13
I have a fuel sender in my kit car that goes from 20R to earth (full) to 200R (empty). Unfortunately the gauge (constant current source, measuring resistance to earth?) expects empty to be 240 ohms, so displays '1/4 full' when empty.
Status Spec. Actual Reading Voltage
------------------------------------------
Full 20 Ω 20 Ω 100% ...
11
This may not be what you are looking for as far as getting insight into a regular multiplier circuit, but you can implement this easily with just a ROM. All you need is 64 memory loctions, each 6 bits wide. You load the ROM with answers to each of the possible 64 input cases, and the rest is a lookup.
This may sound dismissive, but actually small ...
8
Write the multiplication down and you'll know what to do. For example A2A1A0×B2B1B0
A₂ A₁ A₀
× B₂ B₁ B₀
-----------------------
B₀A₂ B₀A₁ B₀A₀
+ B₁A₂ B₁A₁ B₁A₀
B₂A₂ B₂A₁ B₂A₀
As you see, just multiply each bit (which is accomplished by an AND) then add them together using full adders. Here is an implementation
...
8
If you're a EE/CPE student and you're just starting to learn logic, maybe you haven't learned this yet but you will.
Arithmetic and logic functions are essentially realized in circuit form by starting with a truth table and filling in the values that implement the function you want. For 2x2 bit multiplication, this is the truth table:
From the program ...
8
There are two different ways the term "lookup table" are used in FPGA design that might be confusing you.
First, the main building block of combinatorial logic in an FPGA is called a lookup table, but usually abbreviated as LUT. This is just a small RAM element that takes 4 or 5 or 6 inputs (depending on which type of FPGA you have) and uses that to select ...
7
A few companies made stand alone digital multipliers. A long time ago I used one from Weitek (might be spelled differently). If I remember there were two other competing products we considered at the time. One was from Honeywell, but I'm not sure that ever made it to the market. I think the third was Analog Devices.
I designed a 3D graphics processor in ...
7
The MOSFET is acting as a source follower and this generally means that whatever signal is placed on the gate, is also seen on the source AND it is quite resiliant to ripple or movements on the drain.
So the source is the output and the gate is at DC level of 82% of the drain voltage BUT it's filtered (by R1 and C1) to remove any ripple seen on the drain.
...
6
TRW made multipliers such as this 16 X 16 part :-
or this, which would be over 3" long!
6
(Cheat)
The simplest yet most powerful solution is to use a Flash memory as a lookup table for the results. 8-bit A input + 8-bit B input + 3 bits operation code is 19 bits. Pick a 512 k \$\times\$ 16 bit Flash (about $2), use the 19 input lines as address, and program it with the results for each input combination. This way you can have any operation you ...
6
Sure, if you happen to have a 1955 Lincoln Continental with all the options kicking around.. they were used in the automatic headlight dimmer circuit.
OPERATION: With headlights turned on, and the pilot
light indicating automatic control, light striking the
photo-multiplier tube of the control assembly causes
generation of a weak electrical current....
6
You need to ditch Rd from your first schematic, and use a low impedance push-pull output as in your second schematic. However, as you correctly say, 36v will toast the gates of 20v Vgs FETs. There are few fets with Vgsmax greater than 20v, and none to my knowledge with more than 30v.
Amongst the options are to use
a) suitable level shifters to control the ...
6
The Gilbert cell is properly viewed as a differential-inputs EXOR gate. With 200 milliVoltsPeakPeak signals on RF and on the LO, the current-source will be strongly steered to one or the other "mixer" outputs at the top.
If 1 is 200mV higher than 2, and if 3/6 is 200mV higher than 4/5, the current appears on the left output at top.
If 2 is high AND 4/5 is ...
5
The simplest approach would be to decode the x y z inputs into eight lines. Then from these, you implement logic which drives chip select lines to enable the appropriate unit which handles the inputs, as well as any transformations that are needed so that the unit performs the correct operation.
I don't think you can use an adder for your logic operations ...
5
Try loading that circuit with a resistor across C1.
You will find that, when delivering current to a load, its output voltage falls far below the open circuit voltage - especially if you add further voltage multiplication stages.
The problem is that the impedance of those capacitors quickly adds up, and is effectively doubled by each multiplication stage, ...
5
If you want to build an analogue multiplier that is a little off-the-beaten track then consider what happens when you feed an analogue signal through an analogue switch but control the analogue switch with PWM at a high frequency (significantly above nyquist to make life easier).
If the PWM is 50% mark-space then the baseband analogue signal is attenuated ...
5
I think the easiest way would be just getting a new variable resistor. You could use a transistor to switch in an extra resistance but potentiometers are pretty cheap, so swapping out would be the easiest option.
4
You are correct, and the diagram is wrong. You definitely do need wider adders as you move down the tree as shown. Each of the results from the first tier of adders is 34 bits wide, those from the second tier are 36 bits wide, etc. Third tier sums are 40 bits, and fourth tier sums are 48 bits.
By the time you get to the last (fifth) tier, you're adding ...
4
Well I'm not sure exactly what you're asking, but I'll try to be of some help.
First, if you know how K-maps work, it'll make circuit simplification much easier. Jim Pytel does a much better job at explaining them than I could typing on an online message board
With that said, you cannot make an adder or multiplier with AND gates alone. You're going to ...
4
I presume that you are synthesising for an FPGA or CPLD. The simulator will successfully simulate many language constructs which are technically valid Verilog, but are very hard to translate to real programmable logic hardware.
I can't synthesise your block myself as I don't have the part_prod_gen1, count_one and shift modules available, but the problems I ...
4
I believe the component you are looking for is called an RF switch. They are used for example for switching the TX path of bluetooth and wifi chips to a single antenna. RF switches are quite inexpensive (relative term, I know), and come for many purposes.
The terminology for RF switch classifications is similar to regular switches, i.e. SPDT means single-...
4
I made very basic error. Problem was in quality of ±12V power source.
After I added some capacitors, precision of those multipliers improved dramatically.
Precision is within 1% now.
IN1=7.004V
IN2=2.0046V
OUT=1.4082V
So, now it is out by 0.3%. That is better :)
4
The AD633 always divides by 10V. It says that at the very beginning of the datasheet, in figure 1. 10V is not the same as 10 without units.
When doing calculations in mV and V (or indeed any other mixture of units: meters and feet; kilos and grams; seconds and hours) you should convert everything into the same units before doing your maths.
So for two ...
4
Each stage of such a voltage multiplier stacks a voltage the size of the input voltage on top of the circuit input voltage. But also, each stage has substantial losses. So the key to success is starting with an input voltage as high as possible.
Don't have it 1V but at least 200..300V. You can create that input voltage for the voltage doubler ladder using a ...
4
There are arithmetic binary multipliers and then there are clock frequency multipliers.
I'll discuss some of it only, you decide what you want.
For how arithmetic multipliers and dividers work, read here
Frequency multipliers work in several ways;
1) by using a PLL to divide down to the mixer frequency. So the multiplier is actually a divider. (binary, ...
answered Mar 12 '18 at 17:27
Tony Stewart Sunnyskyguy EE75
88.9k2 gold badges33 silver badges131 bronze badges
4
These things exist - Analog devices (used to?) have some multiplier ICs you can (could?) buy. They also have this excellent appnote which I definitely suggest reading.
One of the classic buildingblocks in analog design is the Gilbert Cell, named after Barrie Gilbert. It does what you seek (at least, if I understand your question correctly). Because of this ...
4
You scrambled the order of the input connections on your 4-bit adder. On the submodule, you have
A3 B3 A2 B2 A1 B1 A0 B0
but in your main circuit, you have them wired up as if they were
A3 A2 A1 A0 B3 B2 B1 B0
... Unless there's some magic that I can't see in your diagrams.
3
The circuit is performing the middle addition in long multiplication:
WXYZ
* 11
-------
WXYZ
+ WXYZ0
-------
(result)
answered Jun 10 '13 at 19:09
Ignacio Vazquez-Abrams
46.5k2 gold badges67 silver badges95 bronze badges
3
I'm having a hard time trying to understand the schematic...however: there are several different architectures for binary multipliers, but all of them are based on simple considerations.
1- The multiplication in binary logic is made by the AND operator. Therefore you will need a battery of AND gates that multiply each bit of one factor for each bit of the ...
3
M1 is a k-1 wide to 1 multiplexer with individual enables driven from the ring counter. Where the READY signal forces the output to a '0'. M2 is a 1:k-1 selector with individual enables driven off the ring counter. The ring counter has k stages. MUX1 is a k+1 wide 2:1 multiplexer with a single select input.
Model M1, M2 and MUX1 behaviorially, you have ...
3
If you connect a second one like this:
simulate this circuit – Schematic created using CircuitLab
(only one oscillator is required, so you can use all the inverters in the second chip in parallel if you want).
You'll get about -5V, so the difference between the two will be about 15V.. that will allow you to connect more LEDs in series.
You might ...
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