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Pulse generator/transistor mishap
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xiphocoleon



Joined: Oct 04, 2013
Posts: 40
Location: Brooklyn

PostPosted: Mon Sep 22, 2014 6:00 pm    Post subject: Pulse generator/transistor mishap
Subject description: I think I put the load in the wrong place w/ my transistors
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In the circuit image below, you can see on my outputs that I tried to buffer my outputs with transistors. I haven't rechecked yet, but I think this is the method Ken Stone posted in several of his designs.

However, three (of six) of my transistors have melted. They melted at different times, and I can't remember the patch settings during the melting. But, does anyone see an obvious problem with these output buffers that would cause a ton of current to go through the transistors and melt them?

Thanks.


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elmegil



Joined: Mar 20, 2012
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PostPosted: Mon Sep 22, 2014 10:15 pm    Post subject: Reply with quote  Mark this post and the followings unread

Deleted previous post, wasn't following the schematic correctly. The outputs all go through the 10k pots, so output limiting isn't the problem.

However, you're driving the bases directly, you might want resistors in series between the CMOS and the base of the transistor.
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prgdeltablues



Joined: Sep 25, 2006
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PostPosted: Tue Sep 23, 2014 3:13 am    Post subject: Reply with quote  Mark this post and the followings unread

Unless whatever you are connecting the outputs to has the possibility of shorting to ground. If you have your output pot turned fully up on this module, and if the module you're sending the output to has a pot on its input, and that was turned fully down, you'd have a direct path through the transistor from V+ to GND. Might be best to put a 1k in series after the pot?

Peter
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MoonBase 69



Joined: Sep 16, 2014
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PostPosted: Tue Sep 23, 2014 9:52 am    Post subject: Reply with quote  Mark this post and the followings unread

My vote is for a little of both, something between the collector and +12 and a little something on the base. To keep the transistor with in limits under any foreseeable condition with the output IE shorting the emitter directly to ground through a maxed pot... A bit rusty with my saturated switch calculations to make a educated value recommendation. Working from memory... ~100 ohms on the emitter and 2.7K on the base someone want to double check me?
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elmegil



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PostPosted: Tue Sep 23, 2014 3:01 pm    Post subject: Reply with quote  Mark this post and the followings unread

I did the math for a 2N3904 shorted to ground with my original, deleted post. 120R or higher will keep the current below 100mA, which is half the rated max. 100R ought to work just fine.

For the base, I'm not sure how you'd calculate that.
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MoonBase 69



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PostPosted: Tue Sep 23, 2014 3:39 pm    Post subject: Reply with quote  Mark this post and the followings unread

I used bR=Eon/Ib where Ib=Ic/hfe. Used a 100 hfe, max of 200ma Ic and doubled the base current to be sure it's fullt saturated assuming +12v comming from the gate to turn on and came up with 3K, 2700 should suffice.
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xiphocoleon



Joined: Oct 04, 2013
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PostPosted: Tue Sep 23, 2014 5:24 pm    Post subject: Reply with quote  Mark this post and the followings unread

Thanks everyone here. These suggestions have been super helpful. Yes, I think at heart, the issue has been that a short has occured. I am going to do a "surgical strike" upon the board and insert needed resistors. Thanks again for your help.
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xiphocoleon



Joined: Oct 04, 2013
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Location: Brooklyn

PostPosted: Tue Sep 23, 2014 6:32 pm    Post subject: Reply with quote  Mark this post and the followings unread

Skysurfer wrote:
I used bR=Eon/Ib where Ib=Ic/hfe. Used a 100 hfe, max of 200ma Ic and doubled the base current to be sure it's fullt saturated assuming +12v comming from the gate to turn on and came up with 3K, 2700 should suffice.


Just curious on the math:

appropriate base resistor = full voltage at base (+12V) / current at base (Ib)

Ib = Ic/hfe = 0.2 A/100 = 0.002 A

Double the base current: 0.004 A

so: bR = 12 / 0.004 = 3000 ohms

Was this your method? And how did you decide to double the base current? Thanks again.
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MoonBase 69



Joined: Sep 16, 2014
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PostPosted: Wed Sep 24, 2014 3:32 am    Post subject: Reply with quote  Mark this post and the followings unread

Yep that was about it. I doubled the worst case current to make sure it was well into the saturation region. Just something I remembered doing. Under normal conditions with the load at 10K it insures your still well into the saturation region with ~2.4V on the base which doesnt exceed the 6V max base to emitter voltage.
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xiphocoleon



Joined: Oct 04, 2013
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PostPosted: Wed Sep 24, 2014 5:58 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks. Just wanted to make sure I understood. Will consult my Horowitz and Hill/Art of Electronics as well.
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MoonBase 69



Joined: Sep 16, 2014
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PostPosted: Wed Sep 24, 2014 6:24 am    Post subject: Reply with quote  Mark this post and the followings unread

Not a problem, this is going to be an interesting review for me. Been dealing with more repair for many years not so much component level design stuff. So will require peer review of anything I throw out there until I can remember what I'm doing.
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