I have a 24VDC supply, and a generator that outputs a 1KHz square wave with an amplitude of 10V referred to ground. How can i get a square wave of 24V amplitude (referred to ground) ? The signal will be loaded with 20 or 30 mA.
Thank You AnalogKid,Do you want to buy something, or build a small circuit from scratch. Without more details, this sounds like two transistors, two 15V zener diodes, two resistors, and one capacitor.
How "square" of an output do you need? Risetime, over/undershoot, etc.?
ak
Thank You ScottWang,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
The false load means that it just provide a ground for the output signal, but maybe the value will be too high for the device, if the input of device has its own ground then there is no grounding problem, otherwise the false load should be reduce the values to suits the device.Thank You ScottWang,
very interesting for my needs. What do you mean for "flase load " ?
Regards,
Thank You MikeML,
The load are the digital inputs of a group of PLC i need to test togheter. The load is resistive and is betwwen 4 to 28mA.Both circuits posted have vastly asymmetrical output impedances. The rising edge output impedance is almost zero, while the falling edge impedance is either 1 k or 47 k, or left to the load to discharge itself. We still don't have load details.
ak
I usually use 2SC945, 2SA733, because I can buy they from local stores easily, I labeled the 2N3904 and 2N3906 that it just for the members may easy to buy or already have some.For a resistive input to a plc, it is more than likely that an active high side driver (the level shifter of post #8) is just fine. R1 in my circuit repesents the input to the plc.
If I was building it, I would go to my junk box and pull out a 2n3906 and 2n3904. There are probably 10,000 transistors that would work in that circuit...
ps, I just noticed that Scott posted almost the same circuit.
What is "cross-conduction protection" ? A protection of the: output stage, input stage or load ?This is a symmetrical driver with cross-conduction protection. The output resistor can be adjusted to a compromise between max output current and required voltage across the load.
ak
Very well explained, thank you.The output stage has one transistor to pull the load up and another one to pull it down. This is called a totem-pole output stage. One possible problem is that both transistors might be on, or partly on, at the same time, appearing as a very low impedance or dead short across the circuit power supply. This is called cross-conduction or shoot-through. To prevent this, you want to make sure one transistor is off before the other comes on. Because the two zener diodes add up to more than Vcc, the current path through both emitter-base junctions and both diodes never is forward biased, so cross-conduction is prevented.
ak