Pulse stretcher vhdl

PWM Generator in VHDL with Variable Duty Cycle

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It only takes a minute to sign up. I have a fast clock and a switch called 'ready'. When the switch is flipped ready goes HIGHI would like the output pcEn to produce a pulse that lasts only for one clock cycle. The following code simulates correctly, but is unfortunately not synthesizable. Note that space is an important consideration, so I was wondering if it's possible to implement it without using a state machine which uses a lot of logic elements. My usual technique is to implement a 2-stage synchronizer to bring the asynchronous input in to the clock's timing domain, and then use one more flip-flop as the edge detector.

Depending on the logic you use in the last statement, you can detect rising edges, falling edges or both. If ready is externally driven it's worth cleaning it up to align with a clock edge with a single register stage first. There's no need for the reset. If you are going to count the output, you also need to address switch bounce.

To address Daves's comment: modified to have to registered output pcEn and to trigger on rising edge. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. How to efficiently implement a single output pulse from a long input on Altera? Ask Question. Asked 6 years, 1 month ago. Active 2 years, 8 months ago. Viewed 7k times. Tools mostly encode them to one-hot and use one register per reachable state.

Control logic tends to be dwarfed by data path multiplexers, adders etc. At the moment my adder is consuming tons of logic elements. Is there a away to divert these resources to the RAM or embedded multipliers they are cheap in my case? However, it's just a simple adder with no carry in or out. Basically a bunch of XOR gates.

Active Oldest Votes. Dave Tweed Dave Tweed k 11 11 gold badges silver badges bronze badges. How does this compare to shuckc's modified answer above? It seems to use more logic elements than necessary. This is the minimum required to deal correctly with an asynchronous input. I've never used one before. I'll see how it goes. Here's a good reference on level to pulse converter which also explains other implementations for the above logic. Also, the output pulse will be less than one clock wide.Remember Me?

Simple Pulse Stretcher out of one Flip-Flop. Simple Pulse Stretcher out of one Flip-Flop Hi, I searched for a similar topic, but all I could find were some unanswered threads and because I don't necessarily want to hijack someone's topic and I also have an approach, I started my own.

pulse stretcher vhdl

But now to my problem. I have to stretch an UWB pulse with some hundred ps up to at least 2. The amplitude of the pulse can be easily adjusted, so no problem there. Well I thought, perhaps I can abuse the propagation delay of a flip flop to stretch my incoming pulse to a much wider one. My idea is to feed the outgoing signal back to the reset pin of the flip flop which has a propagation delay of about 3. This would be this one: sn74lvcad by Texas Instruments. The Data Sheet can be easily found over Google I'm sorry, but I cannot post links yet and don't want to be banned because of cheating this link into the post by removing the protocoll prefix.

But honestly I'm a little unsure about the minimum pulse width mentioned in the data sheet which would be 3. On the other hand I have read that this only concerns the clock and not the actual input pulse. I don't have much experience in flip flops, so perhaps could someone tell me if this would even work or otherwise help me obtaining my goal Best, Fiech. Last edited by FIech; 9th November at Reason: Grammar Issues. Why don't explain the original problem and there may be other solution than make an "pulse extender".

I think this problem is possible to solve just using regular logic gates, But I cannot come up with something smart right now. I hava another very simple circuit that generates very short spikes.

How to implement a PWM in VHDL

But it won't be longer than initial pulse so that idea won't help much I'm afraid. Unfortunately there is no way around a pulse stretcher because this is for a project at my university. Also it's not my own project, so I can't just change parameters as I like. You see, I have some fix parameters: For one there is the initial pulse width of some hundred ps and also the minimum pulse width my TDC can handle which is as above 2.

And this I found out shortly after my post could also be a possible solution: The comparator which is an ADCMP, an ADCMP or an ADCMP, all made by Analog Devices and again no links, but this time I will double post to get you the linkshas the ability to latch the output, which means it will stay on the last output for as long as the latch is active. This I can use as described in the link which will be in the upcoming post to stretch the pulse to a varible length.

pulse stretcher vhdl

Perhaps you or someone else could help me with this approach instead. These are the two ways I imagine how I could solve my problem. Last edited by FIech; 10th November at Hi, Could someone suggest me a circuit to get the below ouput from the input? Re: pulse streatcher. One very crude way is to instantiate some delays between in and indlyd. Or declare indyd as register and clock in through some number of cycles.

This depends on whether you have a clock to which in is synchronous or how wide you want the pulse to be etc. If Input is high, set output high. Do not set Output high, if the stored information says, that Input was already high. Hint: In the 2nd. Finally: This is a "pulse compressor", not a "pulse stretcher" Ralf. BB code is On. Smilies are On. Trackbacks are On. Pingbacks are On. Refbacks are On.

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The total will be updated once daily. You may need to clear your browser cache to see the updates. Two different ways to code a shift register in VHDL are shown.

A register stores data i. A shift register has the capability of shifting the data stored in the register from left to right or right to left. Shift registers consist of D flip-flops as shown in the figure below. This is a four bit shift register and therefore consists of four D flip-flops. This shift register is configured to shift data from the left to the right.

Data is fed into the D input of the first flip-flop on the left. This data can be either a 0 or a 1 and will be shifted to the right on each rising edge of the clock pulse.

Mechanics/Redstone/Pulse circuit

Whatever the state of the data input when the rising edge of the clock pulse occurs will be the logic level that is shifted into the first flip-flop. The data in each flip-flop will be shifted to the flip-flop on its right when the rising edge of the clock pulse occurs.

The image below shows an eight bit shift register that is created in VHDL code in this tutorial. It is also possible to shift data from right to left and to use the LSB as an input for serial data. There are two examples of a shift register written in VHDL below. The two different examples create the same shift register using slightly different VHDL code. This video shows the VHDL shift register in action. Can't see the video? These shift registers are both serial to parallel shift registers as they take the serial data from the D input and shift it for display on 8 LEDs in parallel.

This register is initialized with the value of 00h so that when power is switched on to the CPLD board, the register will be cleared. The shift register has a D input for serial data.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. The dark mode beta is finally here. Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information.

UART works fine, it sends the 8 bit message along with a flag to a control unit. My question here is, how can a signal flag in this case trigger a pulse just for one clk period?

The trick to making a single cycle pulse is realising that having made the pulse, you have to wait as long as the trigger input is high before getting back to the start. Essentially you are building a very simple state machine, but with only 2 states you can use a simple boolean to tell them apart. Morten is correct about the need to adopt one of the standard patterns for a clocked process; I have chosen a different one that works equally well.

There are several issues to address in order to make the design for a one cycle pulse using flip flops registers. First, the use of flip flops in hardware through VHDL constructions typically follows a structure like:. Below is a way of creating a signal flag2 that lasts exactly one clock period from a signal flag1 that lasts at least one clock period. I am new to verilog and this is the sample code, which I used for triggering. Hope this serves your purpose. You can try same logic in VHDL.

The way to achieve this is to create a debounce circuit.

pulse stretcher vhdl

If you need a D flip-flop to change from 0 to 1, only for the first clock, just add an AND gate before its input like the image below: So here you can see a D flip-flop and its debounce circuit. Circuit created using this. The Rise, Edge and Fall outputs will strobe for one cycle when those events are detected. Inputs and outputs are synchronised for use with a Finite State Machine. Learn more. ONE clock period pulse based on trigger signal Ask Question.

Asked 6 years, 4 months ago. Active 1 month ago. Viewed 18k times. David Kester David Kester 51 1 1 gold badge 2 2 silver badges 10 10 bronze badges. I only started studying vhdl and there is some magic for me, that's why I answer in the comment. Try remove clk'event from elsif. Sometimes it's helped me.Let the counter clock to be for example 50 MHz.

The clock period is 20 ns. Even if you are very very fast in pushing the button it will be difficult to generate a pulse of 20 ns in order to enable the counter for only one clock cycle. As you can see we need another solution than trying to push the button very very fast!

In this case, we have the opposite problem: the control signal shall remain high at least 2 clock cycle remember Nyquist theorem?

In the simulation of Figure3 is clear that the circuit generates a pulse of only one clock cycle, no matter how long is the control signal. Surfing the web, you can find many examples of rising edge detector as reported in Figure4.

Let see a simulation of this circuits. As you can see, if the edge of the input signal is very close to the internal clock edge the pulse generated is very short as shown in Figure5 and Figure6. Moreover, it is possible to lose the detection due to the internal physical delay of the circuit. As a drawback, this solution will introduce a delay of one clock cycle in the output of the circuits. I appreciated your article as a beginner in vhdl, however I think it could have been made much clearer.

Two pedagogical mistakes: — The vhdl code does not correspond strictly to the preceding diagram.

How to make a 1Hz Clock (VHDL)

When reading the code for the first time we cannot help wonder what is this reset doing there? It just disturbs our mind which is focused on the main issue of the post. In addition having the same signal names than in the diagram would also help a lot. I realized later that you also made another change: you removed the first register and apparently that is the only? Hi, thank you for your feedback! The mistake is not the use of the inverter, but the missing of the first flip-flop as you can see in Figure 5 where the edge is not detected.

I hope I answered your doubts Ciao. Thank you for the post, it was an interesting read! Either the input signal is asynchronous, in which case both circuits suffer from metastability issues, or the input signal is synchronous, in which case either of the two circuits are fine depending on the use case.

The second design may be favorable if you know that the pulse will last long enough for your needs, since less hardware is required and since it decreases the latency of that path which e. Am I right? I read the example there are some issues in the first example.Post a Comment.

Pulse Width Modulation PWM is a very popular modulation technique which is mainly used to control the power delivered to electrical devices such as motors.

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ALL ; -- fpga4student. Recommended VHDL projects : 1. What is an FPGA?

pulse stretcher vhdl

VHDL code for 8-bit Microcontroller 5. VHDL code for 8-bit Comparator 9. VHDL code for counters with testbench How to generate a clock enable signal instead of creating another clock domain VHDL code for Traffic light controller VHDL code for a simple 2-bit comparator No comments:. Newer Post Older Post Home. Subscribe to: Post Comments Atom. Today, f Verilog code for bit single cycle MIPS processor. Verilog code for D Flip Flop. D Flip-Flop is a fundamental component in digital logic circuits.

Verilog code for D Flip Flop is presented in this project. There are tw Verilog code for counter with testbench. In this project, Verilog code for counters with testbench will be presented including up counter, down counter, up-down counter, and r A display controller will be This FPGA project is aimed to show in details how to process an image using Verilog from reading an input bitmap image.

A full Verilog code for displayi Verilog code for D Flip Flop here. There are several types of D Flip Flops such It normally executes logic and arithmetic op