beast. 50 cents.

On 11/24/2011 9:04 PM, John Larkin wrote:

Change the top two transistors to be the compliment of the bottom two transistors. Like this:

Version 4 SHEET 1 1176 680 WIRE -32 -192 -272 -192 WIRE 320 -192 -32 -192 WIRE 544 -192 320 -192 WIRE -272 -144 -272 -192 WIRE -32 -144 -32 -192 WIRE 320 -64 320 -192 WIRE -272 -48 -272 -64 WIRE 256 -16 208 -16 WIRE 544 -16 544 -192 WIRE -32 32 -32 -64 WIRE 144 32 -32 32 WIRE 320 80 320 32 WIRE 320 80 208 80 WIRE 544 80 544 48 WIRE 544 80 320 80 WIRE -32 160 -32 32 WIRE 320 272 320 80 WIRE 752 272 320 272 WIRE 992 272 832 272 WIRE 992 304 992 272 WIRE 320 336 320 272 WIRE 320 336 224 336 WIRE -32 384 -32 240 WIRE 96 384 -32 384 WIRE 160 384 96 384 WIRE 320 384 320 336 WIRE -224 432 -304 432 WIRE -96 432 -144 432 WIRE 256 432 224 432 WIRE 992 432 992 384 WIRE 96 448 96 384 WIRE 224 448 224 432 WIRE -32 512 -32 480 WIRE -32 624 -32 592 WIRE 96 624 96 512 WIRE 96 624 -32 624 WIRE 224 624 224 528 WIRE 224 624 96 624 WIRE 320 624 320 480 WIRE 320 624 224 624 WIRE 96 656 96 624 FLAG -272 -48 0 FLAG -304 512 0 FLAG 992 432 0 FLAG 96 656 0 SYMBOL npn 144 -16 R0 SYMATTR InstName Q1 SYMATTR Value 2N2222 SYMBOL pnp 160 432 M180 SYMATTR InstName Q3 SYMATTR Value 2N2907 SYMBOL npn 256 384 R0 SYMATTR InstName Q4 SYMATTR Value 2N2222 SYMBOL res 208 432 R0 SYMATTR InstName R1 SYMATTR Value 100k SYMBOL res -48 -160 R0 SYMATTR InstName R2 SYMATTR Value 100k SYMBOL npn -96 384 R0 SYMATTR InstName Q5 SYMATTR Value 2N3904 SYMBOL res -48 496 R0 SYMATTR InstName R3 SYMATTR Value 4.7k SYMBOL voltage -272 -160 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value 10 SYMBOL Misc\\signal -304 416 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value SINE(.79 .1 1K) SYMBOL res -128 416 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R4 SYMATTR Value 56k SYMBOL schottky 560 48 R180 WINDOW 0 24 72 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D4 SYMATTR Value 1N5817 SYMATTR Description Diode SYMATTR Type diode SYMBOL pnp 256 32 M180 SYMATTR InstName Q2 SYMATTR Value 2N2907 SYMBOL res 736 256 M90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R5 SYMATTR Value 100 SYMBOL cap 80 448 R0 SYMATTR InstName C1 SYMATTR Value 10p SYMBOL voltage 992 288 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V3 SYMATTR Value 5 SYMBOL voltage -32 144 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V4 SYMATTR Value 1.137 TEXT -328 -216 Left 2 !.tran 0 .01 0

Reply to
John S
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The only thing that makes sense to me is compound pairs up and down, and only two bias diodes. But JT says there are no big PNPs.

Or it is, in fact, not much like the figure.

What do you think?

John

Reply to
John Larkin

I can't argue with JT's comment. I would agree that it is not like the schematic.

John S

Reply to
John S

snip

but then 3 diodes will be too much bias, most like it should have been like figure 6 in this:

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3 diodes for bias, no need for a big pnp

-Lasse

Reply to
langwadt

No, I didn't see that because I only tested the way I did because it was stated things would blow up once the top side would start past 1 diode click, which I showed it did not however, after looking at the circuit, there is a missing inverting transistor at top side not being shown in the block diagram and I don't think any one needs reminding that it's not good business to publish your trade secret s. So either show it wrong or just don't show all of the components.

Conception block diagrams are becoming more common these days and I can think of a couple reasons why.

Jamie

Reply to
Jamie

In the part you snipped, I didn't use diodes nor did I use a big PNP.

Reply to
John S

The fastest way to collapse a magnetic field is to allow the inductor's voltage to go wherever it wants. Unfortunately, there are limits and your driver transistor is one of them. So your assignment, Jamie, if you choose to accept it, is to allow as much voltage as possible to develop across the inductor while protecting the breakdown rating of your driver.

Makes sense?

John S

Reply to
John S

yes I understand and I've come up with a solution that should work fine because as it appears, It seems, I need that back current to help maintain a constant hold. So, instead of using a diode that is aggressive i've made a driver amp with dynamic clamping. I will then later on put the switching bridge on the other side of the coil to take advantage of the returned energy

Have a good day. Jamie

Reply to
Jamie

but used V4 with was ~2 diodes and Q2 would have to be big to since it has to handle the output current

-Lasse

Reply to
langwadt

Yes, I cheated just to get a reasonable output. My only purpose was to see if a reconfiguration of the two top transistors had any hope of salvaging it, at least in theory.

Reply to
John S

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