a problem of level shift

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gdhp

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hi all

i am designing a circuit. The power is 2.5v, adn the input signal level 2.5v with

25M. I want the output signal level is 1.5v.

can anyone give me some suggestion to perform the function?

Now i have some thought. one is that i use the linear regulator to transfer the 2.5v power to 1.5v.

Another is use the source follower.

some suggestion? some materials?
Thanks!!
 

Use a zener diode as a voltage regulator with it's cathode common to the output signal (you wanted to be 1.5V) and a pull-up resistor to your input signal of 2.5V and its anode to the ground.

Another method is to use a buck or step-down converter. It's topology is still source-follower but with added components like inductor, diode and a capacitor.

Simply a source-follower is a common method too. I used that too, in the past for some simple signal conversion.

Linear voltage regulator, most people uses that too.
 

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    gdhp

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can you give me some materials?

Added after 1 minutes:

Use a zener diode as a voltage regulator with it's cathode common to the output signal (you wanted to be 1.5V) and a pull-up resistor to your input signal of 2.5V and its anode to the ground.???

i don't understand!can you explain it?
 

Choosing a zener diode and resistor:

The zener voltage Vz is the output voltage required
The input voltage Vs must be a few volts greater than Vz
(this is to allow for small fluctuations in Vs due to ripple)
The maximum current Imax is the output current required plus 10%
The zener power Pz is determined by the maximum current: Pz > Vz × Imax
The resistor resistance: R = (Vs - Vz) / Imax
The resistor power rating: P > (Vs - Vz) × Imax

Example: output voltage required is 5V, output current required is 60mA.
(There is more informaation about regulators on the Electronics in Meccano website. )

Vz = 4.7V (nearest value available)
Vs = 8V (it must be a few volts greater than Vz)
Imax = 66mA (output current plus 10%)
Pz > 4.7V × 66mA = 310mW, choose Pz = 400mW
R = (8V - 4.7V) / 66mA = 0.05k = 50, choose R = 47
Resistor power rating P > (8V - 4.7V) × 66mA = 218mW, choose P = 0.5W
 

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    gdhp

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thank you skyhigh!

Another can you give me some suggestions about the source follower. because the output has to connect to 10pf Cap, so the delay-time should be considered .
 

It depends how 10pF capacitor is connected to the source and the load.
 

In the interface between different ligic level application , we often open drain output with a pull up res.

The output with 10pF cap. you can tune the NMOS driver capability and pull up res.
 

i know the opendrain ,but the delay is too large. so i give up it!
 

i am designing a circuit. The power is 2.5v, adn the input signal level 2.5v with

25M. I want the output signal level is 1.5v.
Click to expand...

I have questions about ur problem
1. what is 2.5V with 25M ? do u mean 25mV or 25MHz ? input signal level 2.5V is the DC level ?
2. do u have 1.5V power supply ?
3. do ur signal is clock (full swing 0~2.5V), or it is a analog signal ?
4. what's ur load for the 1.5V output ?
5. what application?
 

hi Btrend

1: the input signal is clock signal with full swing of 0- 2.5v, the frequency is 25M.
2: i have no 1.5v POwer. so it is problem.
3:the signal is a clock.
4: the load of output is a PIN, who has a 10pf cap.
5: it is a level shif.
 

level shift from high voltage to low voltage is easy, what u want to do are:
1. produce a 1.5V power supply from 2.5V using LDO like topology
2. cascade two inverters to form a buffer and their VCC connect to 1.5V from (1)
3. 10pF is no problem, if u sizing the final inverter in (2)
 

cascade two inverters to form a buffer ??

why use the cascade two inverters??
 

gdhp said:
cascade two inverters to form a buffer ??

why use the cascade two inverters??
Click to expand...
refer to this
 

hi Btrend

thank you for your suggestion! some problems!
1:at the input , the 2.5 dc is divided half and 1.25v is connected to opamp.

2: so i think the output of v15 is 1.5v.

q:
1: why the value of 1.25 is choosed ? can i connect the 2.5 to opamp,and use two resistance to get the 1.5v?

2: at the output ,you use the pmos. i think it is a wrong, it should be nmos. i don't know why couldn't use the pmos??

3:because the ckin is 25M clock, so a current pulse exist on the buffer. so in my simulation, i found the current pulse will change the current on the two resistance. so the output of v15 is not stable. some suggestion to get out of it?
 

u had better read some documents about LDO or regulator to know what I had stated.
such as
 

thank you btrend

i am a fresh man in analog. i will try my best. thank you !
 

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