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900Mhz Operation

This is a BJT mixer design I found on google

BJT Mixer Design

From my understanding in order to use a BJT amplifier as a mixer you have to be in the 1dB compression point.

Question: Am I wrong ?

If not...here is a BJT I found and here is a low compression point for it.

1dB compression point

Now on to the manufactured Mixers

Mouser Mixers

It looks like on mouser the mixers are design to provide best performance for a specific IF so you are limited by RF_In and FL_LO see Mouser Picture

Question: The BJT Mixer... will it mix anything ? I don't care what IF looks like I just want the product of two signals.

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3 Answers 3

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No, you do not have to use that circuit at the 1dB compression point (or any other compression point) for it to work as a mixer. If you were feeding both signals into the base (or emitter), then yes, you would -- but you're not.

Done properly, that circuit should multiply input at the base by a function of the signal injected at the emitter. In the extreme of "done right" you'd inject a square wave at the desired frequency into the emitter, and the result would be a chopped and amplified version of the input.

Note that I'm not advocating for this circuit, unless you're building something that absolutely needs to be extremely inexpensive, low power drain, or small. Note also that you'd be challenged to make this work at 900MHz with the given component values -- it really wants to be designed around lower impedances.

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  • \$\begingroup\$ wow okay that's interesting \$\endgroup\$
    – Jay
    May 12, 2019 at 18:53
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    \$\begingroup\$ It would be a good exercise (as @BrianDrummond mentions in his answer) to simulate the circuit or (if you feel up to it) analyze it. Note that it's a thoroughly unbalanced mixer, so you'll get mixing of many more harmonics of both input and LO than, say, a diode ring. \$\endgroup\$
    – TimWescott
    May 12, 2019 at 18:57
  • \$\begingroup\$ I'm actually doing that right now using ADS...but my square wave signal is not producing an output I'm going to start another question about it \$\endgroup\$
    – Jay
    May 12, 2019 at 18:59
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    \$\begingroup\$ The BC548B has an \$f_T\$ of 300MHz. The rule of thumb for transistor circuits is that you need an \$f_T\$ that's at least a few times higher than the working frequency of the circuit you're designing, and higher is better. It's also in a TO-92 package, which in and of itself is going to provide you with some challenge at 900MHz -- a transistor that's specifically designed for small-signal amplification at 900MHz (which will mean an \$f_T\$ significantly higher than 900MHz) will bring you far more joy. \$\endgroup\$
    – TimWescott
    May 12, 2019 at 20:56
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Yes, it will indeed mix anything. Unlike a proper mixer the output won't be the product of the two signals you want, but also anything else that came into the RF port. I exaggerate, but only slightly.

You'll get mixing but not just from your RF and LO frequencies. For example if RF = 900 MHz and LO = 1000 MHz you'll get the wanted IF output on 100 MHz.

Trouble is, RF signals on 800MHz will mix with your wanted 900 MHz signal; as will unwanted signals at 801 and 901 MHz, 802 and 902 MHz etc etc, and the sum of all these may very well drown out the signal you are looking for.

Proper mixers will do this too if you operate them at the compression point ... which is basically why you don't.

Now if you don't care about interfering signals, or know there aren't any (or they are much weaker than your wanted signal) go ahead with a simple non-linear mixer. But most people do care, which starts to explain the price of decent mixers.

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  • \$\begingroup\$ Thank youuuuuuuuuuuuuuuuuuuuuuu \$\endgroup\$
    – Jay
    May 12, 2019 at 18:09
  • \$\begingroup\$ I don't believe that you are analyzing that circuit correctly. Taking the emitter-side input as the LO, If the amplitude of that signal is set correctly, then the response to the RF input can be made as linear as you wish, while still having a strongly time-varying component w.r.t the LO input. It would be difficult to make it work nicely, to be sure -- but it's not going to be as bad as you say, by any means. \$\endgroup\$
    – TimWescott
    May 12, 2019 at 18:31
  • \$\begingroup\$ Agreed, to know exactly how bad, and whether that mattered in the asker's context, would take a lot more information, and simulation would be useful. \$\endgroup\$
    – user16324
    May 12, 2019 at 18:40
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The intercept points (IP2 and IP2 are some dB different, maybe 10dB different) for a single bipolar are about -10 or -15 dBv. At the -10dBv level, this requires 1v/sqrt(10) or 0.312 volts across the emitter/base junction.

Notice this is the Intercept Point.

More usefully, that same bipolar given 4 milliVolts across the EB junction will have about 10% distortion, so 4milliVolt total signal (RF + LO) is not a high-conversion-efficiency mixer.

At the 18 millivolt level, the change in current has increased from 10% (1 - 1.1) to e^(0.018/0.026) or a current change ratio of 2:1.

At the 26 millivolt level, the change in current has increased from 10% (1 - 1.1) to e^1, or a ratio of 2.718 and the conversion efficiency continues to improve.

Thus 4 millivolts total drive (RF + LO) is poor mixing efficiency, whereas 18mv or 26mv will produce much stronger 2nd and 3rd order conversion gain. The Taylor Series polynomial model of the diode/bipolar_emitter_base will show this steep improvement in mixer gain.

At the 58 millivolt level (remember this is the sum of RF + LO drives), you get 10:1 ratio.

At the 5 * 18 millivolt level, you get 2.0^5 or 2*2*2*2*2 = 32:1 ratio.

So even at the 4mV level you get some "intermodulation".

At the 32mv or 58mv or HIGHER level, you get strong mixing.

The NE602 IC is double-balanced mixer, working well past 100MHz, with just a few milliAmps power at 5 volts; Rin and Rout are in the KiloOhm range.

Some companies make ICs for GHz up/down conversion, needing dozens of milliAmps power but operating in 50 ohm environment.

If you design a radio-system-on-chip, you don't need 50 ohm environment to handle 0.1 millimeter distances, and you can dial down the power burned.

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