kline@ux1.cso.uiuc.edu (Charley Kline) (06/19/91)
Hi, we're trying to build a power supply for a device that requires 3A at 13.8 volts. The voltage regulator will be an LM350T, a 3A-rated variable voltage regulator. The problem is the power transformer. I can't find anywhere a 3A transformer at 18 volts. My question is, am I guessing the secondary voltage wrong? Would a 14V secondary be okay since that is the RMS value and the rectified/filtered DC out of that would be sqrt(2) * 14 = 20V? ________________________________________________________________________ Charley Kline, KB9FFK, PP-ASEL c-kline@uiuc.edu University of Illinois Computing Services Packet: kb9ffk@ka9mnx 1304 W. Springfield Ave, Urbana IL 61801 (217) 333-3339
mcovingt@athena.cs.uga.edu (Michael A. Covington) (06/19/91)
See THE ART OF ELECTRONICS or the ARRL Handbook. Let me know what circuit you are using and I'll be glad to help you work something out. Remember that the transformer must deliver full voltage _under_load_ and that overvoltage is OK; in fact that's what the regulator is supposed to take care of. I wouldn't use a 3-amp regulator on a load that is drawing 3 amps continuously. It pays to be conservative. -- ------------------------------------------------------- Michael A. Covington | Artificial Intelligence Programs The University of Georgia | Athens, GA 30602 U.S.A. -------------------------------------------------------
luns@spocom.guild.org (Luns Tee) (06/19/91)
kline@ux1.cso.uiuc.edu (Charley Kline) writes: > Hi, we're trying to build a power supply for a device that requires 3A > at 13.8 volts. The voltage regulator will be an LM350T, a 3A-rated > variable voltage regulator. Let me guess, you're trying to power a 50W car stereo right? I've got one in my living room, and I have to get around to building a beefier power supply (the one I'm using was meant for the 5W car stereo that was there before). The only place I know of locally that stocks a good variety of transformers is a surplus store which I haven't visited in a while. Check the surplus stores around you. I got a 3A 12V transformer for $5.00.
myers@hpfcrlm.HP.COM (Bob Myers) (06/19/91)
>My question is, am I guessing the secondary voltage wrong? Would a 14V >secondary be okay since that is the RMS value and the >rectified/filtered DC out of that would be sqrt(2) * 14 = 20V? No, you're probably pretty close. Remember that that "20V" output could only be had from such an arrangement if there is no load - that's the peak voltage of the transformer secondary, and the filter can't maintain that if you're pulling charge out of the caps and sending it to the load. An 18V secondary may be hard to find, but I would guess that there are 24V transformers all over the place. That might be a bit much, but if you don't mind wasting some extra power in regulator you'll be OK. And, as Micheal C. already said, be conservative. If you're really expecting a 3A load, for example, look for a transformer, regulator, etc., rated at, say, 5A. 5A fixed regulators are out there, but may be harder to find. If so, use a smaller one and a pass transistor. Also, don't forget to protect the regulator by putting a hefty diode "backwards" from the regulator's output to the input. (That way, when the input is shut off, you won't wind up with 13V or whatever the "wrong way" across the regulator, as you wait for the output filter caps to discharge.) Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not Ft. Collins, Colorado | those of my employer or any other myers@fc.hp.com | sentient life-form on this planet.
grege@gold.gvg.tek.com (Greg Ebert) (06/20/91)
In article <1991Jun18.175131.8374@ux1.cso.uiuc.edu> kline@ux1.cso.uiuc.edu (Charley Kline) writes: >Hi, we're trying to build a power supply for a device that requires 3A >at 13.8 volts. The voltage regulator will be an LM350T, a 3A-rated >variable voltage regulator. > >The problem is the power transformer. I can't find anywhere a 3A >transformer at 18 volts. > Radio Snack used to (and maybe still does) sell an 18V center-tapped xformer rated at 4 amps. >My question is, am I guessing the secondary voltage wrong? Would a 14V >secondary be okay since that is the RMS value and the >rectified/filtered DC out of that would be sqrt(2) * 14 = 20V? Pretty close. You need to include diode voltage drops (they are about 0.8 volts for currents around 1-3 amps. A full-wave bridge rectifier will bite you twice and gobble up 1.6 volts. You will also experience voltage 'sag' between AC peaks as your filter capacitor drains. If you assume it's linear (a very good approximation), I*t=C*V. For a 3 amp load, 60Hz in, you get a 1volt dip if your filter capacitor is 25,000 uF. Then the voltage regulator will take another 2.5 volts. Let's guess that voltage drops in the Xformer are 0.5 volts. So, 13.8 (load) + 2.5 (LM350) + 1.6 (Rectifier) + 0.5 (Xfmr) = 18.4 volts. That leaves you with 1.6 volts of ripple coming from the filter cap. You will need at least 15,000 uF of filtering to prevent 'dips' below 13.8 volts out. If you use the 18 volt xformer, be aware that your regulator will be taking a larger voltage drop, hence it will need more heat-sinking.
kline@ux1.cso.uiuc.edu (Charley Kline) (06/21/91)
Thanks for all the replies and email. For what it's worth, one response shamed me into actually going and researching this myself, and what I was missing in my original analysis were the 1.4V drop across the full-wave rectifier and the 10% slop required for low AC line voltage. So I thought I would share my results with the net in the hopes that it would be constructive. My analysis goes something like: We want 13.8 volts out at 3 amps. The regulator has a 2-3 volt dropout, so we're conservatively up to 16.8. Let's set the ripple on the filter capacitor to 2 volts at maximum load (more on this later), so now we're up to 18.8V. The full wave rectifier will create a 1.4V drop, giving us 20.2V. That's peak-to-peak voltage, so divide by sqrt(2) to get 14.29V RMS for the transformer secondary. But transformer secondaries are set given the "average" primary voltage of 117V, and it could slip as low as 105V and still be power company legal, so multiply 14.29 * 117/105 to get right about 16Vrms for the secondary. Well, no one stocks 16V transformers, which is why they went up to 18V, I guess. This does not include the resistive drop inside the transformer secondary, which I suppose could be a volt or so. Okay, what size filter capacitor to use? The regulator can be modeled as a constant-current source at maximum load, which is 3A. Capacitance is defined as Q = CV, so differentiating we get I = C dV/dt, or dV/dt = I/C. Since current is constant, the capacitor discharges in a linear ramp, so dV/dt is a constant, the slope of the ramp. The peaks of the full-wave rectified AC from the secondary are 1/120 second apart, so we want to set the capacitor's value such that 1/120 * dV/dt is about our design ripple voltage of 2V. So 2 = dV/dt * 1/120 = I/C * 1/120 = 3/C/120. This gives C=12,500 microfarads, and since electrolytics have very high tolerance ranges, we'll round this up by 20% or so to get 15,000 uF. Many pointed out that running a regulator continuously at its maximum rated current isn't good conservative design, so I've upped the regulator to an LM338 which is rated at five amps. Heat sinking? Well, at the maximum power company voltage of 125VAC, the secondary is now at 19.2V = 27.2V p-p. Power dissipation in the regulator is then P=IV = 3 * (25.8 - 13.8) = 36 Watts. Clearly a largish heat sink will be required on the regulator. And no, this isn't for a car stereo in my house. :) It's for an amateur television transmitter. /cvk
wayned@wddami.spoami.com (Wayne Diener) (06/21/91)
In article <1991Jun18.175131.8374@ux1.cso.uiuc.edu> kline@ux1.cso.uiuc.edu (Charley Kline) writes: >Hi, we're trying to build a power supply for a device that requires 3A >at 13.8 volts. The voltage regulator will be an LM350T, a 3A-rated >variable voltage regulator. > >The problem is the power transformer. I can't find anywhere a 3A >transformer at 18 volts. > >My question is, am I guessing the secondary voltage wrong? Would a 14V >secondary be okay since that is the RMS value and the >rectified/filtered DC out of that would be sqrt(2) * 14 = 20V? > > You also lose the forward voltage drop of the diode, but you're basically correct. Try to find a 12.6 volt "filament" transformer. It should give you about 17 volts of filtered DC... enough overhead to run the regulator, but low enough that you won't need a massive heat sink. Even then, you'll have to dissipate almost 10 watts in the regulator. You're also going to need pretty large capacitors on the output of the rectifier to "hold up" the input voltage during the AC zero crossings. -- |---------------------------------------------------------------| | // Wayne D. Diener | | // Spokane, WA | | \\ // E-mail reply to: | | \X/ To: isc-br!hawk!wddami!wayned@uunet.uu.net | |---------------------------------------------------------------|
rsd@sei.cmu.edu (Richard D'Ippolito) (06/21/91)
In article <1991Jun20.182012.10847@ux1.cso.uiuc.edu> Charley Kline writes: > But transformer secondaries are set > given the "average" primary voltage of 117V, and it could slip as low > as 105V and still be power company legal, so multiply 14.29 * 117/105 > to get right about 16Vrms for the secondary. Minor nit: power-company legal is 120V +/- 10%, so you have to deal with 108 to 132 volts. Rich Good taste is timeless. Why is a good time often tasteless?
amichiel@rodan.acs.syr.edu (Allen J Michielsen) (06/25/91)
In article wayned@wddami.spoami.com (Wayne Diener) writes: >In article kline@ux1.cso.uiuc.edu (Charley Kline) writes: >>Hi, we're trying to build a power supply for a device that requires 3A >>at 13.8 volts. The voltage regulator will be an LM350T, a 3A-rated >>variable voltage regulator. >You also lose the forward voltage drop of the diode, but you're >basically correct. Try to find a 12.6 volt "filament" transformer. >It should give you about 17 volts of filtered DC... enough overhead >to run the regulator, but low enough that you won't need a massive >heat sink. Even then, you'll have to dissipate almost 10 watts >in the regulator. You're also going to need pretty large capacitors >on the output of the rectifier to "hold up" the input voltage during >the AC zero crossings. Depending on your noise requirements, you may well find that the lm350 and straight rectified ac will produce so much noise at high current usage that it is unusable. You can improve this with higher input voltages (smaller zero crossing times) and larger capacitors, but you end up with a pretty big box that you may be able to buy cheaper.... Then, with higher current usage, and higher input voltages, the power dissipated in heat becomes critical to the lm350 and either massive heat sinks or fans are needed. If noise (and we can say, like 5 volts) isn't a problem, then do it. al -- Al. Michielsen, Mechanical & Aerospace Engineering, Syracuse University InterNet: amichiel@rodan.acs.syr.edu amichiel@sunrise.acs.syr.edu Bitnet: AMICHIEL@SUNRISE