At the moment, i am confused about the behaviour of my LTSpice simulation, not by the capacitor.I am surprised at the persistant confusion caused by one capacitor.
If you want to compare the digital and analog signal, maybe you could do some test, you can using a square wave as 1Hz, 10Hz, 100Hz, 1Khz, 10Khz, 100Khz, 1Mhz to adding to the circuit you provided and adding to a 74HC04(2V~6V)) with 6 inverters, but you only need to use two and one in series with another, using the O'Scope to measure the output waveform, and to see what's their difference.I am surprised at the persistant confusion caused by one capacitor.
Ahh, I see what is bothering you. The bottom transistor doesn't turn on for the first few pulses, but the load brings the output voltage back to ground. With no load or the load returned to +24 the pulses will magically disappear. As a matter of fact I suppose you could eliminate the bottom transistor and the cap and just drive it that way.At the moment, i am confused about the behaviour of my LTSpice simulation, not by the capacitor.
Also with an input square wave of 20Hz i can't reproduce the absence of the first two or three periods in the Vout.
Yes ronv, exactly as you explained, here the LTspice result. Now let me put this in a breadboard.Ahh, I see what is bothering you. The bottom transistor doesn't turn on for the first few pulses, but the load brings the output voltage back to ground. With no load or the load returned to +24 the pulses will magically disappear. As a matter of fact I suppose you could eliminate the bottom transistor and the cap and just drive it that way.
Although the analog amplifier can be amplifying the digital signal, but they still have some difference in different frequencies, specially is the waveform.an analog amplifier will amplify a digital signal too. if you want a 50-50 duty cycle 1khz square wave, run a 2 khz square wave thorugh a cmos flip flop first, that will turn it into a 50-50 duty cycle 1 khz.
I apologize for this delay.Yes ronv, exactly as you explained, here the LTspice result. Now let me put this in a breadboard.
Sorry : these are the correct attachments.I apologize for this delay.
I put the circuit in a breadboard (the only difference is that i used an BC337-25 instead of the 2N4401) and supply it at 24V and a 10Vpp input square wave. The output doesn't show any lost cycles at the beginning of the capture. The same at lower frequency then 1KHz. I see that "lost cycles" only when i lower the amplitude of the input signal, 8V and 7V as shown in the pictures.
Final results: the circuit proposed by Analog Kid is fine for 1KHz 10Vpp input square wave: it gives the desired output. It is just a little more difficult to assimilate, but it is stimulating.
Every consideration is welcome.
I know mine works because I simulated it and have built it in various incarnations, lots of times. I would have to analyze the other circuit...
Mine has a higher input impedance (loads the source less).
Mine is shown driving a 1000Ω (24mA load) while the other circuit shows a 47K load.
I put in a breadboard the circuit you proposed, and the output was exactly as espected.The circuit is a standard of voltage converter, the output current calculating as this:
Vce_Q1 = 0.2V
Vbe_Q2 = 0.7V
Vce_Q2 = 0.2V
I_R4 = (24V - Vbe_Q2 - Vce_Q1)/R4
= (24V - 0.7V -0.2V)/4.7K
= 23.1V/4.7K
= 4.91 mA
Ic_Q2 = I_R4 * 10 = 49.1 mA
So if you want to reduce the current then you can increase the value of R4, or you want to increase the current then you have to reduce the value of R4, you just calculate as above formula.
Vo = 24V - Vce_Q2 = 24V - 0.2V = 23.8V
R5 is a false load.
View attachment 86463
I put in a breadboard the circuit you proposed, and the output was exactly as espected.
by Jake Hertz
by Aaron Carman
by Jake Hertz