voltage to current converter

Hi,

I am trying to simulate a DC block circuit in PS PICE. I am using the following practical circuit link1:

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I replaced the voltage to current section with the current source symbol given in PS pice. Please find the circuit diagram in the following link link2:

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I have following questions

  1. How can I see the capacitor charging and discharging in PSPICE? I found the capacitor always charged whenever I run the simulations.

  1. What is the difference between voltage to current converter and constant current source? The INA circuit that I am using could be a constant current source ( plz. refer link1) and can be replaced with the current source given in PSPICE. I tried to use the PSpice models of INA and opa131 but did not work .

  2. I am trying to get rid of the DC leakage current produced by the INA 133 circuit using capacitors. I have discussed this issue in my previous post but need more advice! There problem is

a. At the power up the DAC outputs DC -2.5 volts. The inverting input of the INA133 is grounded and the 470uF capacitor is reduing the -2.5 volts to -0.5 volts accross the 384ohm resistor. There is a volage difference between the inputs which reflects at the output. I solved this problem partly by using the waveforms that ends at zero. So, after sending the waveform out of DAC, the DAC holds the zero and no more leakage. But at power up its an issue.

b. The leakage current is constant and I came up with the circuit that I mentioned in the link2 to get rid of it. Any comments!!

Regards John

Reply to
john
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On a sunny day (Fri, 16 Nov 2007 12:35:18 -0800 (PST)) it happened john wrote in :

None basically.

I would like to understand what exactly you mean by 'DC leakage'. In a previous post I showed you need no capacitor,

So then for a 0 to +ma output current you need -2.5 V on the other input of the INA.

No output cap either, but your load impedance must be low enough so the requested current can flow. If yo uhave a supply of plus and minus 10V, and a 5 k series resistor, then whatever yo udo,m the output current cannot be more thne +or- 10 / 5000. If your load is say 1K, then the output current can never be more then

+or- 10 / (5000 + 1000). So this limits the input voltage range, as I out is U1-U2 / 5000 and can never be greater then Iout max.

Using the 1M makes no sense as for 1mA you need 1000V.

So, practially the calc goes a bit like this:

U1-U2 = 2.5V, R=5000, Iout = 500uA, for 10V supply Rload total should be

10 / .0005 = 20000, you already have 5000 for R, so Rload must be < 15000 Ohm.

If I did not make any errors, but thisis the system.

What is your Rload?

Reply to
Jan Panteltje

Hi,

The DC leakage current will go into the load without the capacitors. The -2.5 volts offset is only there at the power up not after the DAC outputs the waveform. After the DAC outputs the waveform, the voltage at the non inverting input is zero so leakage gets low but still its there. I think that DAC zero and the ground zero has some offset and need trimming but can not do trimming because of PCB issues.

John

Reply to
john

On a sunny day (Mon, 19 Nov 2007 12:28:58 -0800 (PST)) it happened john wrote in :

At power up the caps have charge zero, so the ywill not help, but perhaps even increase startup surges / currents in teh load.

3rd time I ask now: What is your load impedance.

OK I give up.

Reply to
Jan Panteltje

Hi,

The laod is in the range of 10kohm to 200kohm.

John

Reply to
john

Hi,

The resistor in parallel with the current source will not let the capacitor charged upto +/- 15 volts. so when the power is turned off , the capacitor will have less charge to dump in the load. What do you think?

Regards John

Reply to
john

On a sunny day (Mon, 19 Nov 2007 12:53:26 -0800 (PST)) it happened john wrote in :

OK, now think for a minute.

The output was (U1-U2) / R, where R was 5000. Your DAC does 2.5 V or more ful lsignal. the output current is then 2.5 / 5000 = .5 mA .5 mA in 200.000 Ohm = 100 Volt. Your supply voltage is? 10V? Go back to start and pay 100$ :-)

Reply to
Jan Panteltje

On a sunny day (Mon, 19 Nov 2007 13:07:51 -0800 (PST)) it happened john wrote in :

I was thinking about the other capacitor, it is a short at power up. But I was also thinking that none of this can work because see my other post.

Reply to
Jan Panteltje

Reply to
robowaldron

Hello,

Would you please confirm two more things

  1. What is the output impedance of this voltage to current converter? Ideally current sources have infinite output impedance.

  1. If the voltage across the load is 1000 volts and power supply is

10volts. What would likely to happen a. the current will not flow b. the current will flow

Regards John

Reply to
john

On a sunny day (Tue, 20 Nov 2007 10:39:06 -0800 (PST)) it happened john wrote in :

Ideally, yes. The output impedance is set by the open loop gain.

The current CANNOT flow, as you need 100V for it to flow. As you only have a much lower supply vltage, only a lower current will flow, resulting in the amplifier hanging one way against the supply.

Reply to
Jan Panteltje

On a sunny day (Tue, 20 Nov 2007 10:39:06 -0800 (PST)) it happened john wrote in :

Ideally, yes. The output impedance is set by the open loop gain.

The current CANNOT flow, as you need 100V for it to flow. As you only have a much lower supply vltage, only a lower current will flow, resulting in the amplifier hanging one way against the supply.

Reply to
Jan Panteltje

Jan Panteltje snipped-for-privacy@yahoo.com posted to sci.electronics.design:

Gosh, a fixed value source versus a controlled source but no difference? That does not make sense to me. Spice has controlled sources.

Reply to
JosephKK

john snipped-for-privacy@hotmail.com posted to sci.electronics.design:

formatting link

To bad the link got truncated. It makes it hard to figure out what you are trying.

Reply to
JosephKK

On a sunny day (Wed, 21 Nov 2007 04:11:27 -0800) it happened JosephKK wrote in :

look at this in context, he has a controlled source. Any constant current source can be controlled in some way, If you really wanted to be pedantic.

F*ck spice. Reality rules.

Reply to
Jan Panteltje

Jan Panteltje snipped-for-privacy@yahoo.com posted to sci.electronics.design:

:

It seems unlike you to not follow the chain properly.

Reply to
JosephKK

On a sunny day (Fri, 23 Nov 2007 04:48:24 GMT) it happened JosephKK wrote in :

Reality rules.

Reply to
Jan Panteltje

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