1
\$\begingroup\$

Application note 969 derives the voltage sensitivity at zero bias of a detector diode:

\$ \gamma = \frac{V_o}{P_{\rm in}} = \frac{0.52}{I_s(1+\omega^2 C_{j0}^2 R_S R_V)} \cdot \frac{R_L}{R_V+R_L}, \$

where \$\omega\$ is the input frequency in rad, \$C_{j0}\$ the junction capacitance, \$R_V=\frac{nV_{th}}{I_s}=\frac{0.028}{I_s}\$ the nonlinear junction resistance and \$R_L\$ the load.

Their argument is made with small signal circuit approximations. However, I want to "see" this by using an actual .diode element and sweeping over multiple Is values as in 1 (for example, via tran simulation).

I use this simple LTspice simulation which sweeps over the same range of Is as in 1:

enter image description here

As can be seen, I measure the output voltage and input power into the diode and then plot gamma1=Vout/Pin. gamma0 is the formula from above 1. If I set \$R_L=\infty\$ and \$C_{j0}=0\$, I get great matching results beween gamma0=0.52/Is and gamma1:

enter image description here

Note that I have to add some load to the output; I add a capacitor (although not explicitely described in 1). The caveat is I need to wait with until the signal has settled.

However, once I set \$C_{j0}\neq 0\$ or \$R_L\$, I no longer get a match. I tried various ranges of \$C_{j0}, R_L, C_L\$ and ensure the output voltage settles. For example, for \$C_{j0}=0.1\rm{pF}\$, I get:

enter image description here

What am I doing wrong?

\$\endgroup\$
3
\$\begingroup\$

You seem to be plotting the direct voltage output, but that's not how they're describing it. What they defined is a voltage sensitivity, which they give as (\$f\$ in GHz):

$$\gamma_1=\frac{0.52}{I_S(1+\omega^2C_j^2R_SR_V)}\xrightarrow{C_j=0.1p, R_S=50,R_V=\frac{0.026}{I_S}}\frac{10^3}{f^2+2\cdot 10^6I_S}$$

Here there is a datasheet for HSPS8101 where, at page 2, they show both the SPICE parameters, and the equivalent schematic, in which the only variable element is \$R_V\$, as a function of \$I_S\$. Still, I used your values, and this is the result of (I also used your 0.9 GHz value for frequency):

.step dec param Is 10n 100u 5
.meas gamma1 param 1k/(0.9**2+2meg*Is)

gamma1

which seems to be very close to the one in their figure 2 (the 1 GHz trace). If you want to plot \$\gamma_2\$ and \$\gamma_3\$, you'll have to use their formulas; no need for any schematic, by the looks of it, all seem to be usable as .meas directives.

BTW, I applaud your valiant efforts with the .model cards, but you can simplify that to only one:

.model d d Is={Is} N=1.06

and step Is as in the .step command above.


In order to avoid using the formulas from the paper and, instead, use the simulation, then you need to plot the sensitivity, which is given in V/W. Also, in the paper, at page 1, Voltage sensitivity, they say:

A detector diode may be treated as a current generator across the diode video resistance. The voltage sensitivity, \$\gamma\$, is the product of current sensitivity, \$\beta\$, and the video resistance, the inverse of the derivative of current with respect to voltage.

So the sensitivity of the diode could be extrapolated to the current setup as V/W = V/(V*I) = 1/I, or the reverse of the current through the diode. Using:

.meas Iavg avg I(D1)`

results in this plot:

1/Iavg

The actual dynamic resistance would be very similar, (Vin-Vout)/Iavg, as given by:

.meas Vin avg V(in)
.meas Vout avg V(out)
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Awesome, this is already very helpful!! But, what I meant, I would like to reproduce this line using the actual Diode! Your plot basically just implements their equation (which is somewhat derived from a small signal model). If I define my diodes without/small Cjo and no/huge output load, I would expect to be able to reproduce this curve directly by measuring voltage output (and probably normalizing by the input power) of a tran or ac analysis. This is really what I am looking for. \$\endgroup\$ – divB Aug 14 at 20:19
  • \$\begingroup\$ You could use I_s as the result of .meas I_s param -I(D2), instead of the Is above, but you'll find out that, for a 1 V reverse voltage, you'll get just about the same value, so the graph won't change. Using V(out) you certainly won't get the same graph, since it's expressed in V, while \$\gamma\$ in V/W. You could try to cheat with 1/V(out) (since V/W=V/(V*I)=1/I=R/V), but I have doubts. Not lastly, while LTspice will happily simulate 1 THz, that doesn't mean things are that simple. At hundreds of MHz and up, things get very different. \$\endgroup\$ – a concerned citizen Aug 14 at 22:05
  • \$\begingroup\$ About the only thing I see is to simply plot V(out) (or the power across the load resistor), as it is, in .TRAN, and see how the levels increase with Is. That should be proof enough. Try using a 50 Ohm load and a 10 nF capacitor as load, simulate with .tran 0 {9u+100/0.9g} 9u, use a larger input, say 0.1, and maybe add .opt plotwinsize=0 to disable compression (and avoid plotting artifacts). \$\endgroup\$ – a concerned citizen Aug 14 at 22:06
  • \$\begingroup\$ I'm very happy to report that I got it working: snipboard.io/0ZadiL.jpg !! I think the reason is that constant input voltage does not imply constant input power (because input impedance changes indirectly proportional to Is as well). So, I measure the input power through the Rs and normalize by it. As can be seen in the link, there is now a perfect match with the predicted theory! When you incorporate this into your answer I can mark it as Solved. \$\endgroup\$ – divB Aug 14 at 23:40
  • \$\begingroup\$ If the picture you uploaded is used exactly as it is drawn, then you should know that you are simply plotting the magnitude of an integrator. Which makes sense since you left out RL, so the load is now CL, only. It doesn't matter that you are measuring the current through Rs (which should be defined in the diode's .model), I(Rs)=I(D1)=I(CL). It's about transmitting power, so the I/O resistances should be comparable. Also, that inductor won't do anything like that, across a voltage source. With these in mind, I get very different results. \$\endgroup\$ – a concerned citizen Aug 15 at 7:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.