help with volume 6 - simple op-amp

Thread Starter

glopv2

Joined Nov 20, 2008
10
Hi all,

I just built a copy of the simple op-amp design from http://www.allaboutcircuits.com/vol_6/chpt_5/17.html

It seems to be working great, Except one (rather large) problem, which is that although I am giving it +10 VDC and -10 VDC, it cannot swing down past about 1.2V into the negative.

I am new enough to electronics that I have really no idea why that would happen or even where to start debugging, other than 1.2 V looks like 2 diode drops to me....

Any help would be greatly appreciated. Thanks!
 

beenthere

Joined Apr 20, 2004
15,819
If you built it as shown, the measuring point for reference has to be the connection between the batteries. The op amp in the circuit is operating single-ended, between ground and +12 volts. So it will swing from about 1.2 volts above ground to about 1 volt below 12 if you measure from ground. Moving the meter lead to the common battery connection will let you see the output going from +5 to -5 volts.

In your case, that should be about -9 to +9.
 

hgmjr

Joined Jan 28, 2005
9,027
The circuit you are using is only intended to accept input signals that are positive. The output is not intended to go below a volt or two.

hgmjr
 

Thread Starter

glopv2

Joined Nov 20, 2008
10
Sorry --

I've actually done this using a power supply that is giving +10V, -10V, and GND. So my ground is different from the schematic.. (sorry!!)

Also, I've connected the "non-inverting" input to ground, and the inverting input to a function generator, and I'm looking at the output on an oscilloscope.

Any ideas why the circuit wouldn't be able to drop past -1.2 volts? (or, using the reference of the schematic, any reason it wouldn't drop past 4.8 volts?)
 

hgmjr

Joined Jan 28, 2005
9,027
In the case of your altered circuit, the output at the collector of the right most transistor in the diferential pair is constrained not go more negative than a diode drop below the base which you have indicated that you tied to ground (0 volts).

hgmjr
 

Thread Starter

glopv2

Joined Nov 20, 2008
10
Damn, that makes sense...


I grounded that transistor's base because I wanted to mimic the standard "inverting" setup of an op-amp. How would I do that in this case?
 

hgmjr

Joined Jan 28, 2005
9,027
One way to do it is to add a common collector output stage and then use a feedback network from the output of the common collector stage back to the base of the right most transistor. The feedback would be just like the feedback associated with an opamp.

I take it that you have some basic knowledge of opamps.

hgmjr
 

Thread Starter

glopv2

Joined Nov 20, 2008
10
I did what you said, and used a common collector to return feedback like an op-amp. The ratio of resistors should give me a gain of 2 (similar to the bottom diagram from the above link, but including the voltage follower/buffer). Again, I see the output gain of 2 in the positive region, but in the negative region, a gain of 1, presumably because it is limited by the input voltage still.. Is there another work-around?
 

Thread Starter

glopv2

Joined Nov 20, 2008
10
This is my basic schematic. I'll call the terminals the left input (VL), the right input (VR), output (OUT), Vcc, Vee, and Ground (GND).

What I did was connect a 25k resistor from OUT to VR, and a 25k resistor from VR to GND. Then I applied the signal to VL. I see the signal amplified by 2 at OUT, but like I said earlier, there is still the problem that the output is limited by what it sees at VR.
 

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Thread Starter

glopv2

Joined Nov 20, 2008
10
Ok, I've tried to draw the problem that I see when I connect things in the manner you described, using Ra=Rb. I expect a gain of 2.

(BTW sorry for the quality. I'm copying my scope into paint :))

I want a symmetric gain of 2
 

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Thread Starter

glopv2

Joined Nov 20, 2008
10
Yes. However, I tried 2.2k and it didn't make a whit of difference. I also tried 1k, also no difference. I then decided to measure the node between Ra and Rb, and I notice that it tracks the other input exactly. How can it do that? That must mean that little or no current is flowing across Ra during the bottom half of the swing -- right!?
 

Thread Starter

glopv2

Joined Nov 20, 2008
10
It's on the negative rail. I just measured the resistance to the negative rail as 0.1ohms and to ground as 87k.


Also this: I found a way that seems to make the circuit work, although I don't know why. I connected Collector Q2 directly to Vcc and disconnected it from Q5 and Q7. This still doesn't resolve for me why it wasn't working before... How frustrating
 

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