9
\$\begingroup\$

enter image description here

The circuit is taken from Analog Dialogue, Volume 3, March 1969, the article is titled Noise and Operational Amplifier Circuits. The circuit is a ramp generator, but I can't figure out how it works. The transistor at the top is a current source for the zener, but what's the point of the zener here? Also, why does it need the 2 PNP transistors? I assume one is to discharge the capacitor.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ @periblepsis Here it is \$\endgroup\$
    – ganymede
    Commented May 17 at 22:26
  • 1
    \$\begingroup\$ The text tells you what the zener does. It's a zero-TC zener family (cool!) and it has 6.2 V across it. What can you work out from there? It's kind of easy to see. So I'd like you to look closely at the English text description and the schematic surrounding the buffer and charging cap. The amp is a follower. What does its output follow? \$\endgroup\$ Commented May 17 at 22:41
  • \$\begingroup\$ That is an interesting take on a ramp generator. Haven't seen this one before. \$\endgroup\$
    – AnalogKid
    Commented May 18 at 0:58
  • 1
    \$\begingroup\$ RE the 7.5mA current source: The 2N1131 is drawn incorrectly as an NPN, in reality, it is a PNP. I suspect the collector is connected to the output of the op-amp. Checking the other semis: yes they are drawn correctly: the 2N2944 is PNP. 2N930 is NPN. 1N825 is a 6.2V zerner. 1N457 is a silicon diode. \$\endgroup\$ Commented May 18 at 1:09
  • \$\begingroup\$ I think you may have the title incorrect; it seems to be a brief circuit snippet titled "An op-amp and a chopper give precision ramp generator", it just happens to be on page 4, embedded within the article you mentioned. Page 1 of the article says "Cont'd on page 5". \$\endgroup\$ Commented May 18 at 1:38

3 Answers 3

10
\$\begingroup\$

The article published with the schematic explains the operation quite well, I have copied the article below via screenshots from this link:
https://www.analog.com/media/en/analog-dialogue/volume-3/number-1/articles/volume3-number1.pdf#page=1

Written by an instrumentation engineer, I suspect; the bit about "if necessary in an oven or thermo-electric bath" was a bit of a give-away. This is a serious bit of kit, with good precision: "The linearity check showed no deviation from a straight line within the limits of its 1mV tolerance."

Here is my explanation for its operation:
As a starting point, the block diagram published in that article is helpful (but possibly confusing due to the change in polarity, as I will explain later):
enter image description here

The precision current source is set up by a zener, a resistor and an op-amp. The opamp output moves the voltage of one side of the zener in sympathy with the voltage of the capacitor, to keep the voltage across the resistor constant as the capacitor voltage changes. That is what is meant by the term "in a bootstrap fashion". The current into the capacitor is the same as the current in the resistor, assuming the current flowing to the other two nodes is zero (the other two nodes are the switch, and the op-amp input).

Referring now to the schematic: the polarity of the current source in the schematic is such that it pulls the timing ramp node toward the negative rail, not the positive rail, in contrast to the block diagram, which could be a source of confusion.

The rest of the circuit is quite simple: 2N1311 and 2N930 set up the bias current of the zener, not the op-amp, nor the capacitor ramp current. The 2N1131 is drawn incorrectly as an NPN, in reality, it is a PNP (I suspect its collector is connected to the output of the op-amp). This part of the circuit is intended to be adjusted for precision, just look at the bias method of the 2N930: it relies on the leakage current through a reverse-biased silicon diode! Obviously, this is a circuit intended to be used for precision application in temperature controlled environment - not for general use. Refer to the note on the 500-ohm pot: "Adjust to zero out with op-amp out".

Also, why does it need the 2 PNP transistors? I assume one is to discharge the capacitor.

Well, both of the PNPs discharge the capacitor; the inverted PNP is added to bring the ramp voltage closer to 0V compared to what can be achieved by the normally connected PNP. The 2N2944 is characterised for inverted-mode use (emitter and collector swapped, aka "reverse active mode"); the benefit is lower Vce(sat) at the cost of lower current gain. Here is the heading of the datasheet, and a snippet showing the low Vce(sat) (~ 1mV).

enter image description here

enter image description here

Motorolla AN-470 is a good resource on this topic (BJT inverted-mode):

https://dn790000.ca.archive.org/0/items/Motorola-SeminarsandApplicationBooksAN-470BipolarChopperTransistorsAndCircuits/Motorola-SeminarsandApplicationBooksAN-470BipolarChopperTransistorsAndCircuits.pdf


Here are the screenshots from the article:

enter image description here

enter image description here

The caption to the image is also helpful: enter image description here

\$\endgroup\$
5
\$\begingroup\$

The transistor at the top (the PNP drawn incorrectly) is not the current source for the zener- that job is the bottom transistor. They are balancing the currents so that the op-amp need supply almost no current. This is likely a trick to keep chip heating during the ramp from affecting the op-amp Vos and causing distortion of the ramp.

The Zener diode is a reference type with very low tempco when operated at the 7.5mA current.

The zener voltage appears across the trimpot and tempco matched 47.5kΩ resistor to create the constant current that charges the capacitor to a negative voltage.

The use of parallel forward and inverted transistors is, I think, another trick to reset the capacitor to the minimum voltage. The 2N2944 is a "symmetrical" transistor to begin with- note the very high emitter-base breakdown voltage (equal to the collector-base breakdown) and the guaranteed very low 'offset voltage' (Vce(sat) @ Ie = 0).

\$\endgroup\$
4
  • \$\begingroup\$ Hmm, good description. The top transistor is the source of the current and is set up as a constant current source. The bottom transistor is working to set a 'bias' or 'rest' level by using the the slope of the Vce or Vcb line at the chosen Zener current to hold the output node constant when the input voltage is at the non ramping level. A high impedance adjustable current sink. Lots of clever stuff was done to save a few transistors, these days we drop half a million transistors to fix a 2 to 3 transistor problem. Needs some simulation I suppose. \$\endgroup\$
    – KalleMP
    Commented May 19 at 18:22
  • 1
    \$\begingroup\$ @KalleMP The top current source goes directly to the op-amp output pin so to first order has no effect at all. A lot of the cleverness demonstrated here is to deal with not-so-good parts using (high precision) analog techniques. Most of the cleverness has moved onto the chip these days. We don't worry too much about current with a modern shunt reference such as an LM4040 because they are so insensitive to current (both voltage and tempco). If you're looking for ppm-level performance the techniques are still worth keeping in mind. Even ovenizing if you have to. \$\endgroup\$ Commented May 19 at 19:53
  • 1
    \$\begingroup\$ You have a good point, the OP-amp determines the output voltage but the top transistor is trying (except for the top resistor that adds supply voltage sensitivity) maintain a constant current shared between the Zener and the OP-amp The bottom transistor again is trying to set the voltage level of the virtual earth while maintaining a constant current with the Zener and charge resistor (1:500 ratio). In essence the Zenner is held between the two current sources to avoid the OP-amp from having to provide the stable Zener current. \$\endgroup\$
    – KalleMP
    Commented May 19 at 20:36
  • \$\begingroup\$ Come to think of it, that strategy might fail if the amplifier was a type with crossover distortion. \$\endgroup\$ Commented May 19 at 23:37
2
\$\begingroup\$

what's the point of the zener here?

A constant voltage (the zener) across a constant resistance (475K + (0 to 5K)) yields a constant current.

A constant current into/out of a constant capacitance yields a straight-line ramp in capacitor voltage. Apparently, a really Really straight line.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.