How to lower the power consumption for a sine-wave generator?

Nick Long

Joined May 12, 2020
75
Hello friends, I want to shrink the sine-wave voltage signal using the passive resistive voltage divider. And I need to keep the resistances of R5 and R6 equal to 2k and 5k ohms. But by doing this, the power disspation on R5 and R6 is relatively large due to the small resistances of these two resistors. Is there a method that I can obtain the same voltage result while keeping the resistances of R5 and R6 unchanged?

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DickCappels

Joined Aug 21, 2008
10,180
I will be very surprised if somebody comes up with a way to take less current to put the same voltage across the same resistance, if I understand your intent.

crutschow

Joined Mar 14, 2008
34,452
Is there a method that I can obtain the same voltage result while keeping the resistances of R5 and R6 unchanged?
No.
Power is determined by the resistance value and the applied voltage.
You need to either increase the resistance or reduce the voltage to reduce the power.
I need to keep the resistances of R5 and R6 equal to 2k and 5k ohms.
Why do those need to be that low?
If you need a low impedance output, then reduce the voltage with a high impedance divider and then buffer the signal with an op amp voltage follower.

DickCappels

Joined Aug 21, 2008
10,180
You can probably reduce the power in the oscillator itself buy multiplying all of the resistors in that part of the circuit by a constant (for example, 10) and dividing all of the capacitors (except bypass capacitors) by the same constant (like 10), but as @crutschow has confirmed the power consumed as a result of the load will not change.

Ian0

Joined Aug 7, 2020
9,817
Not quite sure what you mean, but if you want to keep Vout the same, then you could generate a sinewave of lower amplitude, and remove R5, which would eliminate the dissipation in R5. Have I understood it correctly?
You don't say what you want the voltage to be at R5, but you could start by putting two diodes in inverse-parallel across R4 which would reduce the output to about 1V.
The present circuit is not guaranteed to oscillate, as the gain is exactly three, and it needs to be more than three to get started.
If you do get it started the output will continue to increase until it reaches the supply voltage, 3V peak.

MisterBill2

Joined Jan 23, 2018
18,519
Now it gets confusing: The voltage at "Vout", the junction of the two resistors, is what you wish to hold constant without changing either of those resistance values?
In this world that is not possible even in simulation.
And I am guessing that the resistances are actually 2000ohms and 5000 ohms, as I am not used to that alternate method of expressing common values.

AND, the power spent in those resistors IS NOT "rather large" unless your simulator is thinking that the values are 2.3 OHMs and 5.3 ohms. Six volts across 7000 ohms is not a lot of power. The unusual method of expressing numerical values may also confuse the simulator..

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Alec_t

Joined Sep 17, 2013
14,314
The unusual method of expressing numerical values may also confuse the simulator..
LTspice is perfectly happy with that method. It comes into its own when values are extremely high or extremely small, e.g. way beyond gigas or picos.

MisterBill2

Joined Jan 23, 2018
18,519
OK on the simulator finding that notation OK.. I am not a simulator and I find it uncommon. I also find that those resistor values with that supply voltage does not lead to anything close to relatively high power levels in the resistors. Six volts across seven thousand ohms is less than one mA, about six milliwatts of power. (6.0 V x 0.001 mA= 0.006W)

Ian0

Joined Aug 7, 2020
9,817
OK on the simulator finding that notation OK.. I am not a simulator and I find it uncommon.
You must have owned a Casio calculator in the 1980s? with a vacuum-fluorescent display?

ericgibbs

Joined Jan 29, 2010
18,849
Bill.
How to lower the power consumption for a sine-wave generator?

He is not concerned that they are 'high' power as such, he just wants to reduce the power consumption of the project.

E

Audioguru again

Joined Oct 21, 2019
6,692
The maximum allowed supply voltage for the OPA340 opamp is only 5.5V but the circuit is damaging it or destroying it with 6V.
The sinewave generator had nothing to limit its output voltage swing which will be clipping.

crutschow

Joined Mar 14, 2008
34,452
The sinewave generator had nothing to limit its output voltage swing which will be clipping.
Opposite polarity, parallel diodes across R4 gives a decent sinewave of about 0.6Vpp.

MisterBill2

Joined Jan 23, 2018
18,519
Bill.
How to lower the power consumption for a sine-wave generator?

He is not concerned that they are 'high' power as such, he just wants to reduce the power consumption of the project.

E
First of a, THAT is NOT what the first post asks. It is addressing the power in the two resistor voltage divider.
Next, based on the information about that particular op-amp device, it is being run above it's rated max supply voltage, according to AG. So using a different component is the simple solution. Parts run within their specifications usually perform as specified.
And it may also be driving into saturation, which, I am guessing, will be causing it to draw excess current.

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crutschow

Joined Mar 14, 2008
34,452
The TS visited here a few hours ago but has not replied to clarify anything.

Papabravo

Joined Feb 24, 2006
21,225
The TS visited here a few hours ago but has not replied to clarify anything.
Some TSs are just not the loquacious type.

Nick Long

Joined May 12, 2020
75
I will be very surprised if somebody comes up with a way to take less current to put the same voltage across the same resistance, if I understand your intent.
Yes, you are right. It seems there is no such way to realize that. Thanks.

Nick Long

Joined May 12, 2020
75
No.
Power is determined by the resistance value and the applied voltage.
You need to either increase the resistance or reduce the voltage to reduce the power.
Why do those need to be that low?
If you need a low impedance output, then reduce the voltage with a high impedance divider and then buffer the signal with an op amp voltage follower.
Thanks for your reply. Actually, R5 is the resistance of a sensor. It varies within a small range. What I want to realize is that the change of Vout reflects the change of R5.

Nick Long

Joined May 12, 2020
75
Not quite sure what you mean, but if you want to keep Vout the same, then you could generate a sinewave of lower amplitude, and remove R5, which would eliminate the dissipation in R5. Have I understood it correctly?
You don't say what you want the voltage to be at R5, but you could start by putting two diodes in inverse-parallel across R4 which would reduce the output to about 1V.
The present circuit is not guaranteed to oscillate, as the gain is exactly three, and it needs to be more than three to get started.
If you do get it started the output will continue to increase until it reaches the supply voltage, 3V peak.
Thanks for your reply. Actually, R5 is the resistance of a sensor. It varies within a small range. What I want to realize is that the change of Vout reflects the change of R5. But now the R5 and R6 draw a lot of current from the op amp. It seems I cannot optimize the power consumption further with respect to the output part.

Nick Long

Joined May 12, 2020
75
Sorry for my late reply. There seems to be no way to solve the problem I raised. Thank you for your reply.

Now I got a new problem. I want to control the amplitude of the output sine-wave signal according to an output voltage of a voltage source. The voltage range of this voltage source is 0~3V. I want the amplitude of the sine wave equal to the control voltage. For example, when the control voltage is 1V, the output amplitude of the oscillating circuit is 1V. When the control voltage is 2V, the output amplitude of the oscillating circuit is 2V. Up to now, I only know I can use an analog multiplier or an amplitude modulation circuit to realize it. Is there a simpler method?

crutschow

Joined Mar 14, 2008
34,452
when the control voltage is 1V, the output amplitude of the oscillating circuit is 1V
Is the AC voltage in RMS, peak, or peak-to-peak value.
I only know I can use an analog multiplier or an amplitude modulation circuit to realize it. Is there a simpler method?
Not that I'm aware of.