# Relaxation oscillator explanation

I am having trouble understanding how this relaxation oscillator works. Can someone explain, why it produces a square wave form, how do the currents flow in this circuit and what is the op amp output each time?

• I'm curious, have you ever seen "I" as "i"? and "..form, how.. " as "..form , how.."? - Either way, have you tried to simulate the schematic? – Harry Svensson Nov 26 '17 at 15:18
• @Harry Svensson yes , but i don't understand the way it works . Thank you for the grammar corrections though... – maverick98 Nov 26 '17 at 15:19
• Cant find a better explanation than en.wikipedia.org/wiki/Relaxation_oscillator .You should learn how to effectively use google. – Mitu Raj Nov 26 '17 at 15:37
• yes , but i don't understand the way it works . Thank you for the grammar corrections though… The irony is strong with this one. – Marcus Müller Nov 26 '17 at 15:43
• @MITU RAJ , I knew how to google stack exchange now, didn't I? – maverick98 Nov 26 '17 at 16:07

How it works:

• On power-up C1 is discharged V- (op-amp inverting input) is a 0. V+ is biased at towards half-supply by R3 and R4 so the output switches high. R2 now pulls V+ a little higher than half-supply.
• C1 now charges up through R1.
• When V- exceeds V+ the output will switch low.
• V+ will now be a little lower than half supply.
• C1 now discharges through R1.
• When V- becomes lower than V+ the output will switch high again.
• Repeat until the power goes off.

What do you mean by switches high and switches low? Figure 1. Internals of the ancient 741 opamp. Source: Wikipedia.

Most opamps will have an output arrangement similar to the push-pull arrangement of the old 741. Others will have FET transistors rather than BJTs. In either case if the top transistor (red oval) is turned on the output will be pulled to positive rail (switches high). If the bottom transistor (green oval) is turned on the output will be pulled to negative rail (switches low). How close they get depends on the exact output configuration and the driving circuitry.

So the op amp chooses if it will be for example +15V(switches high) or -15V (switches low) depending on the input.

Yes, if a +/-15 V supply is used. I suspect that your circuit uses only a +15 V supply because R4 is connected to 0 V (ground).

However what do you mean by saying " V+ is biased at towards half-supply by R3 and R4"? Why is it half supply?

I will explain more below but for now think that R3 is connected to V+ and R4 to 0 V so the connection is half-way or mid-supply, 7.5 V for a 15 V supply.

... and where is the supply in this circuit?

Don't ask me - you posted the schematic! It's not there. Neither are the component values. It is a poor schematic.

The oscillator relies on Schmitt trigger operation. A little "hysteresis" is added by the positive feedback to change the switching point. simulate this circuit – Schematic created using CircuitLab

Figure 2. (a) With a dual +/- supply the hysteresis is provided by R2 and R4 and is connected to ground. (b) With a single-ended supply (+ only) a "mid-supply" must be generated.

• what do you mean by switches high and switches low , my english is not that good – maverick98 Nov 26 '17 at 16:26
• See the update and let me know if you understand. – Transistor Nov 26 '17 at 16:31
• So the op amp chooses if it will be for example +15V(switches high) or -15V (switches low) depending on the input. However what do you mean by saying " V+ is biased at towards half-supply by R3 and R4" . Why is it half supply? and where is the supply in this circuit – maverick98 Nov 26 '17 at 16:35
• See the update. – Transistor Nov 26 '17 at 17:07
• Ok I get what you are trying to say. I have two final questions. First I ran the simulation for the (b) circuit and the op amp Vout was constantly 15V, why is that? And the second question, when the capacitor reaches 7.5V so does the V- , why does the op amp switch low to -15V output? is it the op amps function to do that , or the circuit makes it change?? – maverick98 Nov 26 '17 at 18:24

Maybe you don't know how the op-amp works- here it is intended to be used as a comparator and should be easy to analyze in the two possible output states, however the 'designer' has made a classic error and used an inappropriate component.

The OP-07 is a bipolar design with active bias current cancellation and has 2 series pairs of diodes back-to-back internally across the inputs. So once the differential input voltage exceeds two diode drops the inputs will present about a 1K load between them and the oscillator will work poorly and at a much higher frequency than would be expected.