Impedance matching - Page 1
Forum section: electronics / projects

author: Digibin
author_rank: Regular Contributor
timestamp: August 22, 2015, 08:02:55 pm
content: I have a signal from an opamp (LMV722) and I want to use it to drive the input of theLT5560.The input signal is in the range 20kHz to 600kHz and is 3.3Vpp max.The input impedance of the mixer is 28R so to drive this with the op amp I need some impedance translation. So I'll use a transformer. If I'm using a transformer I may as well use it to convert to differential. The opamp max output current is 10mA so at full 3.3V voltage swing the minimum impedance should be 117R. So I need an impedance ratio of 4:1, or turns ratio of 2:1.Now if I drive the inputs directly with the transformer will there be any filtering effects going on (secondary inductance with the input resistance)? Because if so I need to select a suitable secondary inductance.Essentially my question is how should I interface the two parts together - is this the right approach?

author: nctnico
author_rank: Super Contributor
timestamp: August 22, 2015, 08:42:24 pm
content: At these frequencies I'd use a single ended to differential amplifier. A transformer is likely to get bulky (=large) at these frequencies.

author: Digibin
author_rank: Regular Contributor
timestamp: August 22, 2015, 08:43:48 pm
content: How would you go about the impedance translation?

author: nctnico
author_rank: Super Contributor
timestamp: August 22, 2015, 09:22:06 pm
content: No problem at the frequencies you want to use. Connect the outputs of the differential amplifier directly to the IN+/- inputs. At higher frequencies >100MHz you need to match the complex impedance.

author: KJDS
author_rank: Super Contributor
timestamp: August 22, 2015, 09:31:00 pm
content: When faced with a task such as this, instead of contemplating transformers, change the op-amp to one that can drive the load.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 22, 2015, 10:28:28 pm
content: For what purpose? The LT5560 looks way overkill for anything like that.And were you intending on driving the RF or LO port?Is NE602 out of production these days?Would you be better off with a couple transistors (single or double balanced mixer), or a multiplier IC, or diode gates, or (with transformer coupling usually) a diode double balanced mixer, or...?Tim

author: Digibin
author_rank: Regular Contributor
timestamp: August 25, 2015, 09:15:02 pm
content: Quote from: nctnico on August 22, 2015, 09:22:06 pmI'm only concerned with the real impedance, i.e. the signal input port has a low impedance which will overload the op-amp.Quote from: KJDS on August 22, 2015, 09:31:00 pmBut this would require around 40mA of current - this is a battery powered application so that's not ideal.Quote from: T3sl4co1l on August 22, 2015, 10:28:28 pmI'm talking about the RF port. I chose the LT5560 since it's a 3.3V part. I was using the NE602 but then moved to 3.3V. If you know of a part more suitable...?

author: rfeecs
author_rank: Frequent Contributor
timestamp: August 25, 2015, 10:34:21 pm
content: A few things to consider with this part:The IN+and IN-pins require a DC path to ground. So you will have to use inductors to ground or a transformer. It may be possible to use a differential output amplifier that allows the common mode output voltage to be set to ground, I suppose.How much input power do you need? This part has in input 1dB compression point of around 0dBm or lower. Typically you would operate much lower than that unless you want the mixer to saturate. 0dBm into 28 ohms is a 0.24V peak voltage swing, 0.48V peak to peak. So your approach of trying to use a full 3.3V peak to peak swing doesn't sound right here.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 25, 2015, 10:44:11 pm
content: Could you tell us about the system and application?

author: Digibin
author_rank: Regular Contributor
timestamp: August 28, 2015, 09:57:50 pm
content: Quote from: T3sl4co1l on August 25, 2015, 10:44:11 pmIt's a bat detector system. I made one using the NE602 with direct down conversion but would like to improve it with a superhet receiver. Mostly for the fun of itQuote from: T3sl4co1l on August 22, 2015, 10:28:28 pmYou're probably right. All I want to do is implement a superhet receiver, input signals of 15kHz to 200kHz, up-converted to an IF of 500kHz, bandpass filtered, and down-converted to audible frequencies 0-10kHz.What's the best approach for this? The input signals are coming from a tiny hearing aid microphone that is sensitive into ultrasound (probably drops off entirely around 100kHz), through a pre-amp. I'd like to use 3.3V system power rail but can go to 5V if I need to. Is there a 3.3V double balanced mixer IC out there, something like the NE602? I've looked around for one and the LT5560 was the best I could find.From what I can see analogue multipliers tend to need in the region of +-15V.Whatever it is it doesn't need to be high performance, it's not a demanding application.Perhaps I should just go back to 5V and use the NE602?

author: Zero999
author_rank: Super Contributor
timestamp: August 28, 2015, 10:45:37 pm
content: Do you want to build a bat detector?Have you tried just building an oscillator and using the signal to amplitude modulate it? You could then use a simple envelope detector to convert it down to the audible range.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2015, 11:19:58 pm
content: Heck, might as well do it with transistors...Averse to discrete? Just don't know how?It's quite simple! Here's an example from my Theremin (similar bandwidth I suppose, though at a somewhat higher center frequency):The two collectors at the top carry the output signal currents. You can attach a transformer here, or a resonant circuit (handy for more radio and IF), or ground* one side and use the other (which is as shown; this is the volume mixer, so a resonant tank detects the off-center difference, and when that difference is on top of the resonant frequency, a bias voltage is generated, which goes to the other mixer to control its volume level). In your case, a single resistor would be fine, giving audio output. (Or use two resistors, one for each collector; followed by a differential amplifier, which removes the DC offset and amplifies the desired difference signal. Which can be combined with the needed lowpass filter, so that's handy.)*For AC signals, supplies count as ground. The collector has to draw current from a supply at a higher voltage than the emitters are at, so it's understood that DC is "above", while AC is "ground". Hence, a supply rail.Of course, it's easier with more supply voltage, but signal levels can be ratcheted down a bit, and bias tightened up, without much expense to performance. This is certainly feasible at 3.3V.If you're just detecting bats, that should do fine. (Heck, maybe you even want that resonant amplitude detector part, too!)If you want to "listen", you might not want to do direct conversion. Reason being, if you're expecting any kind of pitch or harmonics*, half the output band will be reversed (the frequencies below the LO), and superimposed on the "correct" half (frequencies above LO). It might be valuable to up-convert fairly substantially, use an "SSB" style IF filter to eliminate the lower sideband (which will be at higher frequencies, this time), then convert this to baseband for listening (perhaps using a BFO and detector, which is traditional).*Note that you'll only see harmonics one at a time (more or less), which isn't very harmonic at all. This is fundamentally why slightly-out-of-tune SSB communications sound so goofy or garbled or alien: your ear expects evenly spaced harmonics, and when they don't line up, it becomes unintelligible. You won't even be hearing sounds as such, if you're listening to the gaps between harmonics!How can this be solved? (FYI, it can't through mixing methods. You need an altogether different kind of nonlinearity for that!) One way is to acquire a big wad of samples, and run some DSP on it to rescale the frequencies (a pitch bend effect). This takes a lot of processing, but is relatively simple when you have the libraries handy (it can possibly be done in real time on an average ARM core, otherwise a cheap DSP should do it no problem; a Fourier transform is needed, followed by resampling the frequency spectrum (a decimation or downsampling function), and inverse FT).It can also be done in analog, for more limited constraints: if the signal is periodic, the fundamental can be detected, and samples taken at a fractional rate, reconstructing the waveform through equivalent time sampling. It's played back in real time, a step at a time.Tim

author: rfeecs
author_rank: Frequent Contributor
timestamp: August 29, 2015, 12:06:10 am
content: A mixer is just a switch, with the LO signal chopping the RF signal. With two switches that are balanced, you can cancel out and suppress the RF or LO from leaking out the IF. With four switches (double balanced) you can suppress both RF and LO.As already mentioned, you can use diodes or transistors as switches, or you could even use analog switches at these frequencies. Using amplifiers with balanced outputs you can do single balanced or double balanced.For your application, you may not need a double balanced mixer. Your mixer could be as simple as a single diode.The Gilbert cell mixer uses transistors. You could make one out of an HFA3101 for example. The datasheet gives examples of using it for an upconverter and downconverter:http://www.intersil.com/content/dam/Intersil/documents/hfa3/hfa3101.pdf

author: tree
author_rank: Regular Contributor
timestamp: August 29, 2015, 12:25:56 am
content: Quote from: nctnico on August 22, 2015, 09:22:06 pmYou need to match the complex impedance to its complex conjugate impedance. This is key!(Ouptut impedance Z0 matched to Z0*)

author: Zero999
author_rank: Super Contributor
timestamp: August 30, 2015, 05:42:36 pm
content: What are your requirements?The simplest way to do this is to change the oscillator frequency, according to what you want to detect. So if you want to detect 15kHz to 200kHz then use an oscillator with that frequency range and mix it with the signal. The problem with that it will detect signals at either side of this frequency, so if you tune to 50kHz, and you get 5kHz, the signal could be 45kHz or 55kHz.An easy way round this is to use a fixed frequency oscillator, say 455kHz (many ceramic resonators are available for this frequency, as it's a common IF in AM radios) and mix the signal with that. You'll now get 255kHz to 440kz (lower side-band) and 470kHz to 655kHz (upper side-band). You can then use a tunable filter (say an LC circuit) to select the frequency of interest (use the upper side-band, not the lower) and detect the signal using envelope detection, like an old crystal radio.

author: Kleinstein
author_rank: Super Contributor
timestamp: August 30, 2015, 09:17:54 pm
content: At these low frequencies CMOS switches als make good mixers too. 74HC40.. can well work on 3 V.

author: Digibin
author_rank: Regular Contributor
timestamp: August 31, 2015, 12:46:15 am
content: Quote from: rfeecs on August 29, 2015, 12:06:10 amThanks for the link. I've looked at this part and at Gilbert Cells in more detail. I understand how they work but I'm not sure how to correctly bias it for my requirements.How do I determine the DC quiescent current I need through the transistor pairs? How do I work out the base bias voltages, the load resistance and the emitter resistors?I don't really need much gain and would prefer more dynamic range at the input since the signals will already be amplified through a pre-amp. I've read the biasing section in the datasheet, I just don't know how to actually determine what gain/bias voltages I need.

author: Zero999
author_rank: Super Contributor
timestamp: August 31, 2015, 12:34:31 pm
content: You could just make the mixer part of the oscillator.Note that whatever method you choose you'll also get harmonics of the input signal and their modulation products. If the input signal is 100kHz, you'll get 200kHz, 300kHz, 400kHz, 500kHz etc. as well as the upper and lower side bands.

author: rfeecs
author_rank: Frequent Contributor
timestamp: August 31, 2015, 05:23:20 pm
content: Quote from: graymp89 on August 31, 2015, 12:46:15 amFor the biasing, if you look at the data sheet, they bias it around 10 mA. You can look at figure 4. The voltage divider on the left sets the base voltages to 1V for the bottom transistors and 2.5V for the top transistors. The bias current is set by the base voltage of the bottom transistor, 1V. (1V - Vbe) is the voltage across the 27 ohm emitter resistor. So if you take Vbe as about 0.7V, then 0.3V/27ohms = 11mA.With the base of the top transistor set at 2.5V, this limits the output voltage swing. If the output swings down to about 2V, the output transistors will saturate. They are using an inductor on the output which sets the output DC voltage to 3V, so you have a 1V peak swing. You would need a huge inductor at your frequencies, so you might want to use a resistor load and set the voltage at 2.5V. Then you will only have a 0.5V maximum swing. That puts your load resistor at 0.5V / 10mA = 50 ohms.This gives you a voltage gain of about 4, or about 12dB. You have a 0.5V swing on the output, so you will have a maximum input voltage of about 0.125V peak.You really want to drive the mixer with a low input power level. The lower power you drive the mixer, the lower the distortion. You may not even need a front end amplifier. If you need more gain, put it after the mixer, not in front.

author: Digibin
author_rank: Regular Contributor
timestamp: September 01, 2015, 09:15:23 pm
content: Quote from: rfeecs on August 31, 2015, 05:23:20 pmThis was really helpful, thank you.I understand the biasing of the bottom two transistors to set the DC quiescent current through each top pair, and I understand how the bias of the top transistor pairs sets the available output voltage swing, as you've described. So really I could use the two bias points given in the example and just go with that, which I probably will. But I don't quite get one thing. Working bottom up, you set the bias of the bottom transistors to set the DC current as described. So we have 10mA total, with 5mA in each differential pair, and 2.5mA in each upper transistor.The bias of the upper pairs needs to be such that you have some voltage swing on the output, but also such that there is enough voltage to drop over Vbe and remain greater than Vb of the bottom transistors to prevent saturation. How do I determine the headroom I need at Vc of the bottom transistors to allow for Vce swings but maintaining Vc > Vb? Is it as simple as this: the max input voltage swing is 0.125V peak, so the min Vb of the upper pairs is 2.5 - 0.125 = 2.375V, which means the min Ve is 2.375 - 0.7 = 1.675V. Since this is also the collector voltage of the bottom transistors this is fine since Vc > Vb. However there is some available headroom here, so could I drop the bias of the top pairs, improving the gain a little?How did you arrive at a gain of 4? The gain of a common emitter amplifier (with emitter degeneration) is typically -Rc/Re, but I suspect this is doesn't hold for this configuration?

author: rfeecs
author_rank: Frequent Contributor
timestamp: September 02, 2015, 01:30:51 am
content: If you lower the upper transistors base voltage, you will have more output voltage swing capability. But as you say, you can't lower too much or you will put the lower transistors into saturation. The lower transistors need a Vce of at least about 0.5 V to avoid saturating, so the base voltage of the upper transistors needs to be at least 0.5V + Vbe, or about 1.2V. So yes, you can lower the upper transistor base voltage for more output swing head room. This shouldn't affect the gain.I got a gain of 4 by starting with the lower transistors are each biased at 5mA. So their transconductance (gm) is gm = Ic / vt. So gm = 5mA / 26mV = 192mS. Since the bottom transistors form a differential pair with only one input driven, the input voltage applied to one transistor is divided in 2. The current from one transistor ends up in the load (although it is chopped by the LO, it all ends up passing through to the load, alternating between two of the upper transistors). So the voltage gain of the differential pair is gm/2*RL = (192mS)/2*50 ohms = 4.8.I played around with this circuit in the circuit simulator and found that yes, you can lower the upper transistor base voltage to 1.5V and it doesn't affect the gain, and the voltage gain is about 3. The gain is lower than my crude calculation because the transistor model has emitter and base resistance that wasn't accounted for among other things.

author: Digibin
author_rank: Regular Contributor
timestamp: September 02, 2015, 09:08:41 pm
content: Quote from: rfeecs on August 31, 2015, 05:23:20 pmFor the single ended output (as in datasheet example) should this not be 5mA, so 0.5V / 5mA = 100R, since there is only 5mA through the load resistor?Quote from: rfeecs on September 02, 2015, 01:30:51 amI've also simulated the circuit. The gain of the RF signal is dependent on the input drive at the LO inputs, so where does this come into the calculations? What drive level are you using at the LO? Is the input dynamic range the same at both LO and RF inputs?

author: rfeecs
author_rank: Frequent Contributor
timestamp: September 02, 2015, 09:42:42 pm
content: Quote from: graymp89 on September 02, 2015, 09:08:41 pmYes, you are right. My mistake.Quote from: graymp89 on September 02, 2015, 09:08:41 pmI used 0dBm for the LO. So .33V peak. I didn't include that in my rough calculation. I assumed the top transistors are just switching on and off. Typically in a mixer you want the LO signal large so that the mixer is acting like a switch. Here we want the top transistors to switch on and off hard. This way the top transistors do not limit the current flow to the load. That gives you the highest conversion gain and the gain is not as sensitive to LO power level.

author: Digibin
author_rank: Regular Contributor
timestamp: September 03, 2015, 09:09:35 pm
content: Quote from: rfeecs on September 02, 2015, 09:42:42 pmI assume you took 0dBm based on experience. How can I verify this is sufficient to drive the transistors hard, i.e. if you hadn't told me to use 0dBm, how could I arrive at that figure (or close to) based on circuit analysis?Thanks for the help, I'll go ahead with this circuit for my application. I probably don't need to go into more detail but this stuff is good to know

author: nctnico
author_rank: Super Contributor
timestamp: September 03, 2015, 09:33:46 pm
content: Signal levels are usually a trade-off between the noise floor, distortion (maximum output swing and slew rate) and power dissipation.

author: Digibin
author_rank: Regular Contributor
timestamp: September 04, 2015, 05:47:49 pm
content: Quote from: nctnico on September 03, 2015, 09:33:46 pmYes, but in this case how do I know what is a 'hard' input level is, i.e. close to distortion?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 04, 2015, 09:51:42 pm
content: Look at the operating range of the device.For a common emitter BJT amplifier, this is the output current range, times whatever voltage range the load impedance generates. But in practice, it's a lot less than that, because of the exponential transconductance; you won't go linearly from full off to full on, with a properly grounded emitter. So there will be distortion produced, even before clipping.The power limit occurs when the device is going into cutoff (current goes to zero) or saturation (current reaches maximum, as determined by load impedance, supply voltage, and input drive).Usually, an emitter degeneration resistor is used to improve linearity, so that the power limits aren't really affected much (worsened if anything, because a fraction of output power is spent in said resistor), but distortion for comparable signal levels is improved. The effect being, if you look at distortion vs. input strength (or output power vs. input power), you get closer to a straight line, for a greater useful range (higher -1dB compression point, 3rd order intercept).The same or similar applies for all types of amplifiers, BJT, MOS, triode or pentode, or even the more unusual kind like negative resistance (though in that case, you don't really have a third terminal to apply feedback to).Tim

author: rfeecs
author_rank: Frequent Contributor
timestamp: September 04, 2015, 10:48:45 pm
content: Quote from: graymp89 on September 03, 2015, 09:09:35 pmTo roughly estimate the LO drive level, you want the upper transistors to switch the current back and forth between them. With no signals applied, the quiescent bias current is 5mA in the bottom transistors and 2.5mA in the top transistors. So the gm of the top transistors is about 2.5mA/26mV = 96mS. The voltage swing required to switch all the 2.5mA from one transistor to the other is roughly 2.5mA / 96mS = 26mV. Since you are applying the LO signal only on one side of the differential pair, it gets divided by two. So you need 52mV peak at least. You will need it to be a bit larger than that because the bottom transistors are modulating the current flowing into the upper transistors and you don't want that to overcome your LO drive. So 0dBm or 330mV should be more than enough to drive the upper transistors.I'm not sure this part is the best way to go for your application since it can work up to about a GHz, or 1000 times higher in frequency than you need. You may have to be careful with your layout to keep it from oscillating. If you really want to make a Gilbert cell you could also put one together with some cheaper transistor arrays, matched pairs or matched quads:http://www.digikey.com/product-search/en?FV=fff40015%2Cfff8007a%2C1c0002%2C1c0003%2C1c0006%2C1c0011%2C128004b%2C1280055%2C128009e%2C12800a2%2C12800a3%2C12800ba&mnonly=0&newproducts=0&ColumnSort=1000011&page=1&stock=1&pbfree=0&rohs=0&quantity=&ptm=0&fid=0&pageSize=25