Question about driving heavy capacitive load

[jordan76]

Hi guys,

Below is an excerpt from the handbook of National Semiconductor:

Driving Capacitive Loads (These hints apply to analog-output sensors).
National’s temperature sensor ICs are micropower circuits, and like most micropower circuits, they generally
have a limited ability to drive heavy capacitive loads. The LM34 and LM35, for example, can drive 50pF without
special precautions, while the LM45 can handle 500pF. .

If heavier capacitive loads are anticipated, it is easy to
isolate or decouple the load with a resistor; see Figure 4.5. Note that the series resistor will attenuate the output
signal unless the load resistance is very high. If this is a problem, you can improve the tolerance to capacitive
loading without increasing output resistance by using a series R-C damper from output to ground as
shown in Figure 4.5

Figure 4.5. Capacitive drive options. The LM34, LM35, and LM45 can drive large external capacitance if isolated
from the load capacitance with a resistor as in (a), or compensated with an R-C network as in (b).
The LM50 and LM60 have internal isolation resistances and can drive any value of capacitance with no stability
problems. Ensure that the load impedance is sufficiently high to avoid attenuation of the output signal.

Could anyone explain the bold and underlined part in detail? Thanks.

regards,
jordan76

 

[jordan76]

Sorry, I forgot to attach the figure.

regards,
jordan76

 

[starguard2527]

i also have the same question. and m a newbie in electronics. please some one explain us....

 

[Tweedle_Dee]

Me too. In general, how does a series resistor isolate the capacitance of a wire or trace? I hear this all the time at work and I've never understood it.

 

[steveelliott]

Me too. In general, how does a series resistor isolate the capacitance of a wire or trace? I hear this all the time at work and I've never understood it.

Good question. Let me explain why capacitance makes (nearly all) negative feedback systems unstable, then I'll explain why the resistance works.

First a lesson in feedback systems:
A negative feedback system is one where an output signal is compared with an input command and the output is adjusted until they match (or as close as the system can make them).

Background of opamp:

The controlling equation for an opamp (not for the circuit, just the op-amp itself) is Vout = (Vnoninverting - Vinverting)*Gain_open_circuit. Now Gain_open_circuit is a BIG number, numbers of 100,000 to 1,000,000 are normal.

Assume you have an opamp with an open circuit gain of 1000000 and without any circuitry around it. If you put 2.000000V on the inverting input and 2.000007V on the non-inverting input, how many volts will you get on the output?
Vout = (Vnoninverting - Vinverting)*Gain_open_circuit
Vout = (2.000007V - 2.000000V)*1000000
Vout = (0.000007V)*1000000
Vout = 7V

If you put 2.000000V on the inverting input and 1.999999V on the non-inverting input, how many volts will you get on the output?
Vout = (Vnoninverting - Vinverting)*Gain_open_circuit
Vout = (1.999999V - 2.000000V)*1000000
Vout = (-0.000001V)*1000000
Vout = -1V

How about if I put 2V on the inverting input and 3V on the non-inverting input, will I get 1,000,000 V on the output? Of course not. How could it without a 1000000V power supply? What's wrong with our equation then?

Vout = (Vnoninverting - Vinverting)*Gain_open_circuit

Nothing's wrong with our equation - when the output voltage reaches the power supply voltage (+/-15V or 5V or 12V or whatever is in your system), the Gain_open_circuit drops to a very low value (like 1 or 10), whatever it takes to make Vout = Supply voltage (more or less)

Example circuit:
OK, so let's use our opamp in a voltage follower circuit now and see why engineers LOVE opamps.

In an op-amp voltage follower, the output of the opamp is connected back to the inverting input of the opamp (the - pin) and the input signal is connected to the non-inverting input of the opamp (the + pin).

So in the follower example, Vout = Vinverting because they are tied together with a wire. If we put that into our equation for the opamp we get:
Vout = (Vnoninverting - Vinverting)*Gain_open_circuit
Vout = (Vnoninverting - Vout)*Gain_open_circuit

Now let's solve for Vout:
Vout = Vnoninverting*Gain_open_circuit - Vout*Gain_open_circuit

Vout*(Gain_open_circuit+1) = Vnoninverting*Gain_open_circuit
Vout = Vnoninverting*(Gain_open_circuit/(Gain_open_circuit+1))

If Vnoninverting = 3.000000V, the Vout=(1000000/1000001)*3V = 2.999997V

Nearly the same -- too small a difference to measure with most instruments! A lot of circuit effects degrade the perfect result, but we can mostly ignore the gain of the opamp - 100k, 1M what's the difference? 2.99997 or 2.999997 - nothing As engineers we always assume the Vnoninv and Vinv input are at the same voltage for quick circuit analysis; this is true as long as the output voltage is normal (not at the positive or negative supply).


Real World and Feedback:

OK now back to the real world and capacitive feedback. The opamp has a measurable output resistance. If it didn't, you would be able to pull millions of amps from a little tiny plastic package. The opamp also takes time to respond internally to changes in the Vnoninv-Vinv signal (it has its own capacitance inside). The internal capacitance also decreases the open circuit gain with increasing frequency; in fact the open circuit gain of most opamps has decreased by a factor of 10 by at 10Hz input frequency. The open circuit decreases linearly with frequency, at 100Hz the gain is one tenth of the 10 Hz gain, at 10kHz, the gain is 1/1000th of the 10Hz gain and so on.

So if we add a little capacitor to the output pin, that capacitor is charged up through the output resistance of the opamp. That means the signal is delayed just a little. No big problem right? The opamp may overshoot the right point, but it will settle down.

But what if the internal delay of the opamp and the added capacitance delay add up to say 50 microseconds? Let's pretend everything is a squarewave (high or low) so I can explain it more easily.

Let's say the capacitor voltage is too high and is detected by the noninverting-interting signals. After a short time due to internal delays in the opamp, the internal output of the opamp (before the opamps's internal output resistance) goes low. During this short delay interval, though, the output voltage on the capacitor has been increasing because the internal output is still high. Now the capacitor finally starts discharging, because the internal output has gone low.

A while later (a little less than 50 microseconds), the capacitor is discharged to the perfect point where the noninv and inv inputs are perfectly balanced, and the opamp trims the internal output voltage to the perfect point as well. But wait, it took some time to set the internal output voltage, so during the delay the internal output voltage and capacitor voltage actually went BELOW the balance point, so now the opamp wants to make the output go up to the perfect point. The output keeps going lower until the noninv-inv error signal makes its way through the opamp, and the output capacitor starts charging UP. But it's going to overshoot again (50 microseconds later), and the cycle repeats. So every 100 microseconds (10 kHz) we get a complete undershoot and overshoot.

So what do we do about this? Well if we get rid of the the output capacitor it won't delay the signal going back to the inverting input, and the opamp can respond much faster to the overshoot. In fact the design of the opamps is such that they have very low open circuit gain near the frequency that corresponds to their internal time delay. The low gain means they just ignore the fast error signal.

What if we put a resistor in series with the capacitor, but take our feedback from the opamp side of the resistor? The opamp doesn't know that the capacitor is there, since its feedback is from someplace else... Happy opamp!!! (If we took our feedback from the capacitor side we would make it MUCH worse - see schmitt trigger logic gate oscillators like Resistor/Capacitor plus 74C14)

OK, so how does the damper work? The resistor is much smaller than the opamp's output resistance, so when the output tries to change quickly, the feedback signal is taken from the output resistance-damper resistance divider and is much smaller than normal. (The capacitor doesn't have an effect on AC, its value is essentially the same). Hopefully the opamp doesn't have enough gain to amplify the signal more than it is divided-down. These are a lot trickier to work with because they have upper and lower limits for both the capacitor and resistor.

Now the LM34:

The LM34 has an internal micropower opamp (its gain is very low and its output resistance is very high compared to a normal opamp). When the output voltage pin is not what it's supposed to be (representing the temperature) the opamp drives the output voltage pin to the right point.

If there is capacitance on the output pin, the opamp signal is delayed (like in the previous example). To prevent the delay, a resistor is added to allow the internal opamp to reach its proper point quickly. If the resistance is too small, the load is too high for the opamp and it can't get to the right point in time and the internal feedback causes oscillation. A high value of resistance, though, lets the LM34 work without oscillation.


Hope this helps.