How to charger an op amp without additional power source?

Thread Starter

circuitnaiver

Joined Oct 1, 2021
22
Hi, All,

I have this circuit as follows. I only have one power source which is +-12Volts. My input and output point has been highlighted in the figure. My question is that, how can I power up the op amp without additional power source? Here we can suppose the op amp only require a single power supply, such as LM358.

I tried to connect the vcc on the 555 timer (the 555 timer I used in the physical circuit is ne555) to op amp vcc directly, but it seems like it does not work in the simulation. I get stucked here. Anyone have any idea for this?

Besides, since my output voltage is lower than 6volts, which is the reason that I decided to add an op amp here. Is there any solution we can make the output on the 555 timer become 6volts?

Note that, in the phsyical circuit I tested, no matter I remove the R8 (2k ohm) or not, it cannot make the output reach 6 volts. I also wonder why that is the case

Thank you.
1634290611828.png

1634291075669.png
 

ericgibbs

Joined Jan 29, 2010
14,130
Note that, in the phsyical circuit I tested, no matter I remove the R8 (2k ohm) or not, it cannot make the output reach 6 volts. I also wonder why that is the case
hi c,
Because there is only 6V on the 555 Vcc.
E

Update:
Check this 2nd image.
What exactly are you trying to do with circuit.?
 

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Thread Starter

circuitnaiver

Joined Oct 1, 2021
22
hi c,
Because there is only 6V on the 555 Vcc.
E

Update:
Check this 2nd image.
What exactly are you trying to do with circuit.?
Hi, Thank you for your reply.

I want to make the input after 1 K resistor become 6volts -12volts square wave and the output on the 555 timer become 6vlots 0volt suqare wave with the duty cycle being changed at the same time.
 

Thread Starter

circuitnaiver

Joined Oct 1, 2021
22
hi c,
Because there is only 6V on the 555 Vcc.
E

Update:
Check this 2nd image.
What exactly are you trying to do with circuit.?
Hi, I noticed one thing: It seems like NE555 and TLC555 would behave very differently in this case. I do not have these two components in my simulation. But I have checked the previous figure you provided.
1634328191582.png

My phsical circuit is exactly as this design. The vout from NE555 is very small. However, by using TLC555, the amplitude could be much higher at the output pin in TLC555. It seems like by replace the NE555 with TLC555, the output of the circuit can be 6vlot without amplification.
 

salihkanber

Joined Oct 16, 2021
7
Apply PWM to the voltage from supply, output voltage will change when you connect the load, simulations vs real life is also another issue :)
 

Irving

Joined Jan 30, 2016
2,308
NE555 is an older BJT design and draws a lot of current so the diode/capacitor D1/C2 can't deliver enough current. And you loose 0.7v across the 1N4001, a schottky diode wouldgive another 0.3 - 0.5v but thats not enough. A CMOS TLC555 draws much less and would give more output, but that 2.7k output load would still pull the output down with a 3k input resistance.

You need to reduce C1 to 500pF and R1 to 27k otherwise they dictate output pulse width rather than C3R3.

Since your input goes to -12v you could use a voltage doubler to increase the voltage/current available - add C4/D2/D3 as shown below - you also need to reduce R2 - ideally to 0, but less than 100ohm else you can't make it work at all. Some experimentation needed.

1634391577103.png
 

Audioguru again

Joined Oct 21, 2019
3,828
The NE555 is very old and draws a power supply current of 400mA (!) each time the output switches.
The LMC555, TLC555 and ICM7555 are all newer Cmos and do not have that huge current draw and have an output of the supply voltage into a fairly high resistance load.

The antique 741 opamp you are using is 53 years old. In your last schematic it is destroyed with a 60V power supply.
 

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Thread Starter

circuitnaiver

Joined Oct 1, 2021
22
NE555 is an older BJT design and draws a lot of current so the diode/capacitor D1/C2 can't deliver enough current. And you loose 0.7v across the 1N4001, a schottky diode wouldgive another 0.3 - 0.5v but thats not enough. A CMOS TLC555 draws much less and would give more output, but that 2.7k output load would still pull the output down with a 3k input resistance.

You need to reduce C1 to 500pF and R1 to 27k otherwise they dictate output pulse width rather than C3R3.

Since your input goes to -12v you could use a voltage doubler to increase the voltage/current available - add C4/D2/D3 as shown below - you also need to reduce R2 - ideally to 0, but less than 100ohm else you can't make it work at all. Some experimentation needed.

View attachment 250457
Hi, Thank you very much for the detailed explanation. From your right figure (oscilloscope), it seems like the maximum amplitude at output is still just around 2Volts (still similar to the result of my physically tested circuit). My expected input after R1 is high level 6volts (lower level of the PWM does not matter) and output is also 6vlot as well.

Besides, the output resistance is 880 ohms, I wrongly put the switch in my circuit. Sorry for that.
 

Thread Starter

circuitnaiver

Joined Oct 1, 2021
22
The NE555 is very old and draws a power supply current of 400mA (!) each time the output switches.
The LMC555, TLC555 and ICM7555 are all newer Cmos and do not have that huge current draw and have an output of the supply voltage into a fairly high resistance load.

The antique 741 opamp you are using is 53 years old. In your last schematic it is destroyed with a 60V power supply.
Hi, Thank you for the info. I feel the TLC 555 might work in this case. I have many 741s, I have already replaced the destroyed one.
 

Irving

Joined Jan 30, 2016
2,308
From your right figure (oscilloscope), it seems like the maximum amplitude at output is still just around 2Volts
Not sure how you got that, looks like 8v to me? In any case, if your load resistor is 800ohm, irrespective of which 555, your output volts will be lower with that higher resistance feed, depending on the relative duty cycle of input to output pulse.
 
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