What is the purpose the capacitor here?

Thread Starter

imbaine13

Joined Oct 6, 2013
67
Hello guys,

I don't entirely get the purpose of C4 in this circuit. Correct me if I'm wrong, but does it supply the collector-emitter current through the PNP transistor when the 555 output is low?
If the output of the 555 ic is on for just as long as it is off, (50%duty cycle), and C4 charged fully, with still some time to spare when pin 3 is high (in other words, if it took the capacitor for example 0.1 seconds to charge, and the output is high for 0.5 seconds), wouldn't that mean that when the output is low, the PNP transistor will conduct the stored charge for 0.1 seconds (and for the remaining 0.4 seconds, there will be no current through the PNP collector-emitter) before the NPN turns back on?
OR is there some kind of play between the capacitor and the transformer primary? Inductor--Capacitor circuits:confused:

Also,
I have come to realize that the bjt astable multivibrator doesn't exactly oscillate at the frequency calculated from the formula f=1/1.4RC (practically). This was in my other post (bjt astable multivibrator oscillating frequency?). I was wondering which is more reliable (maintains frequency with changes in variables i.e. temperature, loading...). Is it the 555 timer IC or the bjt astable multivibrator.
I plan on using one of them for the driver circuit of an inverter.
12-to-240v-inverter.jpg
Appreciate your time.
 

t06afre

Joined May 11, 2009
5,934
The setup is a very crude AC square wave generator from a single supply source. The purpose of C4 is to act as a power supply to T1 when T2 is not conducting. This is possible since the capacitor charges to a dc value of VCC/2 at the connection of the two emitters. You will often see this setup in single supply audio amplifiers.
 

BillB3857

Joined Feb 28, 2009
2,570
The purpose of C4 is to act as a power supply to T1 when T2 is not conducting. This is possible since the capacitor charges to a dc value of VCC/2 at the connection of the two emitters. You will often see this setup in single supply audio amplifiers.
I agree with the first part of the quoted statement. That fact will result in a direction reversal of current in the transformer primary.
I'm not sure I totally agree with the second statement about Vcc/2. Depending upon the value of the capacitor and the frequency, wouldn't it be possible for the voltage on the cap to reach full Vcc?
 

t06afre

Joined May 11, 2009
5,934
Ah yes it will be a lower cut-off frequency. The lower cut-off frequency is determined by the value of the output coupling capacitor C4 and the load caused by the transformer.
 

crutschow

Joined Mar 14, 2008
34,282
As Mike noted, the purpose of the capacitor is to block the DC component of the uni-polar square-wave so the transformer doesn't saturate. The result of this is to give a waveform polarity as to6 noted. The DC value at the C4 input depends upon the duty-cycle of the waveform. It would be Vcc/2 for a 50% duty-cycle. Normally C4 would be large enough so it has no significant effect on the signal amplitude.
 

t06afre

Joined May 11, 2009
5,934
I guess the most correct thing is to say that the C4 serve two purposes here, as mentioned. In this setting I tend to lean against the fact that "power supply" role is the most important design criteria in this circuit.
 

crutschow

Joined Mar 14, 2008
34,282
I guess the most correct thing is to say that the C4 serve two purposes here, as mentioned. In this setting I tend to lean against the fact that "power supply" role is the most important design criteria in this circuit.
On what basis do you consider that the the most important design criteria? :confused: How do you rank importance?
 

t06afre

Joined May 11, 2009
5,934
That is easy! Picture you a single supply audio amp with a push-pull output stage If want to create a DC block a much lower value of the cap could be used in the same postion as C4. But would create a very distorted output. The schematic posted by the OP is one of those crude attempts to make an inverter. Again how well would that design work with a 10 or 100uF cap. Given that the 555 is tuned to give a 50 or 60 Hz output
 

crutschow

Joined Mar 14, 2008
34,282
That is easy! Picture you a single supply audio amp with a push-pull output stage If want to create a DC block a much lower value of the cap could be used in the same postion as C4. But would create a very distorted output. The schematic posted by the OP is one of those crude attempts to make an inverter. Again how well would that design work with a 10 or 100uF cap. Given that the 555 is tuned to give a 50 or 60 Hz output
Of course the cap has to be a large value. The cap value is always made large enough to carry the AC signal essentially undistorted to drive it's output load. That's SOP for any capacitor used as a DC block.
 

AnalogKid

Joined Aug 1, 2013
10,986
A 555 oscillator is way more temperature-stable than the standard 2-transistor multivibrator because it depends on the ratios of components rather than absolute values. The MV freq wanders as the transistor Vbe changes with temperature. By its internal design a 555 removes itslef from the basic stability equations by comparing the ratio of two external components to the ratio of two internal resistors. If the external R and C tempcos are well matched, a 555 can deliver excellent stability over time and temp.

ak
 
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