Maker Pro
Maker Pro

circuit explanation

lotus

Dec 4, 2010
14
Joined
Dec 4, 2010
Messages
14
I have found this circuit in Keith Billings book. Can anybody please explain the working and logic behind this circuit
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Joined
Jan 21, 2010
Messages
25,505
If you posted a link, you might have to try doing it again.
 

lotus

Dec 4, 2010
14
Joined
Dec 4, 2010
Messages
14
I have attached the circuit now
 

Attachments

  • under.pdf
    26.9 KB · Views: 255

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Joined
Jan 21, 2010
Messages
25,505
Weird.

Pass.
 

Resqueline

Jul 31, 2009
2,848
Joined
Jul 31, 2009
Messages
2,848
Definitely a weird circuit by itself, but it needs to be seen in context with both the power supply driving it and the type of load. I don't know what use it could have though.
It's a kind of switched capacitor voltage multiplier. When the voltage is above the treshold, A1 output is high and Q1 does not conduct. R1 charges C1 and R2 charges C2.
When the voltage goes below the treshold, A1 output goes low and Q1 conducts, thereby series-connecting the capacitors, doubling the power supply voltage (for a while).
The process will repeat if the power supply is below the treshold voltage (determined by R3/R4 & Ref V).
There will be a substantial ripple voltage generated, so both the supply and the load will need to be able to deal with this.
 

selva

Apr 26, 2010
25
Joined
Apr 26, 2010
Messages
25
Definitely a weird circuit by itself, but it needs to be seen in context with both the power supply driving it and the type of load. I don't know what use it could have though.
It's a kind of switched capacitor voltage multiplier. When the voltage is above the treshold, A1 output is high and Q1 does not conduct. R1 charges C1 and R2 charges C2.
When the voltage goes below the treshold, A1 output goes low and Q1 conducts, thereby series-connecting the capacitors, doubling the power supply voltage (for a while).
The process will repeat if the power supply is below the treshold voltage (determined by R3/R4 & Ref V).
There will be a substantial ripple voltage generated, so both the supply and the load will need to be able to deal with this.

Hello Resqueline,

I hope u made a mistake in that highlighted text. I think Q1 conducts when A1 output is high. I hope this circuit is to prevent the load from sharp edges, more or less like a basic caps operation, but i don't why they made it clumsy with a op-amp.
Sorry if i have made any mistake.

Warm Regards,
Selva
 

Resqueline

Jul 31, 2009
2,848
Joined
Jul 31, 2009
Messages
2,848
I reviewed the circuit again and I find that I mistook Q1 for a PNP instead of an NPN.. (I would have drawn an NPN the other way around.)
That inverts the function and it could look like it becomes a ripple filter, but..
However; when the voltage is below the treshold, A1 output is low and Q1 does not conduct, R1 charges C1 and R2 charges C2 (to full supply voltage).
When the voltage goes above the treshold, A1 output goes high and Q1 conducts, thereby series-connecting the capacitors, increasing the power supply voltage by 50%.
The process seems not to repeat by itself.
I still don't quite see the point of the circuit, actually even less.. It would be interesting to see some of the text and the rest of the circuits associated with it.
And why don't someone run this in a simulation?
There's no problem in placing questions about anything, even the best can make mistakes, I just rewrote this reply completely more than 3 times! ;)
 
Last edited:

selva

Apr 26, 2010
25
Joined
Apr 26, 2010
Messages
25
Hello Resqueline,

I just made that question to make sure, that operation which i understood is correct :D. I would like to try it in simulation, but i get stuck with pspice simulator, i repetitively get a same error (i forgot that error).

I'll try again with pspice and if i face any error again, i'll let u know. By the mean time can u suggest me any other simulator for windows as well as ubuntu.

Warm regards,
Selvam Anand:)
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Joined
Jan 21, 2010
Messages
25,505
Yeah, I was going to try to simulate it, but I have not had the time.

I would recommend that you have a non-zero impedance of your source voltage. This may be modelled in the simulation, or you can put a resistor in series with the power supply.

Start with 1 ohm.

You will also probably want a load on the output. Start with 100 ohms.

Maybe add some ripple to your input voltage too.
 

lotus

Dec 4, 2010
14
Joined
Dec 4, 2010
Messages
14
when i use square wave(8-14V) as source then it dont remove ripple
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Joined
Jan 21, 2010
Messages
25,505
What is the impedance of your source?
 

Resqueline

Jul 31, 2009
2,848
Joined
Jul 31, 2009
Messages
2,848
Note that we didn't say it would remove ripple, only it might be interesting to see what might happen to the ripple (on top of a varying voltage).
Also, as the circuit most likely outputs squares/spikes it would be more advantageous to test it with a sine or a triangle wave.
And lastly, what did you use for component values? Did you make certain that the divider voltage exceeded the reference voltage?
 

lotus

Dec 4, 2010
14
Joined
Dec 4, 2010
Messages
14
I hav used these values
R1 = R2 = 100 ohms
C1 = C2 = 10uF
R3 = 50k ohms
R4 = 106.4k ohms
Vref = 6.8V
Load = 100 ohms
Vinput = 8-14 V (pulse with Tr = 1ms)
 

Resqueline

Jul 31, 2009
2,848
Joined
Jul 31, 2009
Messages
2,848
Ok, so R1*C1 charge (& discharge) time constant is 1ms, and the switchover point for the comparator is 10.0V. I'd say the load is a little on the heavy side, I'd try 1k.
Also you didn't reply to steve's question about the source impedance. If that is zero then no response will be had no matter what. Put a diode in series with Vinput.
Vinput should also not be a square (or any pulse form) in order to see what might happen. Make it a relatively slow sine or triangle varying around 10V (+/- 2-4V).
 

lotus

Dec 4, 2010
14
Joined
Dec 4, 2010
Messages
14
I hav attached the simulation results
 

Attachments

  • sim.zip
    18.9 KB · Views: 97

Resqueline

Jul 31, 2009
2,848
Joined
Jul 31, 2009
Messages
2,848
How come you have put a capacitor across the load, being 10 times larger than C1 & C2? It was not there in the original diagram..
The red & blue curves shows the same thing of course, but I don't understand why the two yellow curves follow each other perfectly..
And then there's the source impedance, being extremely low. How about using a diode there also?
 

lotus

Dec 4, 2010
14
Joined
Dec 4, 2010
Messages
14
that capacitor across load is the filter capacitor. there is only one yellow curve which is Vgs
 
Top