how to simulate PSRR in bandgap reference?

[leebluer]

how to simulate psrr

how to simulate PSRR in bandgap reference? I think it is different from amplifier. But I don't know the detailed method. Who can give me a clue?

 

[hylas]

Vdd as AC source,
do .ac analysis (hspice)
probe vdb(vbg)

 

[hawkin]

set VDD ac 1mv
PSRR=dB20(VDD/vout) + DC gain of amp

 

[billchen]

set VDD ac 1mv
PSRR=dB20(VDD/vout) + DC gain of amp

to hawkin
i am confused with u expression，why the DC gain of amp was added to PSRR.i think the dB20(VDD/vout) is ok.

 

[Hughes]

set VDD ac 1mv
PSRR=dB20(VDD/vout) + DC gain of amp

VDD's ac value can be set to 1.0V directly.
PSRR=-dB20(vout)

 

[billchen]

sorry
i dont understand the difference between them
can u explain it for me
thx very much

 

[electronrancher]

bill, a simpler way is to just add a 1v AC source on top of your dc source. now your dc sim still works and you can sim AC stuff as well.

the reason you set VAC to 1v is so you can read dB off the plot directly. Now if you have 60dB of PSRR you will see the trace flatten out at -60dB. If you use 1mV then spice gives you this fake, -60mdB number just to make the ratios work out.

it's all relative - you're really not pumping a 1v sine wave into your little circuit, you're just plotting the circuit's response to AC, magnitude = 1.

 

[staric]

You can refer to the book:CMOS Analog Circuit Design ,Pillips Allen. In the chapter 6, there are many measurement for simulation.

 

[billchen]

thx to electronrancher and staric!

 

[leebluer]

hello staric, i refer to the book:CMOS Analog Circuit Design by Pillips Allen. but i think the method proposed is for amplifier not for bandgap circuit. the unity-gain struction is suitable for amplifier to measure PSRR, is it true for bandgap circuit?

 

[nkchenliy]

In fact, in the bandgap circiut, VDD is the input, so you can Vdd as AC source. I always do it as this.

 

[hanjiemy]

set VDD ac 1mv
PSRR=dB20(VDD/vout) + DC gain of amp

PSRR+ PSRR- (need measure?)

 

[melodyseu]

hi,Hughes, I think to set ac value to 1v is too large.Although it is convenient to read dB off the plot directly,it may cause other problems especially in low_Vdd cicuit.

 

[jordan76]

it's all relative - you're really not pumping a 1v sine wave into your little circuit, you're just plotting the circuit's response to AC, magnitude = 1.

Just as electronrancher explained above.

regards,
jordan76

 

[piao]

I think you can refer to Allen's book for the definition of PSRR.

 

[Hughes]

hi,Hughes, I think to set ac value to 1v is too large.Although it is convenient to read dB off the plot directly,it may cause other problems especially in low_Vdd cicuit.

Hi melodyseu,

It doesn't matter. The AC value and DC value can be set independly. In low_Vdd circuit, the DC value of VDD is low, but its AC value can be any value. The simulator use DC value to calculate the operating point of the circuit. Then all elements are linearized at this operating point. AC simulating is run with the linearized model which will not suffer from saturation problem and the like.

You may do a test. Simulate a simple amplifier with 1mV of AC input value and get the gain: VDB(out)-VDB(in). Then change the AC value of the input to 1000V (This will certainly saturate the amplifier in a REAL amplifier). Do the AC simulation again and get the gain: VDB(out)-VDB(in). You can find the gains obtained from two runs are the same.

 

[melodyseu]

Hughes,thanks. I think your explain is veyr clear and I got it,
but I still have another question,how to set ac value in the real test?

 

[yorande]

Hi, Hughes.
no gain compression in operation amplifiers or bandgaps as it's in the RF circuits?
As I simulated an LNA, it shows the gain compression.

 

[always@smart]

dear all,

I've followed what u guys suggested method to measure the bandgap PSRR:

vdd ac=1
.print ac par('vdb(vdd)-vdb(vref)') $plot PSRR

I got the PSRR shown in the graph, is it correct? what is the PSRR value is good for a bandgap?

thanks in advance

 

[radiohead]

I have no experience in this, but I don't think that's a great result. 26dB means around times 20 suppression. So a 100mV pulse on VDD (this is possible!) will result in a 5mV pulse on your voltage reference. This does not seem good to me.