Bad Circuits - Test Your Knowledge

[ranger]

Hey berkeman,

I don't have a scanner right now. Could you post circuit B (x100 op-amp output stage for audio amp) and circuit G (x100 audio amp. (single supply)) so we can get some thoughts on it? You should of course post yours first.

thanks.

 

[berkeman]

Hey berkeman,

I don't have a scanner right now. Could you post circuit B (x100 op-amp output stage for audio amp) and circuit G (x100 audio amp. (single supply)) so we can get some thoughts on it? You should of course post yours first.

The order of the figures must have changed in the 2nd edition of H&H, but I think these are what you're asking for. They're getting a bit harder now...

[G] x100 opamp output stage for audio amp (with BJT input amp)

[J] x100 audio amplifier (single supply)

 

[antonantal]

For the first circuit, I don't know what is the range of the input voltage but if it can reach a higher value than the potential on the base of the BJT, that electrolytic capacitor is not a good idea.

In the second circuit, there is no dc path to the ground for the inverting input.

 

[ranger]

Circuit G:

It seems a coupling capacitor is missing from the output of the bjt amplifier to the input of op-amp. There should also be resistor to ground to cater for the bias current on the non-inverting input. Also the resistor will set the f_3dB frequency; I guess we don't want to amplify all frequencies.
Shouldn't we also want to add a cap in series with the 1K to have unity gain at DC? We'd of course choose C to have the appropriate f_3dB frequency.

 

[berkeman]

For the first circuit, I don't know what is the range of the input voltage but if it can reach a higher value than the potential on the base of the BJT, that electrolytic capacitor is not a good idea.

Good point about the cap, but they are probably assuming a pretty small signal (no more than 15V/100 one would hope), centered around ground as the audio input.

In the second circuit, there is no dc path to the ground for the inverting input.

You don't need a DC path to ground, per se, you just need a DC source for the input bias current. The opamp output DC connection could supply that. The issue with the 2nd circuit is something different. Hint -- be sure to double-check the datasheet for the LM358. It has some unique characteristics that the author of this bad circuit was apparently trying to use...

 

[berkeman]

Circuit G:

It seems a coupling capacitor is missing from the output of the bjt amplifier to the input of op-amp. There should also be resistor to ground to cater for the bias current on the non-inverting input. Also the resistor will set the f_3dB frequency; I guess we don't want to amplify all frequencies.
Shouldn't we also want to add a cap in series with the 1K to have unity gain at DC? We'd of course choose are to have the appropriate f_3dB frequency.

Good stuff, ranger. Yes, the input offset at the V+ input of the opamp would peg the output of the x100 amp. And adding a cap and ground bias resistor will mostly fix that. Adding a cap to the 1k pulldown resistor to ground on the V- input will roll off the DC gain as you say, but it would also help fix one last problem that this circuit would have if you just left the opamp's gain setting resistor as they are right now. Can you see what that would be?

 

[ranger]

Unfortunately, I cannot quite put my finger on the last problem, berkeman

But does it have to do with trimming the off-set voltage?

 

[berkeman]

Unfortunately, I cannot quite put my finger on the last problem, berkeman

But does it have to do with trimming the off-set voltage?

Not exactly. Since the gain is x100, and the supply rail is 9V, try putting in a conservative signal to give you a 5Vpp output signal, so 5V/100 = 50mVpp. What will the output look like?

 

[berkeman]

Unfortunately, I cannot quite put my finger on the last problem, berkeman

But does it have to do with trimming the off-set voltage?

Oh wait, you said "last problem", so I thought you meant the 2nd circuit. But if you're referring to my last comment for the first circuit, then yes, the problem is with the input offset voltage being amplified by x100. What do you typically do to minimize the input offset voltage component generated by the input offset current for the opamp? And assuming that you do that here, what are you left with as an input offset voltage for, say, an LM741 opamp? So what do you get for an output error after the x100 amplification?

 

[ranger]

Oh wait, you said "last problem", so I thought you meant the 2nd circuit. But if you're referring to my last comment for the first circuit, then yes, the problem is with the input offset voltage being amplified by x100. What do you typically do to minimize the input offset voltage component generated by the input offset current for the opamp? And assuming that you do that here, what are you left with as an input offset voltage for, say, an LM741 opamp? So what do you get for an output error after the x100 amplification?

Yes, I meant the last problem with circuit G.

Well to minimize such offset voltage, one could use a trimming network or simply use an op-amp with very small V_OS or a FET amplifier which has little input current. Another way I see to eliminate the input-offset voltage due to offset current is to make the the resistances of both the inverting and non-inverting input equal. In this way, both inputs will have negligible offset voltage due to offset current becuase they see the same equivalent resistance.

According to the datasheet for the 741, we should expect a typical offset voltage of about 1mV. I guess this also applies to a properly trimmed op-amp?

 

[berkeman]

Yes, I meant the last problem with circuit G.

Well to minimize such offset voltage, one could use a trimming network or simply use an op-amp with very small V_OS or a FET amplifier which has little input current. Another way I see to eliminate the input-offset voltage due to offset current is to make the the resistances of both the inverting and non-inverting input equal. In this way, both inputs will have negligible offset voltage due to offset current becuase they see the same equivalent resistance.

According to the datasheet for the 741, we should expect a typical offset voltage of about 1mV. I guess this also applies to a properly trimmed op-amp?

You've pretty much got [G] figured out, ranger. Good job. You might be able to trim out the remaining offset voltage, but I honestly don't know how stable that would be over temperature and aging. Usually you'll just try to do something that minimizes the offset, and be careful that your gain is not too high. The gain rolloff capacitor in the V- leg to ground is one way to get rid of the DC input offset voltage issue, if you can afford the loss of low-frequency gain, and time constant at circuit turn-on.

So how about the 2nd circuit...?

 

[dlgoff]

Well I didn't look at the specs for the op-amp on the last circuit, but it looks like the non-inverting input needs to be biased up a bit. Wouldn't the output be clipped on the low side?

 

[berkeman]

Well I didn't look at the specs for the op-amp on the last circuit, but it looks like the non-inverting input needs to be biased up a bit. Wouldn't the output be clipped on the low side?

Yes, absolutely. The LM358 is used when the inputs are going to be near ground, but it certainly can't drive down below ground when amplifying an AC signal. So you would need to do something like bias up the V+ input to half of the supply or something in order to keep the output signal centered between the top rail and ground.

So to summarize the answers for the two circuits in Post #62,

[G] As drawn, there will be an input DC offset that will peg the output of the x100 amp. And even with a fix to the input offset from the BJT stage, a V- side capacitor should be used to eliminate the input offset voltage of the opamp, to avoid about a 100mV output offset error.

[J] The output will clip at ground as drawn. The input needs to biased to half of the supply to keep the output nominally centered between the upper rail and ground.

 

[berkeman]

Okay, how about a couple last opamp circuits, and then we can switch gears. Here are two more from H&H Chapter 3 on opamps. I especially like the 2nd one [F], which was published in another textbook as an example of a "good" circuit idea...

[D] Voltage-controlled current source

[F] 200mA "current source"

 

[EugP]

Okay, how about a couple last opamp circuits, and then we can switch gears. Here are two more from H&H Chapter 3 on opamps. I especially like the 2nd one [F], which was published in another textbook as an example of a "good" circuit idea...

[D] Voltage-controlled current source

[F] 200mA "current source"

I'll give it a shot.
[D] I think it has something to do with R.

[F] Quick question about F. That "zener" thing in the middle with 9v next to it, does that mean there's a 9v voltage drop across it?

Sorry if I'm wrong, I'm still an undergrad with almost no real life experience.

 

[NoTime]

I'll give it a shot.
[D] I think it has something to do with R.

[F] Quick question about F. That "zener" thing in the middle with 9v next to it, does that mean there's a 9v voltage drop across it?

Sorry if I'm wrong, I'm still an undergrad with almost no real life experience.

Yes a zener is supposed to show a constant voltage.
It was intended as a 9v reference. Note that there is a maximum current that can flow thru the 240 ohm resistor before the voltage will drop below 9v.

In [D] there is something you can do with R that would fix the major issue.
What do you think it is?

 

[EugP]

Yes a zener is supposed to show a constant voltage.
It was intended as a 9v reference. Note that there is a maximum current that can flow thru the 240 ohm resistor before the voltage will drop below 9v.

In [D] there is something you can do with R that would fix the major issue.
What do you think it is?

For [D] I think the R and "load" should be switched.

I'm still thinking about [F] though, I've never seen that before.

 

[AlephZero]

D won't work because the opamp is supposed to be operating in a linear mode, but its two inputs are are different voltages. For example think about the case where the load has zero resistance. One input is at 0V, the other is at the "control voltage".

Swapping R and the load seems to fix that problem.

 

[AlephZero]

F: I'm assuming the op amp has a +15/-15 power supply, not +15/0.

If you took away the 240 resistor, this would "sort of" work, except that all the load current would flow through the zener. A typical small zener wouldn't take 200mA current and 1.8W dissipation though.

I suppose the 15V / 240R is meant to supply 25 mA through the zener which seems a sensible design value.

If that's identified the problem, somebody else can have a go at fixing it.

 

[ranger]

For circuit F, I think we need a resistor from the non-inverting input to ground to deal with offset voltage due to the small input bias current. Or are we intentionally leaving it this way to amplify the [offset] voltage difference to produce an output?

For circuit D, I also think that swapping the positions of the R and the load will work. With the circuit we'd want to fix it by having R connected to the inverting input to ground. And have the load connected to R and the output of the op-amp.

 

[berkeman]

Good job folks on circuit [D]. Yep, the reference resistor and the load were swapped. Doh!

On [F], if the opamp is doing its job and holding its V- input at ground, what is the voltage on top of the "9V" Zener (just calculate the resistor voltage divider first, ignoring the Zener diode). So what does this mean the current through the load really is?

It looks like they were trying to make an opamp circuit that would take a current 9V/45 Ohms and pass that current through the load. Maybe altering the 240 Ohm resistor value might make this circuit work, but it's dumb to brute force push extra current through the 9V Zener (and have to use a bigger, more expensive Zener part), when the opamp is supposed to be the amplifying device. Do you see a way to use the topology of circuit [D] instead (after being fixed with the swap that we talked about)?

 

[NoTime]

Maybe altering the 240 Ohm resistor value might make this circuit work

Not if a 200ma load current is desired.
200ma * 45 Ohms is a 9v drop.
At least they got the values right even if they wired it wrong.

 

[ranger]

Good job folks on circuit [D].
It looks like they were trying to make an opamp circuit that would take a current 9V/45 Ohms and pass that current through the load. Maybe altering the 240 Ohm resistor value might make this circuit work, but it's dumb to brute force push extra current through the 9V Zener (and have to use a bigger, more expensive Zener part), when the opamp is supposed to be the amplifying device. Do you see a way to use the topology of circuit [D] instead (after being fixed with the swap that we talked about)?

So our problem now is to not push 200mA thru the zener? If we used the fixed version of circuit D, we would apply the reference voltage of 9V to the non-inverting input. We'd of course need to bias the zener properly by choosing an appropriate resistor. Since the fixed version of circuit D uses negative feedback, we would have the zener voltage at the inverting input. So if we use R to be 45ohms, we'd have a current of 200mA thru the load. IN this configuration we don't have to worry about excess current thru the zener because we are using the resistor to get the appropriate zener current for voltage regulation.

 

[NoTime]

Looks like you have the right idea.

we would apply the reference voltage of 9V to the non-inverting input.
...
we would have the zener voltage at the inverting input.

Yes, the negative feedback voltage will approximate the reference voltage.
Why won't it be exact?

We'd of course need to bias the zener properly by choosing an appropriate resistor.

The choice of zener bias resistor (the 240 ohm) will affect the final load current.
Why?

 

[ranger]

Yes, the negative feedback voltage will approximate the reference voltage.
Why won't it be exact?

Because the feedback loop tends to subtract some of the output voltage proportional to Z*Vout; where Z is impedance and is obtained by applying the voltage divider to resistors on the inverting input (feedback network)?

The choice of zener bias resistor (the 240 ohm) will affect the final load current.
Why?

Does it have to do with the fact that the inverting and non-inverting inputs see different resistance? And becuase of the high bias current on the non-inverting input, we have offset voltage due to this?

 

[NoTime]

Because the feedback loop tends to subtract some of the output voltage proportional to Z*Vout; where Z is impedance and is obtained by applying the voltage divider to resistors on the inverting input (feedback network)?



Does it have to do with the fact that the inverting and non-inverting inputs see different resistance? And becuase of the high bias current on the non-inverting input, we have offset voltage due to this?

No. The answers have to do with the properties of real op-amps and zeners.

You are vaguely headed in the right direction with the first answer.

Your second answer makes me wonder just where did you put the 240 ohm resistor and zener?

 

[ranger]

No. The answers have to do with the properties of real op-amps and zeners.

You are vaguely headed in the right direction with the first answer.

Your second answer makes me wonder just where did you put the 240 ohm resistor and zener?

Well, let me recap [IIRC] the properties of real op-amps. They have offset voltage, bias and offset [input] current, output current limit, and gain degradation with increasing frequencies.

With regards as to why the inverting input will not see the exact zener voltage - the best answer I can come up with is that the feedback network subtracts some of the output voltage.

The 240ohm is connected to the 15V power supply, the zener to ground (anode:ground; cathode:non-inverting input), and the non-inverting input is hooked up to the zener to get the reference voltage. No?

 

[NoTime]

The 240ohm is connected to the 15V power supply, the zener to ground (anode:ground; cathode:inverting input), and the inverting input is hooked up to the zener to get the reference voltage. No?

Ouch! No!

Where did you lose this statement?

we would apply the reference voltage of 9V to the NON-inverting input.

Take another look at the revised circuit [D]

 

[ranger]

Ouch! No!

Where did you lose this statement?Take another look at the revised circuit [D]

That was a typo, sorry. I'll fix it in post #87 so others don't get confused.

Care to give anymore hints?

 

[NoTime]

That was a typo, sorry. I'll fix it in post #87 so others don't get confused.

Care to give anymore hints?

The statement might have been a typo on your part.
But, it was correct
And the reason I thought you had this figured out.