How do flyback diodes protect FETs in switching power supply circuits?

[Shahil]

Help! Full Wave Bridge AGAIN!

https://www.physicsforums.com/showthread.php?t=66883

I posted this in the Homework help zone and nobody helped me

Please can someone here help me! I would appreciate it!

Thanks!

 

[faust9]

can you post your problem as a pdf? I don't us ms and graphics conversions don't work that well.

 

[Shahil]

Don't have a PDF converter on this PC (campus PC)! Sorry about that faust9. Thanks anyway!

 

[faust9]

Are you trying to make a constant voltage source or a constant current source. As it stands, the circuit you've presented probably won't work the way you intended it to. Essentially, you have 3.9 volts applied to the positive terminal of an opamp wired as a simple comparitor. The opamp will try to adjust its output in order to keep the input terminals at a zero volt differential; however, the circuit as you have it drawn will not work. The base of one of you transistors is not connected. Additionally, you have the Vss of the opamp tied to the GND bus (was this intentional) and the and the Vdd terminal tied to the positive bus.

If you're trying to make a voltage/current regulator look into the 78xx series voltage regulators(google 7805 and read the fairchild pdf that comes up).

If you explain the intent of your design I'm sure I(or some one) could help you out.

If you want to build an opamp based voltage/current reg, I'd suggest googleing it. You'll save yourself hours and hours of t/s'ing a circiut someone already designed. If your going into engineering one of the things you need to learn is how to look stuff up. Don't waste your time trying to reinvent the wheel. If you come across a problem that was previously unanswered then sitdown and work out a solution; otherwise consult a circuit encyclopedia. Also, you should be able to find an analog ckt for just about every use under the sun but you will usually have to modify and thus recalculate all R,C,L,Gain and Q points in the ckt for your needs.

My 2 cents. Tell us what you need and we'll help you develop a working ckt.

Good luck.

 

[Shahil]

Sorry!

My bad...bad circuit diagram I'm guessing. The corrections as you pointed out:
The base of the top 3904 is connected to the collector of the bottom 3904.
The opamp is wrong way round - overlooked that when using the sim
package.

Actually, my friend and I designed and built this circuit and it works perfectly giving a constant 5V DC output from a 7.2V AC input. Problem is that we CALCULATED values from the opamp onwards but guessed values for the capacitor and resistor in series with the Zener. Hence, the part of the circuit after these components can be just treated as a load or something to that extent. (hence the inverted colour).

What I wanted was some pointers and, preferably, equations that can be used to predict values for the cap and zener resistor as I've used. All the equations I've formulated are giving some way out values and I've found no useful info on the net or library about this - though I've only been looking since I posted the thread.

 

[Antiphon]

I'll help you out if you replace Che' with Reagan in your avatar.

 

[Shahil]

I'll help you out if you replace Che' with Reagan in your avatar.

:yuck:

After watching Motorcycle Diaries, how bout no...

C'mon man - you aren't going to get a little thing like politics get in the way of helping a struggling little human being right// :shy:

Please help! My lecturer, when I showed him this diagram, grilled me all the way to hell (or whatever after-existence us communists believe in ) about why I put in a 1k1 reisitor in series with the zener and why the hell do I have a 100ohm output resistance with the opamp. I need help please

 

[Ouabache]

Okay - I know that this is a straight-forward intermediate level circuit theory problem but uh...I've forgotten how to do it so I need help!

I have a full-wave rectifying bridge connected to a rectifying cap and a regulating zener diode. I do have values for the cap/zener/resistor (worked out through trial and error) but: What equations do you use to work out values for this circuit? I've looked all over the net and couple textbook but can't seem to find any good information on it! If it does help,
Please help people - I really need to figure this out!

Note on pic:
Zener = 3.9V
R = 1100 ohms
C = 1000uF

Only the non-inverted coloured part of the circuit is the part that is concerend in my question.

I have not designed a supply as you proposed, but have constructed other PS circuits using a full-wave rectifier.

Your choice of cap has to do with how much ripple you are willing to live with, on your rectified AC. If your load resistance is constant, increasing the capacitor value will decrease the ripple voltage. Remember $$\tau = RC$$ A quick method for finding the appropriate capacitor value is
to allocate 3000 µF for every ampere of load current.

Generally your load (circuit) should be in parallel with just the Zener, not in parallel with the resister in series with a Zener, as you have. The resister is used to limit the current, in the event the load was removed. That would protect your Zener from getting toasted. The voltage you want applied to your load should be the Voltage Rating of your Zener.

the following primer offers lots of diagrams, descriptions and equations--->
http://students.rcc.on.ca/~schlaak/courseNotes/ef200/06_power_supplies_2.htm

here is another one just on Zener-Voltage Regulation
---> http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/zenereg.html

 

[Ouabache]

(personal communication)
Hi!

Same circuit - I'm a bit confused on why there is a 100ohm resistor in series with the op-amp?

When I take it out or change the value - the ripple is monstrous so I left it in. Any ideas why it's there??

Sorry, I looked back to your original design bridge.doc I don’t see what 100ohm series resistor you are referring to. I see an 11.1K
resistor, not quite in series, b4 your op-amp.

 

[willib]

i have a question?
the circuit diagram below is from http://www.irf.com/technical-info/whitepaper/syncrecapec2005.pdfthe top one in figure 11..
the irf website..even though it looks backwards ( the diode direction) i realize that they are FETs.. but i tried it the way it is ( with just diodes ) and it works..so my question is (why or how) could you use FETs in this sync rectifier??what i mean is the waveform looked well NICE! and there didnt seem to be any way it could be improved by using FETs ..? i mean there were no dead spots in the wave??

 

[faust9]

Those diodes are called flyback diodes. FETS are very sensative to over voltage/over current conditions. Inductors and/or capacitors can easilty cause overvolt/overcurrent contitions when switched thus those flybacks are installed to give the excess current a path to ground or back to the source.

FETS are used because you can control the on/off time of the fet thus control the voltage output. Just using a diode will result in a predictible output; however, the magnitude of the output voltage will always be the same--or you'd have to set up a voltage divider circuit which consumes current and produces excess heat. That circuit is one of the many switching power supply circuits floating around. When the FET is switched of, no current flows (no excess heat generated). When the FET is on current flows but the FET consumes very little power (FETS have an on resistance in the ohm to milliohm range usually, 1ohm or less). Basically that circuit is very efficient at controlling the magnitude of the voltage output.

Well, hope this helped.

 

[Shahil]

Updated circuit diagram

Okay - here's the diagram redrawn to include everything.

Sorry for the MSWord format but. as I said, got no PDF converter at campus...

Anyway - the resistor I'm referring to is R2. My friend attempted answering it by suggesting that the op amp is, basically, acting as a current source and to protect the transistors (max. rating 200mA) the resistor is put in. It kinda makes sense I think though I'll trust anything that sounds remotely convincing

So is this right? Hope you can help!

 

[faust9]

That ckt shouldn't work.

Lets look at like this--assume the ckt is past the intial cap transient(the FW and that cap should produce a very flat DC o/p). Current flows throught the zener leg thus applying 3.9V to the (-) of the opamp. the (+) is still zero(nothing has gated T1 on yet). If (-)>(+) then the op-amp will drive the output to the negative rail. T1 will never get gated on because your using an NPN for T1. NPN's need a votage greater than the collector voltage applied to the base to get them to turn on. A PNP on the other hand will turn on with a positive voltage applied to the emitter and the base grounded.

 

[Ouabache]

I just had a chance to view your latest circuit diagram.
Faust9 has an interesting analysis of your circuit.

I don't have any particular idea as to why R2 (100ohms) should or should not be there.. Is your output at R6 what you are expecting? You say when you change R2 you get larger ripple voltage. Did it make any difference as you make R2 larger versus smaller? With the ripple becoming larger, it sounds like you were affecting the overall impedance which is in parallel with C1.

 

[Averagesupernova]

https://www.physicsforums.com/showthread.php?t=66053

Shahil, you've posted a similar schematic in the above thread. I explained to you just about everything. The last schematic you posted in this thread has the + & - inputs reversed on the op-amp and will the circuit will not work as faust has pointed out. The 100 ohm series resistor is part of the over current protection. When T2 is turned on it shunts current away from T1. The 100 ohm resistor has to be there otherwise you would be shorting the output of the op-amp to ground. Sorry to sound like an a$$, but can we keep this circuit confined to a single thread? We've got 3 threads going with this circuit or a slight variation of it (in error I have to assume) and jumping back and forth between things gets old.

 

[willib]

Those diodes are called flyback diodes. FETS are very sensative to over voltage/over current conditions. Inductors and/or capacitors can easilty cause overvolt/overcurrent contitions when switched thus those flybacks are installed to give the excess current a path to ground or back to the source.

FETS are used because you can control the on/off time of the fet thus control the voltage output. Just using a diode will result in a predictible output; however, the magnitude of the output voltage will always be the same--or you'd have to set up a voltage divider circuit which consumes current and produces excess heat. That circuit is one of the many switching power supply circuits floating around. When the FET is switched of, no current flows (no excess heat generated). When the FET is on current flows but the FET consumes very little power (FETS have an on resistance in the ohm to milliohm range usually, 1ohm or less). Basically that circuit is very efficient at controlling the magnitude of the voltage output.

Well, hope this helped.

Yes it helped and thanks for replying !