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https://analog.intgckts.com/rf-mixer/diode-mixer/

Source: https://analog.intgckts.com/rf-mixer/diode-mixer/

Preface: I am not an engineer.

Let the diplexer just be two resistors sharing a lead to Cb. This means that at Cb, the currents induced by Vrf and Vlo accross the diplexer resistors are added together into an AC waveform with positive and negative half-cycles. I will call this current-waveform our "signal". Additionally, let Vdo << breakdown voltage of the diode.

From my understanding, a forward biased diode has very low forward impedance. If we assume that Vdo forward biases the diode, the positive cycle of our signal would see the impedance of the IF filter in parallel to a low impedance path to ground through the diode.

Thus, the positive half-cycle of our signal gets shorted to ground and does not appear at the output of Cb. Is this a correct statement?

What happens during the negative half-cycle of our signal when the ground/Vdo polarity flips?

I would like to guess that the current of our signal would pass through the diode "the wrong way" as leakage current and see a high impedance path to (-Vdo) from ground via the RF choke. Therefore, it would instead pass through Cb to the IF filter. This also where our signal's components get multiplied instead of added and the sum and difference frequencies of RF and LO appear as IF.

If that is true, wouldn't the magnitude of voltage created by the IF leakage current over Rl be absolutely tiny? Are there re-arrangments of this circuit that may lead to a larger magnitude IF voltage present at Rl such as making Vlo and Vrf low impedance current sources?

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3 Answers 3

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An attempt to simplify the OP's question "how does it work", and avoid the math in the OP's reference document...

I asked LTSpice to characterize a bog-standard 1N4148 silicon diode to serve as the single mixer, with a DC scan of its current-voltage curve. Of interest was its dynamic impedance at two different bias points:

  1. where no diode current flows, and voltage across its terminals is zero.

  2. where about 10mA flows (actually near 11mA) and diode voltage is 0.7V.

Ten milliamps is a reasonable guess where peak current caused by the local oscillator mixer input signal might flow. Dynamic impedance of the diode is the slope of the V versus I curve when biased at these two points. LTSpice resuslt at 27 degrees C:

  1. At zero bias, dynamic impedance is 16.44 MegOhms (virtually open circuit)
  2. At 11mA bias, dynamic impedance is 4.6124 ohms

Let me take the liberty of re-arranging (simplifying) OP's circuit to eliminate the bias inductor. I have made the local oscillator signal source a current source, (I1) so that all the signal power is sent either to diode (into GND) or to load resistor RL. Coupling capacitors are gone as well. I have added source resistance RS to the R.F. low-level signal input (V1) so that power into the load resistor RL can be found. The basic shunt-type mixer operates much the same as OP's circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

The local oscillator is now a square wave current source at 1.1 kHz. while the small signal "RF input" at 1kHz is a sine wave. Local oscillator amplitude is set so that the diode is either biased at about 11mA (0.7V) or it is biased at 0mA (0V). In one state, almost all signal power is delivered to RL, while in the other state most signal power is shunted through the diode to GND.
To get 11mA diode current, 26mA peak current is required of the current source: 11mA flows into the diode, 15mA is shared by RS & RL.
This is a wide-band mixer. You could regard the mixing output to be either 100 Hz (1100Hz - 1000Hz) and/or an output at 2100 Hz (1100Hz + 1000Hz). The amplitudes of these two outputs at 100 Hz and/or 2100 Hz should be the same.


A LTspice transient run was made, looking for voltage amplitude across RL at 100 Hz: result is 1.0305mV RMS when a peak signal amplitude of V1 was set to 10mV. LTspice's FFT was used to measure amplitude at 100 Hz because there are large-amplitude signals at many other frequencies. One of these at 2100 Hz is also about 1.035mV RMS.
When the local oscillator current source is removed, the very high diode impedance is insignificant - nearly half of V1's voltage is delivered to RL: 3.5346mV RMS.

So the take-away is that this passive mixer's gain is 1.0305/3.5346, which is -10.7 dB power gain. Should a sine wave source be substituted for I1, gain would be even less. Is this the best achievable from this simple shunt mixer?

As a check, the diode is substituted with a voltage-controlled SPST switch, performing the same function. Its leakage resistance when open-circuit was set to 16.44MEGohms, and its contact resistance when short-circuit was set to 4.6124 ohms (same as the 1N4148 diode). It opened and closed 1100 times-per-second. The LTspice was run again, with the same result: mixer power gain of -10.7 dB.

schematic

simulate this circuit LTspice transient run, switching mixer 100 Hz mixer output from 1000Hz signal mixed with 1100 Hz localoscillator

Edit:

Where did the R.F. signal go? (-10.7 dB gain)...
This resistive network stores no energy in reactive components - the diplexer is missing - it can only dissipate power, or deliver it to RL.
Nearly half the available R.F. signal is shunted to ground through the diode while it is conducting (half the time). The tiny fraction of the R.F. signal that finds its way to RL (the diode isn't a perfect short-circuit) subtracts R.F. signal from the desired mixing outputs at sum&difference frequencies.

In addition to these losses, R.F. signal is spread over many frequencies. Not only is it divided between sum&difference (100Hz, 2100Hz for this example), but also divided amongst all the odd harmonics of the local oscillator (above&below). Only ONE of these is usually selected as the desired output - that is one of the functions of the diplexer.

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  • \$\begingroup\$ Fantastic analysis! I fiddled with LTspice to try and get some intuition on how this circuit worked but no matter the topology/component values etc. my IF amplitudes were always abysmal. I thought that I was missing something, but it seems that single diode mixers are just lossy. \$\endgroup\$ Mar 18, 2023 at 3:43
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From my understanding, a forward biased diode has very low forward impedance.

Correct. So we need to approach this circuit under conditions where it doesn't simply short out the source.

To be clear, we could put enough VD0 into it that the signal and LO are effectively shorted out, and nothing happens. That's not interesting, so let's examine the case where something interesting does happen.

Also, to be more specific, ID moreso than VD0. We would almost never apply a fixed, absolute voltage to a diode in practice, and anyway the purpose is to set some bias current. Most likely VD0 will be supplied from a bypass capacitor and bias resistor from some other (higher, fixed) supply voltage.

If you're versed in calculus, the small signal derivation is worth going through:

Roughly speaking (i.e., for present purposes), a diode has the transfer curve \$I = I_s e^\frac{V_f}{V_{th}}\$, where \$I_s\$ and \$V_{th}\$ are component parameters (normally \$I_s\$ is in the ballpark of 1 nA, and \$V_{th}\$ is ~52 mV and proportional to temperature). If we have a small change in \$I_f\$, we have a logarithmically small change in \$V_f\$: the diode looks like an incremental resistance of \$R_i = \frac{dV}{dI} = \frac{V_{th}}{I_f}\$.

If we apply a somewhat larger signal, such that the ratio of peak to valley currents is significant (say, about \$e\$), we will find that the incremental resistance varies over the cycle. We can better approximate the transfer curve as a parabolic section, for which we need the power series of the exponential function (the values of which aren't important, just to say it's easily approximated this way). In that case, we will have not just the linear (resistive) component, but a quadratic component as well, which means the signal will be multiplied by itself.

Suppose we take two sine waves of differing frequencies, \$i_f(t) = A \sin \omega_1 t + B \sin \omega_2 t \ \ (\omega_1 \neq \pm\omega_2\$). The voltage resulting from this applied signal current is:

$$v_f(t) = \left( A \sin \omega_1 t + B \sin \omega_2 t \right)^2 = A^2 \sin^2 \omega_1 t + A B \sin (\omega_1 t) \sin (\omega_2 t) + B^2 \sin^2 \omega_2 t$$ $$= A^2 \left(\frac{1 - \cos (2 \omega_1 t)}{2} \right) + AB \left(\sin (\omega_1 t) \sin(\omega_2 t) \right) + B^2 \left(\frac{1 - \cos (2 \omega_2 t)}{2} \right)$$ The middle term further simplifies via trig identity, $$AB \left(\sin (\omega_1 t) \sin(\omega_2 t) \right) = \frac{A B}{2} \cos((\omega_1 - \omega_2) t) + \frac{-A B}{2} \cos((\omega_1 + \omega_2) t) $$

Which means, in terms of AC steady-state signals, we've turned a sum of two sinusoids, into the sum of (up to) four sinusoids, two of which are 2nd harmonics of the inputs (twice the ω), and two of which are the sum and difference frequencies of them (sum and difference ω).

This is called frequency mixing.

At still higher signal levels, the higher terms of the exponential's power series become significant, and we generate higher order harmonics and mixing products (generally, trig polynomials like the above, which will always reduce to a sum of sinusoids, at various harmonic and sum/difference frequencies, and the mixing products of those in turn, etc.).


This section does not assume calculus, and uses more of a hand-waving explanation.

At the highest signal levels, we might even fail in applying the exponential function in the first place (i.e., we're braking the assumption that the diode stays forward-biased, or other effects dependent on circuit particulars not shown here). In this case, we might prefer a switching sort of argument.

Consider two source signals, one of which is extremely intense:

diode modulation waveforms
(please excuse the crudity of the hand sketch)

Maybe these aren't to scale, so \$i_2\$ is much, much larger than \$i_1\$.

Effectively, \$i_1\$ passes unimpeded during part of the phase (say when \$i_2(t) < -I_D\$), and otherwise, is at the very least greatly reduced, if not effectively switched off entirely.

If you've not understood any of the proceeding sections -- this still won't be very convincing to you, but suffice it to say, when a sine wave is switched on and off, it ain't a sine wave anymore, but we can describe it as a sum of other sine waves (via the Fourier transform), and these other waves are what we are talking about when we filter and tune a radio signal.

As example, amplitude modulation is a simple mixing (product) operation, so this method applies directly to that type of radio system.

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  • \$\begingroup\$ Interesting! I appreciate the derivation. I wonder if the switching case would provide higher magnitude signals of the sum and difference frequences on an FFT; To get the switching case, would (for example) Vlo peak to peak > Rf and Vfd of the diode? \$\endgroup\$ Mar 17, 2023 at 17:08
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    \$\begingroup\$ Yes, that is a necessary part. And Vlo / Rlo > ID. Note that, since Vlo is being rectified in this case, VD0 may be driven negative (in which case bias can be a resistor to GND, no voltage supply needed at all). Relative amplitudes don't get much larger than 1/4 or so, due to the signals being split into multiple frequencies; this is referred to as "mixer loss". Any (passive) mixer approaching 6dB loss is a good mixer. \$\endgroup\$ Mar 17, 2023 at 18:15
  • \$\begingroup\$ A "diode ring mixer" is a more complex crude mixer, sometimes (often?) used with a square-wave signal and it literally switches the polarity of the input signal half the time. \$\endgroup\$
    – user253751
    Mar 18, 2023 at 22:14
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Thus, the positive half-cycle of our signal gets shorted to ground and does not appear at the output of Cb. Is this a correct statement?

Actually, because of the DC bias, all half-cycles are positive half-cycles for the diode. The signal is always positive.

The signal and bias level should be set so that the diode shows some variation in its forward voltage depending on the signal current. It's no good to have the diode all the way on, or all the way off. It needs to be operated in the middle where it is not fully on and not fully off. It makes the output signal into a non-linear function of the input signal, something like the natural log of the signal (and bias), although it's not important to know precisely which function it is.


To understand the mixing action, it's not about the half-cycles. This circuit should be analyzed by thinking about frequencies.

What happens when you pass an RF signal through some random non-linear function is that all the frequencies get jumbled up every-which-way and mix with each other. Suppose you put in a 2Hz and a 10Hz sine wave; if you analyze the frequencies of the output, you see there is still a lot of 2Hz and a lot of 10Hz, but also a whole bunch of combinations like 12Hz and 8Hz and 22Hz and 38Hz and 4Hz. (The simpler combinations like 10+2 and 10-2 have much higher amplitudes than the more complex combinations like 10+10+10+10+2+2+2, so you don't have to consider all the infinite number of combinations)

Then the filter selects the one you want.

When using a mixer based on the principle of "jumble them up and select one", the frequencies have to be carefully selected so that only one frequency combination is the one you want. For example you probably couldn't convert 100Hz into 101Hz by mixing it with 1Hz with this type of mixer, because 100+1 is too close to 100-1 and you couldn't pick one of them separately. If the input contains many frequencies (e.g. many radio stations) you also have to consider all the combinations of input frequencies. The advantage of this type of mixer is that it's very simple and cheap. Some designs use several of these mixers to get the needed frequency conversion.

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  • \$\begingroup\$ thanks for the insight; would the bias voltage have to be less than the forward voltage if the diode should be neither on/off? Would half of the forward voltage be sufficient? \$\endgroup\$ Mar 17, 2023 at 16:44
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    \$\begingroup\$ @YousifAlniemi well, the diode itself is going to keep the bias voltage in that range by resisting all attempts to set it outside of that range. I think you rather select the bias current. I am not sure how it should be set, though. \$\endgroup\$
    – user253751
    Mar 18, 2023 at 22:13

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