Jung problems

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What's your input voltage? Are you sure about it? You say you expect 12V, but with R1=890, R2=1.13k, you design for Vout = 15.6V. You should have at least 18VDC at the input. Is that right?

If not, the opamp will increase its output voltage in an attempt to get to 15.6V, and the collector of Q2 rises until its Q-B junction starts to reverse conduct, and the voltage across R4 goes up until there is no more voltage (or even reverse voltage) across the LED, it goes off.

Jan Didden
 
My input voltage is at least 20V off-load. I'm using a PCB mount Amveco toroidal transformer (two 7v windings in series) with a bridge rectifier. On load it should give a little higher than 18V.

Also, when you use the equation R8 = R9*(Vout-Vref) / Vref , and substitute 890 for R8, 1130 for R9, and 6.0 for Vref; you get 12.33V for Vout.

Thanks for the help, can you guys think of anythig else that could be wrong? Seems like the voltage isn't being regulated as the output is about the same as the input voltge.
 
rhymeswthorange said:


Also, when you use the equation R8 = R9*(Vout-Vref) / Vref , and substitute 890 for R8, 1130 for R9, and 6.0 for Vref; you get 12.33V for Vout.


The LM329DZ as VRef will yield 6.9V, not 6.0V --

While it isn't a problem here, Jung pointed out in the TAA 2000 issue -- that there are occasionally dud LM329's so make sure that it is actually working.

PS -- Jan wrote one of the original series (1995) articles.
 
Could a failed opamp cause this? The one I ordered was a surface mount type (SOIC-8), and after some effort I eventually got single strands of wire from my stranded hookup wire soldered to each contact. I wonder if in the process of doing this, I damaged the amp. The soldering iron wasn't held to the leads for very long (didn't need to be), but how sensitive are these chips to heat? Could, say, a second of soldering undo the chip? I also wiped some excess flux away with some alcohol (and let it dry before it was fired up). Would a failed opamp cause this type of behavior?

P.S. I have yet to check the other suggestions - have to get home first 🙂
 
Yes, you really should measure the in/output of the opamp. Second guessing and groping around in the dark is VERY inefficient.

You're still off on Vout. Across the 890 ohms there is 6.9 V. Therefore, across the 1130 there is 1130/890 * 6.9 = 8.76 volts. Therefore, Vout = 15.66V. Since the opamp needs some volts to operate, and you also loose some in the series transistor, you need probably a minimum of 20VDC input under load. Please check.

Jan Didden
 
janneman said:
Yes, you really should measure the in/output of the opamp. Second guessing and groping around in the dark is VERY inefficient.

You're still off on Vout. Across the 890 ohms there is 6.9 V. Therefore, across the 1130 there is 1130/890 * 6.9 = 8.76 volts. Therefore, Vout = 15.66V. Since the opamp needs some volts to operate, and you also loose some in the series transistor, you need probably a minimum of 20VDC input under load. Please check.

Jan Didden


you also lose across the LM317 pre-regulator -- (the version which I built did not incorporate this.)
 
As for the Vout, switch the two resistor values you're using and you'll get my 12.33V. Across the 1130 ohms there is 6.9 V. Therefore, across the 890 there is 890/1130 * 6.9 = 5.43 volts. Therefore, Vout = 12.33V. 🙂

I'm using a 1n5235 zener to the collector of Q2.

I'll measure voltages as soon as I get home.

Thanks again for all the help!
 
We may still talk about different things.
I'm not using R1, R2, you are.
In the Jung schematic where there are R1 and R2, I see a 1N4148 diode, not a zener. And in that diagram, Vout = 15,66, sorry.
How many volts is that zener?

What else is different?

Can you post a diagram of the one YOU buiilt? We may catch something that went unnoticed so far.

Jan Didden
 
Pin 6 (output) has 13.5V (w/ respect to ground)on it with the 250 ohm load. There is a 4.35 difference between +In and Output.

With no load, Pin 6 has 17.9V and there is 6V between +In and Output.


The voltage difference btween +In and Ouput are the same polarity - one isn't negative.

?
 
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