12 dB and that is A.  Divide A by 20 which
> > equals 0.6.  Negate that and raise 10 to that power which is 0.251.
> Add
> > 1 to that which is 1.251 and take the Log of that which is 0.973.
> > Multiply by 20 which is 1.945 which is ~2 dB.  This is B which is the
> > contribution to the output IMD3 due to the input IMD3.  Therefore the
> > output IMD3 is really -34 dB if there was no contribution from the
> input.
> >
> > I hope that explains it.  If not get back to me and I'll give it
> another
> > try.  If I confused you, I am sure I confused someone else.
> >
> > 73,   Tom   W0IVJ
> >
> > On 5/6/2012 1:49 PM, Gary Schafer wrote:
> >> Hi Tom,
> >>
> >> I can't seem to make your math work for finding input IMD
> contribution?
> >> Where does the divide by 20 come from in finding B?
> >>
> >> 73
> >> Gary  K4FMX
> 
> _______________________________________________
> Amps mailing list
> Amps at contesting.com
> http://lists.contesting.com/mailman/listinfo/amps