+++++++++++++++++++ +++++++++++++++++++ Date: Tue, 30 Apr 2002 22:37:32 GMT From: na5n at zianet.com To: n2cqr at clix.pt Cc: qrp-l at lehigh.edu Subject: [125830] Re: Biasing MOSFETS? Message-ID: <20020430223732.19795.qmail at zianet.com> Mime-version: 1.0 Content-type: text/plain; charset="us-ascii" Bill Meara writes: > I'm using a little amplifier from Ramsey. Two MOSFETS in push-pull. > Broadband linear RF Amp. I'm using it on 17 meters. I am not familiar with this specific amplifier, but the principles for using MOSFET's is the same. > Problem is that it is getting very hot. I've increased the size of the > heatsink, but it still is getting too hot. There are two types of MOSFET's: RF mosfet's and switching mosfets. The RF mosfets are $30-40 each, so seldom seen in amateur work. That leaves the cheaper switching mosfet's, such as the IRF510. These are designed to be a switch ... ON or OFF. In between these two states is a small linear region. This is where most QRP PA's are supposedly designed to operate. In the case of the IRF510, the mosfet "turns on" with a gate voltage around 3.5-4.0v, and goes fully on (called the "full ohmic on region") with around 7-8v on the gate. The exact ON and OFF voltages on the gate will vary tremendously from one IRF510 to the next. When you build an IRF510 PA, you should characterize that particular mosfet so you know exactly at what gate voltage drain current starts to flow, and if possible, at what gate voltage the drain current becomes maximum, or the point of the full-ohmic-on region. If you spend too much time in the full-ohmic-on region, you are drawing excessive current not being transformed to output power, which is dissipated across the source-drain junction ... that is, excessive HEATING of the IRF510. > I'm wondering if there would be any benefit to moving the biasing diode so > that it would be in physical contact with the heat sink. This worked > wonders in stabilizing a BJT amp, and in keeping it cooler -- I'm wondering > if this would also work with MOSFETS. Per the above, it is more important to characterize the biasing points of your IRF510's. You'll need to monitor BOTH the drain and gate voltages simultaneously with either a DVM and/or an oscope. With no gate voltage, the IRF510 will be OFF, such that you will have +12V on the drain (or whatever your raw Vcc is). Connect a 10K or so pot between ground and +12, such that the wiper then will go from 0V to +12v. Connect the wiper to the gate. Initially set it to 0v and record the drain voltage. Say it is +12v. Now increase the voltage on the gate until the drain voltage begins to drop by the smallest amount you can detect. Usually 100-200mV should be obvious on most DVM's or scopes. Say this happens with the gate voltage at 3.8v. This is the gate turn-ON voltage: 3.8v Now increase the gate voltage and watch the drain voltage go down ... 10v, 8v, 6v, etc. Be quick at this. You may need to turn the gate voltage to 0v if the IRF510 starts getting too hot doing this before doing it again. What you are looking for is increasing the gate voltage and watching for the point where the drain voltage no longer gets any lower. Say this happens with a gate voltage of 7.8v and a drain voltage of 1v. If you increase the gate voltage to 8v, the drain voltage stays at 1v. You are defining the dynamic range of the LINEAR region. In this case, the linear region is a gate voltage from 3.8v to 7.8v ... producing a drain voltage that varies from +12v down to +1v. Another way to look at it is the mosfet turns ON with a gate voltage of 3.8v (which means it is OFF below 3.8v), and the "switch" (the IRF510) is fully ON with a gate voltage of 7.8v. This is the gate voltage range you MUST operate within! But remember, with the gate voltage at 7.8v, producing a drain voltage of 1v, you are now drawing FULL CURRENT, which for the IRF510 is around 5A! (Assuming your power supply can provide that). For accuracy, there is one additional test you need to perform. For a typical 5W QRP PA running class C (50% efficiency), a drain current of about 1A is usually required. In this case, 12V x 1A = 12W input power, yielding an efficiency of 5W/12W = 42% (fairly typical inspite of the amazing efficiency claims made by some). How much current does your rig draw to make 5W? Figure out the efficiency yourself using the above. Subtract the receive current from the transmit current to be a purist. But this 1A is rms, meaning the IFR510 will be drawing about 1.5A on the peaks (1A x 1.414). So you need to find out at what gate voltage produces a drain current flow of 1.5A. To do this, you must monitor the drain current flow (either in the drain lead, or between the power supply and the rig, providing you subtract out the receive or standby current on key up). Let's say you parallel some resistors to achieve 1-ohm in the drain lead. This means when you drop 1.5V across this 1-ohm resistor, your drain current is about 1.5A. Now increase your pot/gate voltage again until you see your 1.5V drop across the 1-ohm resistor, identifying 1.5A of drain current. Say this occurs at 7.4v on the gate. This identifies the true dynamic range you need to operate your IRF510 inside of ... a gate voltage from 3.8V (turn ON) to 7.4v for 1.5A peak of drain current. HERE'S THE BIGGEST MISTAKE MOST IRF510 CIRCUITS MAKE: ===================================================== You go keydown for transmit, and adjust the bias for 5W output. When you go keyup, the IRF510 should TURN OFF. But if the bias is set too high, above 3.8v gate turn on in this example, your IRF510 will be drawing current, perhaps several hundred mA of current if you don't check. This, of course, is wasted power and needlessly heating up your IRF510. IRF510's are SWITCHING MOSFET'S ... there is absolutely NO NEED to set this so-called "idling current" in a mosfet. On keyup, there should be NONE, ZERO, ZIP drain current flowing in the IRF510. If you set your bias for 5W out, and you still have drain current flowing on key-up, your problem is simple: INSUFFICIENT GATE DRIVE -- not biasing. You do not have enough peak-to-peak RF drive going to the gate. This is the biggest problem I've seen with most published IRF510 circuits ... insufficient gate drive! Back to our example, the gate turns ON at 3.8v and the drain draws 1.5A at 7.4V. This is 3.6v difference. Your RF drive needs to be AT LEAST twice this! (or 7.2Vpp). MY RECOMMENDATION with the above example, is to bias the gate around 0.5v below the gate turn on voltage ... or 3.8 - 0.5v = 3.3V. With no input signal, this will ensure the IRF510 is completely turned off (with the gate turn on voltage of 3.8v). The input RF drive signal should be capacitively coupled, so that with no RF drive, the gate voltage is this 3.3v, and the IRF510 is turned OFF. For proper class C biasing, your input drive should be twice your gate input biasing range, that is, twice the 3.8v (turn ON) to 7.4v (Id=1.5A), or about 7.2Vpp RF drive at the driver output, feeding the coupling capacitor between the drive and gate input circuit. The first 3.8v will do nothing (except help charge up the gate-source capacitance ... about 150pF). Then when it exceeds the 3.8v bias, the IRF510 will begin to conduct. When the input drive reaches the maximum of 7.2v, this will be the MOMENTARY point of maximum 1.5A drain current, corresponding to about 1A rms drain current. So to properly bias an IRF510, you need to know at what gate voltage drain current BEGINS to flow, at what gate voltage produces about 1.5A of drain current, and at what gate voltage causes FULL current flow. In most IRF510's, the difference in gate voltage between 1.5A and full drain current is 0.5V or even less. Fairly critical. The last problem, of course, is if you have too much gate drive, say 8-9V peak. This is because (in our example), once the gate voltage exceeds the 7.4v (1.5A drain current point), much above that, the IRF510 will be drawing it's full 5A of current over some portion of the input sinewave, usually 10-25% of the cycle. This excess power does two bad things: 1) It saturates the toroidal core being used in the drain circuit for the RFC or transformer primarily (usually a 1:4 bifilar wound transformer). This inductance can not get rid of all of its stored up current into the output filter before the next drain current cycle begins if it goes into core saturation. A T50-43 core will saturate around 2A peak current. 2) The excess current has no place to go except to dissipate in the form of heat across the drain-source junction. This will make the IRF510 VERY, VERY HOT in short order. The kind of heat that removes your finger print! In this full-drain current (full-ohmic on) condition, another bad thing needs to be pointed out: We know that the approximate output resistance of any PA transistor or mosfet is the famous RL=Vcc^2/2Po, which equates to about 14 ohms at 12V and 5W output. When you enter the full ohmic-on region, the output resistance then becomes the Rds of the IRF510, which is about 0.4ohms. Drop nearly 12V at 5A across a 0.4 ohm resistor, and buddy -- you're making some major heat. (P=EI = 12V x 5A = 60Watts!). The maximum power dissipation of an IRF510 is about 40W. Ooops ... something has got to give. Poof - there goes your IRF510 after about 5 seconds of keydown. So, for each IRF510 you build a PA out of, characterize the gate voltage required for gate turn on, 1.5A of drain current, and where it goes into it's full ohmic-on region (full drain current flow). This defines your effective dynamic range of RF drive, and the safe operating area you should remain within. If you built an IRF510 from somebody's article that says to set 50mA or so of idle current on key-up, immediately drop the gate bias a bit until NO DRAIN CURRENT flows. And if you can't get 5W out at this point, then increase your RF drive. Or, if it's getting too hot, then check that you don't have too much drive, drawing more than 1A rms, or that the gate is not exceeding 8V (or whatever you measured) on the peaks. MOSFET PA's make neat amplifiers, but due to the large variations from one device to the next, they are not as reliable to build as a BJT PA using a 2SC2078 or something. This is why each IRF510 should be characterized before building the PA to establish the proper bias point. > What do you say? Would this be a cool move or what? By setting the PROPER BIAS POINT AND RF DRIVE, you shouldn't have to piddle around with diode temperature compensation or bias feedback networks to make the IRF510 work properly. It's OK for the IRF510 to get fairly hot after 5-10 seconds of keydown with a class C 5W PA, but if you're leaving finger prints ... then it's way too hot and it is not properly biased or driven. 72, Paul NA5N +++++++++++++++++++ Date: Wed, 01 May 2002 23:42:46 GMT From: na5n at zianet.com To: qrp-l at lehigh.edu Subject: [125876] Biasing MOS-FET Karl F. Larsen writes: > For CW where the signal is on or off, you should bias the devices > for no drain current in the off condition, TRUE > and drive the devices hard enough to saturate drain current when on. > This will result in the highest efficeincy and of course, least power to > heating. FALSE. This is one of the major points of yesterday's post. You want to AVOID driving a mosfet so hard that the drain current saturates, because in the case of the IRF510, you will suddenly be drawing 5-6 amps and your output load resistance drops to 0.4 ohms - making an impedance match to your 50-ohm filter network impossible. The 5-6A of drain current, even if only 10-20% of the RF cycle, raises your average drain current over rms by that same percentage, and lowers your efficiency by the same percentage. And with 5-6A flowing through the IRF510, even 10-20% of the time, is still a lot of power being converted to nothing but heat inside the IRF510. You want to avoid saturating the drain current at all costs. Except, of course, in the case of Class E, which exploits operating the IRF510 in the saturated drain current region. But here, you are not driving the mosfet with a sinewave, but rather fairly narrow "pulses" to convert the 5-6A of drain current to the desired 1A average current. Actually, it ends up being less than that for efficiencies in the 80-90% range. But that is also due to doing business a bit different in the output filter and how one impedance matches the IRF510 to the 50-ohm filter. But in any case, you never drive a mosfet in class D, E or F with a sinewave. > need a good filter on the output. If your using push-pull your filter job > is easier because even harmonics are very low. TRUE. Filtering does need to be more precisely matched to a mosfet PA depending upon the exact class of operation it is in. > I would be careful not to overdrive the amp and flat top. Well, this is exactly my point, and seemingly opposite of what is said above about "driving the amp into the saturated drain current region." There is no way to avoid flat topping when you are in the saturation region. This is not in any way an attack on Karl, but to emphasize and explain why a mosfet class c amplifier, being fed with a sine wave, should never be allowed to drive the mosfet into saturated drain current. This is a common mistake made by those trying to build mosfet PA's, and a major reason why the mosfet's go poof, or get extremely hot, or sound chirpy, because when you enter the saturated region, it's trying to draw almost everything your power supply has to offer. Even batteries can groan when the load changes from 1A to 5A almost instantaneously. 72, Paul NA5N +++++++++++++++++++ Date: Wed, 1 May 2002 00:55:29 -0700 From: "Bob Okas" To: , , Subject: [125852] Re: Biasing MOSFETS? Hi Paul, Kudos for such a lucid description of Class C mosfet biasing. I've squirreled it away in my na5n folder. However, I do wish to point out that the original poster was describing a mosfet push pull *linear* amp. Remembering Bill Meara's proclivity for SSB operation on 17m, I suspect he's looking to boost his phone signal. If, say, the Ramsey amp was designed for operation in the Class AB-ish region, then some current would be drawn by the finals during those moments of non-speech keydown. I've read W1HUE's article, , which dispells the notion I've held that Mosfets have an absolute negative temperature coefficient that reduces the device gain as the junction temperature increases. So, thermally attaching the biasing diode to the heatsink should help, provided that the bias is not initially too high. I wonder if Bill could tell us what the gate bias voltage is and the part number of those devices? Is there a bias adjustment? 73, Bob - W3CD +++++++++++++++++++ Date: Wed, 01 May 2002 14:25:04 -0400 From: "Jim Kortge, K8IQY" To: WD8CIV at worldnet.att.net Cc: qrp-l at lehigh.edu Subject: [125861] Re: Biasing Mosfets At 01:38 PM 5/1/02 -0400, you wrote: >Folks, > >Seeing the discussion on MOSFETs has got me thinking (always a dangerous >activity) - what is the practical upper frequency limit of an IRF510 (or >two) used as a final? > >Dave Dave, Using properly designed feedback networks, they will easily go up to 17 meters. I use a pair of MTP3055E MOSFETs in a small linear amplifier that I designed and built for 40 and 17. I get about 12 watts out on 40 and about 10 on 17. The feedback circuits were optimized for 17 meter operation. I believe the QRP Plus rig used an IRF510 in the final up to 6 or 10 meters, but I suspect the power output suffers terribly up that high. 72, Jim, K8IQY +++++++++++++++++++ Date: Wed, 01 May 2002 19:11:49 -0100 From: Bill Meara To: vintage2 at earthlink.net Cc: qrp-l at lehigh.edu Subject: [125862] Re: Biasing MOSFETS? At 12:55 AM 5/1/02 -0700, you wrote: >Hi Paul, > > Kudos for such a lucid description of Class C mosfet biasing. I've >squirreled it away in my na5n folder. However, I do wish to point out that >the original poster was describing a mosfet push pull *linear* amp. >Remembering Bill Meara's proclivity for SSB operation on 17m, I suspect he's >looking to boost his phone signal. If, say, the Ramsey amp was designed for >operation in the Class AB-ish region, then some current would be drawn by >the finals during those moments of non-speech keydown. > > I've read W1HUE's article, , >which dispells the notion I've held that Mosfets have an absolute negative >temperature coefficient that reduces the device gain as the junction >temperature increases. So, thermally attaching the biasing diode to the >heatsink should help, provided that the bias is not initially too high. I >wonder if Bill could tell us what the gate bias voltage is and the part >number of those devices? Is there a bias adjustment? Bob: The devices in the Ramsey Amp are MTP3055E. I'm guessing that they are switiching MOSFETS similar to those described in Paul's excellent post. Yea, biasing is more complicated once you leave Class C. The Ramsey manual advises you to monitor the current and adjust bias (no signal in) for 250 milliamps. They have a test point in the bias circuit, and with 250 ma, they say that point should read 3.2 to 3.5 volts. I found that the voltage reading I got was a bit higher -- about 4.4 volts. This confirms what Paul said about variations in bias points. I notice that as the amp gets warm, the idle current starts going up. Mine goes to about 340 milliamps after a standard length CQ. I now have a larger heat sink and a big old fan (tube type!) on the amp, so it is under control, but I'd be happier if the idle current stayed put. I'm wondering how low I can go with the idle current before I start getting non-linear. Lower idle current would obviously keep things cooler. I also saw the temp stabilization article you cited. Looked kind of complicated to me. I wonder if just putting the diode on the heat sink would work just as well. I may give this a try. The amp is working well. Takes me from about 1 watt PEP to about 15-18 watts PEP. "Stand by while I kick in the lin-yer OM!" Lots of fun. 73 Bill CU2JL N2CQR +++++++++++++++++++ Date: Wed, 1 May 2002 14:57:46 -0600 (MDT) From: "Karl F. Larsen" To: qrp-l at lehigh.edu Subject: [125867] Biasing MOS-FET To my way of thinking the bias point is defined by whether the amplifier is linear or not. By linear I mean if the input is 10% 50% or 100% of maximum output the output follows that with good accuracy. For CW where the signal is on or off, you should bias the devices for no drain current in the off condition, and drive the devices hard enough to saturate drain current when on. This will result in the highest efficeincy and of course, least power to heating. Your for sure going to need a good filter on the output. If your using push-pull your filter job is easier because even harmonics are very low. Now you start to bias the devices into conduction when you need a linear amp. This increases loss to heat and the devices get hot. 500 Ma at 12 volts is 6 watts of constant heat dissipation in the device. If you think you need 2 amps of drain bias then the linearity is great but the amp gets real hot. If I were doing this I would try Class B operation. Here there is just a tiny drain current when no signal is present. I would be careful not to overdrive the amp and flat top. This is very possible to set up. -- Yours Truly, - Karl F. Larsen, k5di at arrl.net (505) 524-3303 - http://www.zianet.com/k5di/ +++++++++++++++++++ Date: Wed, 1 May 2002 19:15:35 -0500 From: Nick Kennedy To: "'na5n at zianet.com'" , Low Power Amateur Radio Discussion Subject: [125881] RE: Biasing MOS-FET -----Original Message----- From: na5n at zianet.com [SMTP:na5n at zianet.com] Sent: Wednesday, May 01, 2002 6:43 PM To: Low Power Amateur Radio Discussion Subject: Biasing MOS-FET "You want to AVOID driving a mosfet so hard that the drain current saturates, because in the case of the IRF510, you will suddenly be drawing 5-6 amps and your output load resistance drops to 0.4 ohms - making an impedance match to your 50-ohm filter network impossible." Hmmm ... I always thought you don't impedance match the output device to the load in Class C. And that to do so would reduce the efficiency to 50%. Or at least, restrict it to no greater than that value. And if you don't drive the Class C amp such as its conduction is approximately a square wave, aren't you dwelling in the ohmic region you said you wanted to avoid? 72--Nick, WA5BDU ... confused in Arkansas +++++++++++++++++++ Date: Thu, 2 May 2002 05:45:03 -0600 (MDT) From: "Karl F. Larsen" To: na5n at zianet.com Cc: Low Power Amateur Radio Discussion Subject: [125895] Re: Biasing MOS-FET Let me address just one issue Paul. On Wed, 1 May 2002 na5n at zianet.com wrote: > > Karl F. Larsen writes: > > > For CW where the signal is on or off, you should bias the devices > > for no drain current in the off condition, > > TRUE > > > and drive the devices hard enough to saturate drain current when on. > > This will result in the highest efficeincy and of course, least power to > heating. > > FALSE. This is one of the major points of yesterday's post. You want to > AVOID driving a mosfet so hard that the drain current saturates, because > in the case of the IRF510, you will suddenly be drawing 5-6 amps and your > output load resistance drops to 0.4 ohms - making an impedance match to > your 50-ohm filter network impossible. The 5-6A of drain current, even > if only 10-20% of the RF cycle, raises your average drain current over > rms by that same percentage, and lowers your efficiency by the same > percentage. And with 5-6A flowing through the IRF510, even 10-20% of the > time, is still a lot of power being converted to nothing but heat inside > the IRF510. You want to avoid saturating the drain current at all costs. > I agree that if you saturate a fet the current is a maximum amount the device can draw but this is a DC analysis of the output impedance. Looking at the RF impedance it's actually defined from the median impedance between cutoff which is very high and effective zero in saturation. Actually you need a set of load lines like we had in the old tube days. Might be fun to make them and see how they look. If the IRF510 has a gate type responce to it's input then it won't work very well in RF service. I'm pretty sure that's what the device is used for in the mass market. Until we can get some AC analysis of this device I'm not ready to say that you can't match the IRF510 output because I'm doing that with my SMK-1 5 watt amp which uses this device... and it works. Doesn't get hot at all. -- Yours Truly, - Karl F. Larsen, k5di at arrl.net (505) 524-3303 - +++++++++++++++++++ Date: Thu, 2 May 2002 12:53:11 -0700 From: "Glen Leinweber" To: "David Hinerman" Cc: "qrp-l" Subject: [125904] HF MOSfet amps - capacitance Message-ID: <000801c1f212$fd54a2c0$07ea7182 at mcmaster.ca> MIME-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit Dave Hineman mentions that not all MOSfets are equal - for example, the IRF840 has an input capacitance at the gate of 1300 pf, while the IRF510 that we commonly see has 135 pf. Don't take these numbers as absolute. Capacitance changes with bias, and doesn't include Miller capacitance from gate to drain. Why does capacitance change from FET to FET? These devices have many hundreds or thousands of FETs all running in parallel, right on that single chip. The heftier FETs have more in parallel than the less-powerful ones. You can usually tell by the full-ON channel resistance....when this number is a small fraction of an ohm, its a BIG device, with a lotta fets in parallel. The IRF510 has a channel resistance (fully on) of about half an ohm. Not so many fets. More parallel fets means more capacitance - capacitances in parallel add. Its that simple. We should use MOSfets in HF amps with low capacitance - they're easier to drive. The IRF510 on-resistance of half an ohm is plenty low enough for QRP with 12v supply. For a really powerful amp, you'd have to use a higher supply voltage, or a heftier MOSfet (and be prepared to drive its larger input capacitance with a more powerful driver stage). Logic-level MOSfets can help too - they turn on with lower bias voltages, and are FULLY on when the gate is only +5v (instead of +10v for non-logic-level MOSfets). If you only have to charge gate capacitance to 5v instead of 10v, less drive power is required. So a IRFL510 is better than standard IRF510. +++++++++++++++++++ Date: Thu, 2 May 2002 19:09:02 -0700 From: "Glen Leinweber" To: Cc: "qrp-l" Subject: [125915] Re: Biasing MOSFETS? Paul suggest some great ways to characterize a MOSfet's biasing characteristics, when used as a RF power amp. Paul also suggests some bad things that can happen when over-driving a MOSfet...one of which is saturation of the RFC choke feeding in DC power. Yes, that's bad, wasting power in heat (in the choke). But you won't waste much power heating the FET. When you overdrive a MOSfet (gate voltage swinging up over approx +7v), it does turn on hard, and looks like a small, fractional-ohm resistor. But this only happens for a fraction of a microsecond. During that time, the MOSfet dissipates very little power, since its drain voltage is very nearly zero - same as the source. No voltage across the FET?...no power dissipated - even though there may be a lotta current running thru. Yes, if you LEAVE the gate at +8v, current builds after a few microseconds and something soon gives up the smoke - maybe the FET, maybe the choke. But that's not how it runs in an amp - the FET only asks for current in short, sub-microsecond pulses. So Paul's smoky scenario when overdriving is not likely, if you stay within reasonable limits.... As long as the choke can stand the current, overdriving a MOSfet likely improves efficiency, making the FET run cooler. There'll be a bit more current in the choke, and a bit more current in the FET, but that's turned over to the filter between FET and antenna to be turned into useful RF power. The real neat thing about a switching type PA, is that when the fet is hard ON, with its drain clamped firmly down at zero volts, its not dissipating much power. Yes, current is building in the choke, and you can't stay here too long. We'll only be here for half a cycle. Then the FET goes OFF - bang - current drops to zero and the drain voltage starts swinging up. But again, the FET dissipates no power, because there's no current thru it. Beauty! When its ON - little power dissipated - when its OFF - no power dissipated. Things aren't THAT perfect - the FET takes a bit of time to switch ON & OFF. During that switching time it dissipates power. And some of the drive power coming in the gate goes into heat too. And when its ON, that small ohmic resistance dissipates a bit of power (an amp thru half ohm is half watt, and for only half the time - a total of quarter watt). Overdriving has limits: your gate voltage swing must stay below a swing from +20 to -20 v, or you'll puncture the gate oxide (blown fet). And overdriving too far wastes drive power too. +++++++++++++++++++ Date: Fri, 03 May 2002 15:15:14 GMT From: na5n at zianet.com To: qrp-l at lehigh.edu Subject: [125927] More MOSFET stuff Gang, Glen's post is excellent, and it addresses some of the discussion I have been having with several of you privately, worth repeating here. This discussion began about questioning why a certain MOSFET PA was running so darn hot. This is usually an indication of poor biasing or overdriving the gate ... that is, for some percentage of the input RF, you are drawing full drain current. If this is a relatively small part of the input sinewave, it's usually not a problem. Helping people with their mosfet PA's over the years, particularly those who built the 38-Special add-on mosfet PA (which was initially poorly biased), the main problem I see is people increasing the gate bias to achieve maximum output power. Problem with this approach is when the input drive goes to zero, the gate may be sitting with 4v or so bias, sufficient to keep the mosfet drawing significant current. With full input drive, it may drive the gate into the 7-8v region allowing full drain current. With a SINEWAVE, if this full-drain current is occuring for too long each cycle, you can dissipate a lot of power within the mosfet, which if severe enough, will of course cause damage to the device. Therefore in my original post, I was trying to describe how to determine the useful dynamic range of a mosfet gate drive, since every mosfet tends to have it's own parameters. A second point: while I mentioned that when entering the full drain current region, the drain-source resistance drops to about 0.4 ohms for the IRF510, this occurs ONLY for the period of time of full drain current, not all the time. The overall output resistance of the mosfet will be the AVERAGE of the resistances seen over each cycle, and thus will be a bit lower than the RL=Vcc^2/2Po value, not 0.4 ohms. Of course the longer you stay in the full drain current region, the lower your AVERAGE output resistance will be. Likewise, when in the full drain current region, drawing around 5A, the AVERAGE drain current for each cycle will be closer to the 1A (at 5W) level. Again, the longer you stay in the full drain current region, the higher the AVERAGE drain current will be, but will never be excessively above 1A average. The problem, however, is the momentary "spikes" of 5A drain current can put a large strain on the RFC feeding dc to the drain, possibly causing core saturation, and another set of problems for the PA and output filtering. Keep in mind, the above is based on feeding the mosfet with a SINEWAVE input. The more efficient mosfet PA's, such as those discussed by Glen, are based on driving the mosfet with SQUAREWAVES. This helps assure that when the gate signal goes positive, driving the device into full drain current, the drain voltage is very close to 0V for very little heat dissipation in the device. This is difficult to achieve with a sinewave input, as there is a delay between the gate drive signal and the drain signal due to charging up the relatively large (150pF) gate input capacitance. With a square wave, you are charging up this gate capacitance very quickly, while with a sinewave input, you are charging it more slowly, causing a longer delay, such that it is difficult to get maximum drain current to coincide with zero drain voltage for a very long. With a squarewave drive, max. drain current at 0V drain voltage can exist for 20-30% of the cycle, in effect, expending virtually no power. With a sinewave drive, you vary the mosfet output power by controlling the peak-peak voltage of the gate drive. The more drive, the more output power (until you saturate in the full drain current region). With a squarewave drive, you control output power by controlling the duty cycle of the squarewave. That is, if the drive signal is "hi" for 10% of the cycle, you are drawing full drain current for only 10% of the time and a relatively low average output power results. Increasing the input duty cycle to 30%, means you are drawing full drain current for 30% of the time, for significantly more output power. This is called PWM, or "pulse width modulation" technique. In this case, we are not changing the pulse width fast for the purpose of modulating the carrier, but changing the pulse width on very slow time scales to change the output power. On a typical class E PA I have built using an IRF510, output power varies from about 3W to 9W by varying the input duty cycle from about 10-40%. Less than 10%, the mosfet gets unstable with high phase noise -- above 40%, drain current starts flowing so much of the time, you can stray away from having full drain current at 0V drain voltage, and thus the efficiency starts to suffer. In the 10-40% drive, efficiencies in the 80% range are seen. This means the IRF510 is comfortably warm even with several minutes of solid keydown. It also means it's drawing far less battery current to make 5W than a PA running 50%, obviously. I'm glad many of you are finding this discussion useful and informative. Feel free to share with us your experiences. Mosfet's are fun devices, and capable of making a nice QRP PA in a properly designed circuit. And many of these points discussed by myself and others is why some people claim they are making 5W and the IRF510 is scarcely warm, while others are cooking eggs at 3W. 72, Paul NA5N ++++++++++++++++ Date: Mon, 6 May 2002 12:48:52 -0500 From: "Walter AG5P" To: "Low Power Amateur Radio Discussion" Subject: [126222] FETs and building circuits Message-ID: <015f01c1f526$ac599940$a3476ad8 at default> MIME-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit Hi Everyone, the recent FET thread has been great for getting me into more building. Doing some searching to find out different ways and means to use FETs in the QRP projects has lead me to the following website. All I can say is WoW, these folks are doing some serious transmitting with FETs. Now, before you yell at me, just remember that they are using modules composed of 4 FET units and then combining them into serious high power Ham radio. Just using one FET in the QRP projects with the Fairchild FQA11N90 which in small quantities cost around $2.50 is pretty QRP to my shack budget. An interesting comment from the author, "Right up front, I need to say Don't build one of these if you don't have a good oscilloscope! You will absolutely need a dual-trace scope to really ensure that your transmitter is working correctly." That just amplifies what Paul Harden, NA5N, has been telling us, is to get the circuit components properly selected. Take a look at http://www.netway.com/~stevec/ham/classe.htm#200w there are also some great ideas at monitoring your FET circuits and keeping them happy. 72 / 73 Walter, AG5P +++++++++++++++